The post NCERT Class 12 Maths Solutions Chapter 5 – Continuity and Differentiability appeared first on Babaji Academy.

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**The topics and sub-topics included in the Continuity and Differentiability chapter are the following:**

- Continuity and Differentiability
- Introduction
- Algebra of continuous functions
- Differentiability
- Derivatives of composite functions
- Derivatives of implicit functions
- Derivatives of inverse trigonometric functions
- Exponential and Logarithmic Functions
- Logarithmic Differentiation
- Derivatives of Functions in Parametric Forms
- Second-Order Derivative
- Mean Value Theorem
- Summary

Prove that the functionis continuous at

Therefore, *f* is continuous at *x* = 0

Therefore, *f *is continuous at *x* = −3

Therefore, *f* is continuous at *x* = 5

Examine the continuity of the function.

Thus, *f* is continuous at *x* = 3

Examine the following functions for continuity.

(a)

(b)

(c) (d)

(a) The given function is

It is evident that *f* is defined at every real number *k* and its value at *k* is *k* − 5.

It is also observed that,

Hence,* f *is continuous at every real number and therefore, it is a continuous function.

(b) The given function is

For any real number *k* ≠ 5, we obtain

Hence, *f* is continuous at every point in the domain of *f* and therefore, it is a continuous function.

(c) The given function is

For any real number *c* ≠ −5, we obtain

Hence, *f* is continuous at every point in the domain of *f* and therefore, it is a continuous function.

(d) The given function is

This function *f* is defined at all points of the real line.

Let *c* be a point on a real line. Then, *c* < 5 or *c* = 5 or *c* > 5

Case I: *c* < 5

Then, *f *(*c*) = 5 − *c*

Therefore, *f* is continuous at all real numbers less than 5.

Case II : *c* = 5

Then,

Therefore, *f *is continuous at *x* = 5

Case III: *c* > 5

Therefore, *f* is continuous at all real numbers greater than 5.

Hence,* f *is continuous at every real number and therefore, it is a continuous function.

Prove that the function is continuous at *x* = *n*, where *n* is a positive integer.

The given function is *f* (*x)* = *x*^{n}

It is evident that *f* is defined at all positive integers, *n*, and its value at *n* is *n*^{n}.

Therefore, *f *is continuous at *n*, where *n* is a positive integer.

Is the function *f* defined by

continuous at *x* = 0? At *x* = 1? At *x* = 2?

The given function *f* is

At *x* = 0,

It is evident that* f *is defined at 0 and its value at 0 is 0.

Therefore, *f* is continuous at *x* = 0

At *x* = 1,

*f *is defined at 1 and its value at 1 is 1.

The left hand limit of* f *at *x* = 1 is,

The right hand limit of *f *at *x* = 1 is,

Therefore,* f *is not continuous at *x* = 1

At *x *= 2,

*f *is defined at 2 and its value at 2 is 5.

Therefore, *f* is continuous at *x *= 2

Find all points of discontinuity of *f*, where *f* is defined by

The given function *f* is

It is evident that the given function *f* is defined at all the points of the real line.

Let *c* be a point on the real line. Then, three cases arise.

(i) *c* < 2

(ii) *c* > 2

(iii) *c* = 2

Case (i) *c* < 2

Therefore, *f* is continuous at all points *x*, such that *x* < 2

Case (ii) *c* > 2

Therefore, *f* is continuous at all points *x*, such that *x* > 2

Case (iii) *c* = 2

Then, the left hand limit of *f *at* x *= 2 is,

The right hand limit of *f* at *x* = 2 is,

It is observed that the left and right hand limit of *f* at *x* = 2 do not coincide.

Therefore, *f* is not continuous at *x* = 2

Hence, *x* = 2 is the only point of discontinuity of *f*.

Find all points of discontinuity of *f*, where *f* is defined by

The given function *f* is

The given function *f* is defined at all the points of the real line.

Let *c* be a point on the real line.

Case I:

Therefore, *f* is continuous at all points *x*, such that *x* < −3

Case II:

Therefore, *f* is continuous at *x* = −3

Case III:

Therefore, *f* is continuous in (−3, 3).

Case IV:

If *c* = 3, then the left hand limit of *f *at* x *= 3 is,

The right hand limit of *f *at* x *= 3 is,

It is observed that the left and right hand limit of *f* at *x* = 3 do not coincide.

Therefore, *f* is not continuous at *x* = 3

Case V:

Therefore, *f* is continuous at all points *x*, such that *x* > 3

Hence, *x* = 3 is the only point of discontinuity of *f*.

Find all points of discontinuity of *f*, where *f* is defined by

The given function *f* is

It is known that,

Therefore, the given function can be rewritten as

The given function *f* is defined at all the points of the real line.

Let *c* be a point on the real line.

Case I:

Therefore, *f* is continuous at all points *x* < 0

Case II:

If *c* = 0, then the left hand limit of *f *at* x *= 0 is,

The right hand limit of *f *at* x *= 0 is,

It is observed that the left and right hand limit of *f* at *x* = 0 do not coincide.

Therefore, *f* is not continuous at *x* = 0

Case III:

Therefore, *f* is continuous at all points *x*, such that *x* > 0

Hence, *x* = 0 is the only point of discontinuity of *f*.

Find all points of discontinuity of *f*, where *f* is defined by

The given function *f* is

It is known that,

Therefore, the given function can be rewritten as

Let *c* be any real number. Then,

Also,

Therefore, the given function is a continuous function.

Hence, the given function has no point of discontinuity.

Find all points of discontinuity of *f*, where *f* is defined by

The given function *f* is

The given function *f* is defined at all the points of the real line.

Let *c* be a point on the real line.

Case I:

Therefore, *f* is continuous at all points *x*, such that* x* < 1

Case II:

The left hand limit of *f *at* x *= 1 is,

The right hand limit of *f *at* x *= 1 is,

Therefore, *f* is continuous at *x* = 1

Case III:

Therefore, *f* is continuous at all points *x*, such that* x* > 1

Hence, the given function *f *has no point of discontinuity.

Find all points of discontinuity of *f*, where *f* is defined by

The given function *f* is

The given function *f* is defined at all the points of the real line.

Let *c* be a point on the real line.

Case I:

Therefore, *f* is continuous at all points *x*, such that* x* < 2

Case II:

Therefore, *f* is continuous at *x* = 2

Case III:

Therefore, *f* is continuous at all points *x*, such that* x* > 2

Thus, the given function *f* is continuous at every point on the real line.

Hence, *f *has no point of discontinuity.

Find all points of discontinuity of *f*, where *f* is defined by

The given function *f* is

The given function *f* is defined at all the points of the real line.

Let *c* be a point on the real line.

Case I:

Therefore, *f* is continuous at all points *x*, such that* x* < 1

Case II:

If *c* = 1, then the left hand limit of *f* at *x* = 1 is,

The right hand limit of *f* at *x *= 1 is,

It is observed that the left and right hand limit of *f* at *x* = 1 do not coincide.

Therefore, *f* is not continuous at *x* = 1

Case III:

Therefore, *f* is continuous at all points *x*, such that* x* > 1

Thus, from the above observation, it can be concluded that *x* = 1 is the only point of discontinuity of *f*.

Is the function defined by

a continuous function?

The given function is

The given function *f* is defined at all the points of the real line.

Let *c* be a point on the real line.

Case I:

Therefore, *f* is continuous at all points *x*, such that* x* < 1

Case II:

The left hand limit of *f *at *x* = 1 is,

The right hand limit of *f* at *x *= 1 is,

It is observed that the left and right hand limit of *f* at *x* = 1 do not coincide.

Therefore, *f* is not continuous at *x* = 1

Case III:

Therefore, *f* is continuous at all points *x*, such that* x* > 1

Thus, from the above observation, it can be concluded that *x* = 1 is the only point of discontinuity of *f*.

Discuss the continuity of the function *f*, where *f* is defined by

The given function is

The given function is defined at all points of the interval [0, 10].

Let *c* be a point in the interval [0, 10].

Case I:

Therefore, *f* is continuous in the interval [0, 1).

Case II:

The left hand limit of *f *at *x* = 1 is,

The right hand limit of *f* at *x *= 1 is,

It is observed that the left and right hand limits of *f* at *x* = 1 do not coincide.

Therefore, *f* is not continuous at *x* = 1

Case III:

Therefore, *f* is continuous at all points of the interval (1, 3).

Case IV:

The left hand limit of *f *at *x* = 3 is,

The right hand limit of *f* at *x *= 3 is,

It is observed that the left and right hand limits of *f* at *x* = 3 do not coincide.

Therefore, *f* is not continuous at *x* = 3

Case V:

Therefore, *f* is continuous at all points of the interval (3, 10].

Hence, *f *is not continuous at *x *= 1 and *x *= 3

Discuss the continuity of the function *f*, where *f* is defined by

The given function is

The given function is defined at all points of the real line.

Let *c* be a point on the real line.

Case I:

Therefore, *f* is continuous at all points *x*, such that* x *< 0

Case II:

The left hand limit of *f *at *x* = 0 is,

The right hand limit of *f* at *x *= 0 is,

Therefore, *f* is continuous at *x* = 0

Case III:

Therefore, *f* is continuous at all points of the interval (0, 1).

Case IV:

The left hand limit of *f *at *x* = 1 is,

The right hand limit of *f* at *x *= 1 is,

It is observed that the left and right hand limits of *f* at *x* = 1 do not coincide.

Therefore, *f* is not continuous at *x* = 1

Case V:

Therefore, *f* is continuous at all points *x*, such that* x* > 1

Hence, *f *is not continuous only at *x *= 1

Discuss the continuity of the function *f*, where *f* is defined by

The given function *f* is

The given function is defined at all points of the real line.

Let *c* be a point on the real line.

Case I:

Therefore, *f* is continuous at all points *x*, such that* x *< −1

Case II:

The left hand limit of *f *at *x* = −1 is,

The right hand limit of *f* at *x *= −1 is,

Therefore, *f* is continuous at *x* = −1

Case III:

Therefore, *f* is continuous at all points of the interval (−1, 1).

Case IV:

The left hand limit of *f *at *x* = 1 is,

The right hand limit of *f* at *x *= 1 is,

Therefore, *f* is continuous at *x* = 2

Case V:

Therefore, *f* is continuous at all points *x*, such that* x* > 1

Thus, from the above observations, it can be concluded that *f* is continuous at all points of the real line.

Find the relationship between *a* and *b* so that the function *f* defined by

is continuous at *x *= 3.

The given function *f* is

If *f* is continuous at *x* = 3, then

Therefore, from (1), we obtain

Therefore, the required relationship is given by,

For what value of is the function defined by

continuous at *x* = 0? What about continuity at *x* = 1?

The given function *f* is

If *f* is continuous at *x* = 0, then

Therefore, there is no value of λ for which *f* is continuous at *x* = 0

At *x* = 1,

*f* (1) = 4*x* + 1 = 4 × 1 + 1 = 5

Therefore, for any values of λ, *f* is continuous at *x* = 1

Show that the function defined by is discontinuous at all integral point. Here denotes the greatest integer less than or equal to *x*.

The given function is

It is evident that *g* is defined at all integral points.

Let *n* be an integer.

Then,

The left hand limit of *f *at *x* = *n* is,

The right hand limit of *f* at *x *= *n* is,

It is observed that the left and right hand limits of *f* at *x* = *n* do not coincide.

Therefore, *f* is not continuous at *x* =* n*

Hence, *g* is discontinuous at all integral points.

Is the function defined by continuous at *x *=

π?

The given function is

It is evident that *f* is defined at *x *=

π.

Therefore, the given function *f* is continuous at *x *= π

Discuss the continuity of the following functions.

(a) *f* (*x*) = sin *x* + cos *x*

(b) *f* (*x*) = sin *x* − cos *x*

(c) *f* (*x*) = sin *x* × cos x

It is known that if *g *and *h *are two continuous functions, then

are also continuous.

It has to proved first that *g* (*x*) = sin *x *and *h* (*x*) = cos *x* are continuous functions.

Let *g *(*x*) = sin *x*

It is evident that *g* (*x*) = sin *x* is defined for every real number.

Let *c *be a real number. Put *x* = *c* + *h*

If *x* → *c*, then *h* → 0

Therefore, *g* is a continuous function.

Let *h* (*x*) = cos *x*

It is evident that *h* (*x*) = cos *x* is defined for every real number.

Let *c *be a real number. Put *x* = *c* + *h*

If *x* → *c*, then *h* → 0

*h *(*c*) = cos *c*

Therefore, *h* is a continuous function.

Therefore, it can be concluded that

(a) *f* (*x*) = *g* (*x*) + *h* (*x*) = sin *x* + cos *x* is a continuous function

(b) *f* (*x*) = *g* (*x*) − *h* (*x*) = sin *x* − cos *x* is a continuous function

(c) *f* (*x*) = *g* (*x*) × *h* (*x*) = sin *x* × cos *x* is a continuous function

Discuss the continuity of the cosine, cosecant, secant and cotangent functions,

It is known that if *g *and *h *are two continuous functions, then

It has to be proved first that *g* (*x*) = sin *x *and *h* (*x*) = cos *x* are continuous functions.

Let *g *(*x*) = sin *x*

It is evident that *g* (*x*) = sin *x* is defined for every real number.

Let *c *be a real number. Put *x* = *c* + *h*

If *x*

*c*, then *h*

0

Therefore, *g* is a continuous function.

Let *h* (*x*) = cos *x*

It is evident that *h* (*x*) = cos *x* is defined for every real number.

Let *c *be a real number. Put *x* = *c* + *h*

If *x* ® *c*, then *h* ® 0

*h *(*c*) = cos *c*

Therefore, *h* (*x*) = cos *x* is continuous function.

It can be concluded that,

Therefore, cosecant is continuous except at *x *= *n*p, *n *Î **Z**

Therefore, secant is continuous except at

Therefore, cotangent is continuous except at *x *= *n*p, *n *Î **Z**

Find the points of discontinuity of *f*, where

The given function *f* is

It is evident that *f* is defined at all points of the real line.

Let *c* be a real number.

Case I:

Therefore, *f* is continuous at all points *x*, such that* x *< 0

Case II:

Therefore, *f* is continuous at all points *x*, such that* x* > 0

Case III:

The left hand limit of *f* at *x* = 0 is,

The right hand limit of *f* at *x* = 0 is,

Therefore, *f* is continuous at *x* = 0

From the above observations, it can be concluded that *f* is continuous at all points of the real line.

Thus, *f* has no point of discontinuity.

Determine if *f* defined by

is a continuous function?

The given function *f* is

It is evident that *f* is defined at all points of the real line.

Let *c* be a real number.

Case I:

Therefore, *f* is continuous at all points *x *≠ 0

Case II:

⇒-x2≤x2sin1x≤x2

Therefore, *f* is continuous at *x* = 0

From the above observations, it can be concluded that* f* is continuous at every point of the real line.

Thus, *f* is a continuous function.

Examine the continuity of *f*, where *f* is defined by

The given function *f* is

It is evident that *f* is defined at all points of the real line.

Let *c* be a real number.

Case I:

Therefore, *f* is continuous at all points *x*, such that* x *≠ 0

Case II:

Therefore, *f* is continuous at *x* = 0

From the above observations, it can be concluded that* f* is continuous at every point of the real line.

Thus, *f* is a continuous function.

Find the values of *k *so that the function *f* is continuous at the indicated point.

The given function *f* is

The given function *f* is continuous at, if *f* is defined at and if the value of the *f* at equals the limit of *f* at.

It is evident that *f *is defined at and

Therefore, the required value of *k* is 6.

Find the values of *k *so that the function *f* is continuous at the indicated point.

The given function is

The given function *f* is continuous at *x* = 2, if *f* is defined at *x* = 2 and if the value of *f* at *x* = 2 equals the limit of *f* at* x* = 2

It is evident that *f *is defined at* x* = 2 and

Therefore, the required value of.

Find the values of *k *so that the function *f* is continuous at the indicated point.

The given function is

The given function *f* is continuous at *x* = p, if *f* is defined at *x* = p and if the value of *f* at *x* = p equals the limit of *f* at* x* = p

It is evident that *f *is defined at* x* = p and

Therefore, the required value of

Find the values of *k *so that the function *f* is continuous at the indicated point.

The given function *f *is

The given function *f* is continuous at *x* = 5, if *f* is defined at *x* = 5 and if the value of *f* at *x* = 5 equals the limit of *f* at* x* = 5

It is evident that *f *is defined at* x* = 5 and

Therefore, the required value of

Find the values of *a* and *b* such that the function defined by

is a continuous function.

The given function *f *is

It is evident that the given function *f* is defined at all points of the real line.

If *f* is a continuous function, then *f* is continuous at all real numbers.

In particular, *f* is continuous at *x *= 2 and *x *= 10

Since *f* is continuous at *x *= 2, we obtain

Since *f* is continuous at *x *= 10, we obtain

On subtracting equation (1) from equation (2), we obtain

8*a* = 16

⇒ *a* = 2

By putting *a* = 2 in equation (1), we obtain

2 × 2 + *b* = 5

⇒ 4 + *b* = 5

⇒ *b* = 1

Therefore, the values of *a* and *b* for which* f* is a continuous function are 2 and 1 respectively.

Show that the function defined by* f *(*x*) = cos (*x*^{2}) is a continuous function.

The given function is *f *(*x*) = cos (*x*^{2})

This function *f* is defined for every real number and *f* can be written as the composition of two functions as,

*f* = *g o h*, where *g* (*x*) = cos *x* and *h* (*x*) = *x*^{2}

It has to be first proved that *g *(*x*) = cos *x* and *h* (*x*) = *x*^{2} are continuous functions.

It is evident that *g* is defined for every real number.

Let *c* be a real number.

Then, *g* (*c*) = cos *c*

Therefore, *g* (*x*) = cos *x* is continuous function.

*h* (*x*) = *x*^{2}

Clearly, *h* is defined for every real number.

Let *k* be a real number, then* h *(*k*) = *k*^{2}

Therefore, *h* is a continuous function.

It is known that for real valued functions *g *and *h*,such that (*g *o *h*) is defined at *c*, if *g *is continuous at *c *and if *f *is continuous at *g *(*c*), then (*f *o *g*) is continuous at *c*.

Therefore, is a continuous function.

Show that the function defined by is a continuous function.

The given function is

This function *f* is defined for every real number and *f* can be written as the composition of two functions as,

*f* = *g o h*, where

It has to be first proved that are continuous functions.

Clearly, *g* is defined for all real numbers.

Let *c* be a real number.

Case I:

Therefore, *g* is continuous at all points *x*, such that* x *< 0

Case II:

Therefore, *g* is continuous at all points *x*, such that* x* > 0

Case III:

Therefore, *g* is continuous at *x* = 0

From the above three observations, it can be concluded that *g* is continuous at all points.

*h *(*x*) = cos *x*

It is evident that *h* (*x*) = cos *x* is defined for every real number.

Let *c *be a real number. Put *x* = *c* + *h*

If *x* → *c*, then *h* → 0

*h *(*c*) = cos *c*

Therefore, *h* (*x*) = cos *x* is a continuous function.

It is known that for real valued functions *g *and *h*,such that (*g *o *h*) is defined at *c*, if *g *is continuous at *c *and if *f *is continuous at *g *(*c*), then (*f *o *g*) is continuous at *c*.

Therefore, is a continuous function.

Examine that is a continuous function.

This function *f* is defined for every real number and *f* can be written as the composition of two functions as,

*f* = *g o h*, where

It has to be proved first that are continuous functions.

Clearly, *g* is defined for all real numbers.

Let *c* be a real number.

Case I:

Therefore, *g* is continuous at all points *x*, such that* x *< 0

Case II:

Therefore, *g* is continuous at all points *x*, such that* x* > 0

Case III:

Therefore, *g* is continuous at *x* = 0

From the above three observations, it can be concluded that *g* is continuous at all points.

*h *(*x*) = sin *x*

It is evident that *h* (*x*) = sin* x* is defined for every real number.

Let *c *be a real number. Put *x* = *c* + *k*

If *x* → *c*, then *k* → 0

*h *(*c*) = sin *c*

Therefore, *h* is a continuous function.

It is known that for real valued functions *g *and *h*,such that (*g *o *h*) is defined at *c*, if *g *is continuous at *c *and if *f *is continuous at *g *(*c*), then (*f *o *g*) is continuous at *c*.

Therefore, is a continuous function.

Find all the points of discontinuity of *f *defined by.

The given function is

The two functions, *g* and *h*, are defined as

Then, *f* = *g *− *h*

The continuity of *g* and *h *is examined first.

Clearly, *g* is defined for all real numbers.

Let *c* be a real number.

Case I:

Therefore, *g* is continuous at all points *x*, such that* x *< 0

Case II:

Therefore, *g* is continuous at all points *x*, such that* x* > 0

Case III:

Therefore, *g* is continuous at *x* = 0

From the above three observations, it can be concluded that *g* is continuous at all points.

Clearly, *h* is defined for every real number.

Let *c *be a real number.

Case I:

Therefore, *h* is continuous at all points *x*, such that* x *< −1

Case II:

Therefore, *h* is continuous at all points *x*, such that* x* > −1

Case III:

Therefore, *h* is continuous at *x* = −1

From the above three observations, it can be concluded that* h* is continuous at all points of the real line.

*g* and *h* are continuous functions. Therefore, *f *= *g* − *h *is also a continuous function.

Therefore, *f *has no point of discontinuity.

Differentiate the functions with respect to *x*.

Let f(x)=sinx2+5, ux=x2+5, and v(t)=sint

Then, vou=vux=vx2+5=tanx2+5=f(x)

Thus, *f* is a composite of two functions.

**Alternate method**

Differentiate the functions with respect to *x*.

Thus, *f *is a composite function of two functions.

Put *t* = *u* (*x*) = sin *x*

By chain rule,

**Alternate method**

Differentiate the functions with respect to *x*.

Thus, *f *is a composite function of two functions, *u* and *v*.

Put *t* = *u* (*x*) = *ax* + *b*

Hence, by chain rule, we obtain

**Alternate method**

Differentiate the functions with respect to *x*.

Thus, *f *is a composite function of three functions, *u, v*, and *w*.

Hence, by chain rule, we obtain

**Alternate method**

Differentiate the functions with respect to *x*.

The given function is, where *g* (*x*) = sin (*ax* + *b*) and

*h* (*x*) = cos (*cx *+ *d*)

∴ *g *is a composite function of two functions, *u* and *v*.

Therefore, by chain rule, we obtain

∴*h* is a composite function of two functions, *p* and *q*.

Put *y* = *p* (*x*) = *cx *+ *d*

Therefore, by chain rule, we obtain

Differentiate the functions with respect to *x*.

The given function is.

Differentiate the functions with respect to *x*.

Differentiate the functions with respect to *x*.

Clearly, *f *is a composite function of two functions, *u *and* v*, such that

By using chain rule, we obtain

**Alternate method**

Prove that the function *f *given by

is notdifferentiable at *x* = 1.

The given function is

It is known that a function *f* is differentiable at a point *x* = *c* in its domain if both

are finite and equal.

To check the differentiability of the given function at *x* = 1,

consider the left hand limit of *f* at *x* = 1

Since the left and right hand limits of *f* at *x* = 1 are not equal, *f* is not differentiable at *x* = 1

Prove that the greatest integer function defined byis not

differentiable at *x* = 1 and *x* = 2.

The given function *f* is

It is known that a function *f* is differentiable at a point *x* = *c* in its domain if both

are finite and equal.

To check the differentiability of the given function at *x* = 1, consider the left hand limit of *f* at *x* = 1

Since the left and right hand limits of *f* at *x* = 1 are not equal, *f* is not differentiable at

*x* = 1

To check the differentiability of the given function at *x* = 2, consider the left hand limit

of *f* at *x* = 2

Since the left and right hand limits of *f* at *x* = 2 are not equal, *f* is not differentiable at *x* = 2

Find

:

The given relationship is

Differentiating this relationship with respect to *x*, we obtain

Find

:

The given relationship is

Differentiating this relationship with respect to *x*, we obtain

Find

:

The given relationship is

Differentiating this relationship with respect to *x*, we obtain

Using chain rule, we obtainand

From (1) and (2), we obtain

Find

:

The given relationship is

Differentiating this relationship with respect to *x*, we obtain

Find

:

The given relationship is

Differentiating this relationship with respect to *x*, we obtain

[Derivative of constant function is 0]

Find

:

The given relationship is

Differentiating this relationship with respect to *x*, we obtain

Find

:

The given relationship is

Differentiating this relationship with respect to *x*, we obtain

Using chain rule, we obtain

From (1), (2), and (3), we obtain

Find

:

The given relationship is

Differentiating this relationship with respect to *x*, we obtain

Find :

We have,y = sin-12×1 + x2put x = tan θ ⇒ θ = tan-1xNow, y = sin-12 tan θ1 + tan2θ⇒y = sin-1sin 2θ, as sin 2θ=2 tan θ1 + tan2θ⇒y = 2θ, as sin-1sin x=x⇒y = 2 tan-1x⇒dydx = 2 × 11 + x2, because dtan-1xdx=11 + x2⇒dydx = 21 + x2

Find

:

The given relationship is

It is known that,

Comparing equations (1) and (2), we obtain

Differentiating this relationship with respect to *x*, we obtain

Find :

The given relationship is,

On comparing L.H.S. and R.H.S. of the above relationship, we obtain

Differentiating this relationship with respect to *x*, we obtain

sec2y2.ddxy2=ddxx

⇒sec2y2×12dydx=1

⇒dydx=2sec2y2

⇒dydx=21+tan2y2

∴

dydx=21+x2

Find

:

The given relationship is

Differentiating this relationship with respect to *x*, we obtain

Using chain rule, we obtain

From (1), (2), and (3), we obtain

**Alternate method**

⇒

Differentiating this relationship with respect to *x*, we obtain

Find

:

The given relationship is

Differentiating this relationship with respect to *x*, we obtain

Find

:

The given relationship is

Differentiating this relationship with respect to *x*, we obtain

Find

:

The given relationship is

Differentiating this relationship with respect to *x*, we obtain

Differentiate the following w.r.t. *x*:

Let

By using the quotient rule, we obtain

Differentiate the following w.r.t. *x*:

Let

By using the chain rule, we obtain

Differentiate the following w.r.t. *x*:

Let

By using the chain rule, we obtain

Differentiate the following w.r.t. *x*:

Let

By using the chain rule, we obtain

Differentiate the following w.r.t. *x*:

Let

By using the chain rule, we obtain

Differentiate the following w.r.t. *x*:

Differentiate the following w.r.t. *x*:

Let

Then,

By differentiating this relationship with respect to *x*, we obtain

Differentiate the following w.r.t. *x*:

Let

By using the chain rule, we obtain

, *x* > 1

Differentiate the following w.r.t. *x*:

Let

By using the quotient rule, we obtain

Differentiate the following w.r.t. *x*:

Let

By using the chain rule, we obtain

Differentiate the function with respect to *x*.

Taking logarithm on both the sides, we obtain

Differentiating both sides with respect to *x*, we obtain

Differentiate the function with respect to *x*.

Taking logarithm on both the sides, we obtain

Differentiating both sides with respect to *x*, we obtain

Differentiate the function with respect to *x*.

Taking logarithm on both the sides, we obtain

Differentiating both sides with respect to *x*, we obtain

Differentiate the function with respect to *x*.

*u *= *x*^{x}

Taking logarithm on both the sides, we obtain

Differentiating both sides with respect to *x*, we obtain

*v* = 2^{sin }^{x}

Taking logarithm on both the sides with respect to *x*, we obtain

Differentiating both sides with respect to *x*, we obtain

Differentiate the function with respect to *x*.

Taking logarithm on both the sides, we obtain

Differentiating both sides with respect to *x*, we obtain

Differentiate the function with respect to *x*.

Differentiating both sides with respect to *x*, we obtain

Differentiating both sides with respect to *x*, we obtain

Therefore, from (1), (2), and (3), we obtain

Differentiate the function with respect to *x*.

*u *= (log *x*)^{x}

Differentiating both sides with respect to *x*, we obtain

Differentiating both sides with respect to *x*, we obtain

Therefore, from (1), (2), and (3), we obtain

Differentiate the function with respect to *x*.

Differentiating both sides with respect to *x*, we obtain

Therefore, from (1), (2), and (3), we obtain

Differentiate the function with respect to *x*.

Differentiating both sides with respect to *x*, we obtain

Differentiating both sides with respect to *x*, we obtain

From (1), (2), and (3), we obtain

Differentiate the function with respect to *x*.

Differentiating both sides with respect to *x*, we obtain

Differentiating both sides with respect to *x*, we obtain

From (1), (2), and (3), we obtain

Differentiate the function with respect to *x*.

Differentiating both sides with respect to *x*, we obtain

Differentiating both sides with respect to *x*, we obtain

From (1), (2), and (3), we obtain

Find of function.

The given function is

Let *x*^{y} = *u* and *y*^{x} = *v*

Then, the function becomes* u *+ *v* = 1

Differentiating both sides with respect to *x*, we obtain

Differentiating both sides with respect to *x*, we obtain

From (1), (2), and (3), we obtain

Find of function.

The given function is

Taking logarithm on both the sides, we obtain

Differentiating both sides with respect to *x*, we obtain

Find of function.

The given function is

Taking logarithm on both the sides, we obtain

Differentiating both sides, we obtain

Find of function.

The given function is

Taking logarithm on both the sides, we obtain

Differentiating both sides with respect to *x*, we obtain

Find the derivative of the function given by and hence find.

The given relationship is

Taking logarithm on both the sides, we obtain

Differentiating both sides with respect to *x*, we obtain

If *u*, *v* and *w* are functions of *x*, then show that

in two ways-first by repeated application of product rule, second by logarithmic differentiation.

Let

By applying product rule, we obtain

By taking logarithm on both sides of the equation, we obtain

Differentiating both sides with respect to *x*, we obtain

If *x* and *y* are connected parametrically by the equation, without eliminating the parameter, find.

The given equations are

If *x* and *y* are connected parametrically by the equation, without eliminating the parameter, find.

*x* = *a* cos *θ*, *y* = *b* cos *θ*

The given equations are *x* = *a* cos *θ* and *y* = *b* cos *θ*

If *x* and *y* are connected parametrically by the equation, without eliminating the parameter, find.

*x* = sin *t*, *y* = cos 2*t*

The given equations are *x* = sin *t* and *y* = cos 2*t*

If *x* and *y* are connected parametrically by the equation, without eliminating the parameter, find.

The given equations are

If *x* and *y* are connected parametrically by the equation, without eliminating the parameter, find.

The given equations are

If *x* and *y* are connected parametrically by the equation, without eliminating the parameter, find.

The given equations are

If *x* and *y* are connected parametrically by the equation, without eliminating the parameter, find.

The given equations are

If *x* and *y* are connected parametrically by the equation, without eliminating the parameter, find.

The given equations are

If *x* and *y* are connected parametrically by the equation, without eliminating the parameter, find.

The given equations are

If *x* and *y* are connected parametrically by the equation, without eliminating the parameter, find.

The given equations are

If

The given equations are

Hence, proved.

Find the second order derivatives of the function.

Let

Then,

Find the second order derivatives of the function.

Let

Then,

Find the second order derivatives of the function.

Let

Then,

Find the second order derivatives of the function.

Let

Then,

Find the second order derivatives of the function.

Let

Then,

Find the second order derivatives of the function.

Let

Then,

Find the second order derivatives of the function.

Let

Then,

Find the second order derivatives of the function.

Let

Then,

Find the second order derivatives of the function.

Let

Then,

Find the second order derivatives of the function.

Let

Then,

If, prove that

It is given that,

Then,

Hence, proved.

If findin terms of *y* alone.

It is given that,

Then,

If, show that

It is given that,

Then,

Hence, proved.

Ifshow that

It is given that,

Then,

Hence, proved.

If, show that

It is given that,

Then,

Hence, proved.

If, show that

The given relationship is

Taking logarithm on both the sides, we obtain

Differentiating this relationship with respect to *x*, we obtain

Hence, proved.

If, show that

The given relationship is

Then,

Hence, proved.

Verify Rolle’s Theorem for the function

The given function,, being a polynomial function, is continuous in [−4, 2] and is differentiable in (−4, 2).

∴ *f* (−4) = *f* (2) = 0

⇒ The value of *f* (*x*) at −4 and 2 coincides.

Rolle’s Theorem states that there is a point *c* ∈ (−4, 2) such that

Hence, Rolle’s Theorem is verified for the given function.

Examine if Rolle’s Theorem is applicable to any of the following functions. Can you say some thing about the converse of Rolle’s Theorem from these examples?

(i)

(ii)

(iii)

By Rolle’s Theorem, for a function, if

(a) *f* is continuous on [*a*, *b*]

(b) *f* is differentiable on (*a*, *b*)

(c) *f *(*a*) = *f* (*b*)

then, there exists some *c* ∈ (*a*, *b*) such that

Therefore, Rolle’s Theorem is not applicable to those functions that do not satisfy any of the three conditions of the hypothesis.

(i)

It is evident that the given function *f* (*x*) is not continuous at every integral point.

In particular, *f*(*x*) is not continuous at *x *= 5 and *x *= 9

⇒ *f* (*x*) is not continuous in [5, 9].

The differentiability of *f* in (5, 9) is checked as follows.

Let *n *be an integer such that *n* ∈ (5, 9).

Since the left and right hand limits of *f* at *x* = *n* are not equal, *f* is not differentiable at *x* = *n*

∴*f *is not differentiable in (5, 9).

It is observed that *f* does not satisfy all the conditions of the hypothesis of Rolle’s Theorem.

Hence, Rolle’s Theorem is not applicable for.

(ii)

It is evident that the given function *f* (*x*) is not continuous at every integral point.

In particular, *f*(*x*) is not continuous at *x *= −2 and *x *= 2

⇒ *f* (*x*) is not continuous in [−2, 2].

The differentiability of *f* in (−2, 2) is checked as follows.

Let *n *be an integer such that *n* ∈ (−2, 2).

Since the left and right hand limits of *f* at *x* = *n* are not equal, *f* is not differentiable at *x* = *n*

∴*f *is not differentiable in (−2, 2).

It is observed that *f* does not satisfy all the conditions of the hypothesis of Rolle’s Theorem.

Hence, Rolle’s Theorem is not applicable for.

(iii)

It is evident that *f*, being a polynomial function, is continuous in [1, 2] and is differentiable in (1, 2).

∴*f *(1) ≠* f* (2)

It is observed that *f* does not satisfy a condition of the hypothesis of Rolle’s Theorem.

Hence, Rolle’s Theorem is not applicable for.

If is a differentiable function and if does not vanish anywhere, then prove that.

It is given that is a differentiable function.

Since every differentiable function is a continuous function, we obtain

(a) *f* is continuous on [−5, 5].

(b) *f *is differentiable on (−5, 5).

Therefore, by the Mean Value Theorem, there exists *c* ∈ (−5, 5) such that

It is also given that does not vanish anywhere.

Hence, proved.

Verify Mean Value Theorem, if in the interval, where and.

The given function is

*f*, being a polynomial function, is continuous in [1, 4] and is differentiable in (1, 4) whose derivative is 2*x* − 4.

Mean Value Theorem states that there is a point *c* ∈ (1, 4) such that

Hence, Mean Value Theorem is verified for the given function.

Verify Mean Value Theorem, if in the interval [*a*, *b*], where *a* = 1 and *b* = 3. Find all for which

The given function *f* is

*f*, being a polynomial function, is continuous in [1, 3] and is differentiable in (1, 3) whose derivative is 3*x*^{2} − 10*x* − 3.

Mean Value Theorem states that there exist a point *c* ∈ (1, 3) such that

Hence, Mean Value Theorem is verified for the given function and is the only point for which

Examine the applicability of Mean Value Theorem for all three functions given in the above exercise 2.

Mean Value Theorem states that for a function, if

(a) *f* is continuous on [*a*, *b*]

(b) *f* is differentiable on (*a*, *b*)

then, there exists some *c* ∈ (*a*, *b*) such that

Therefore, Mean Value Theorem is not applicable to those functions that do not satisfy any of the two conditions of the hypothesis.

(i)

It is evident that the given function *f* (*x*) is not continuous at every integral point.

In particular, *f*(*x*) is not continuous at *x *= 5 and *x *= 9

⇒ *f* (*x*) is not continuous in [5, 9].

The differentiability of *f* in (5, 9) is checked as follows.

Let *n *be an integer such that *n* ∈ (5, 9).

Since the left and right hand limits of *f* at *x* = *n* are not equal, *f* is not differentiable at *x* = *n*

∴*f *is not differentiable in (5, 9).

It is observed that *f* does not satisfy all the conditions of the hypothesis of Mean Value Theorem.

Hence, Mean Value Theorem is not applicable for .

(ii)

It is evident that the given function *f* (*x*) is not continuous at every integral point.

In particular, *f*(*x*) is not continuous at *x *= −2 and *x *= 2

⇒ *f* (*x*) is not continuous in [−2, 2].

The differentiability of *f* in (−2, 2) is checked as follows.

Let *n *be an integer such that *n* ∈ (−2, 2).

Since the left and right hand limits of *f* at *x* = *n* are not equal, *f* is not differentiable at *x* = *n*

∴*f *is not differentiable in (−2, 2).

It is observed that *f* does not satisfy all the conditions of the hypothesis of Mean Value Theorem.

Hence, Mean Value Theorem is not applicable for .

(iii)

It is evident that *f*, being a polynomial function, is continuous in [1, 2] and is differentiable in (1, 2).

It is observed that *f* satisfies all the conditions of the hypothesis of Mean Value Theorem.

Hence, Mean Value Theorem is applicable for.

It can be proved as follows.

Using chain rule, we obtain

Taking logarithm on both the sides, we obtain

Differentiating both sides with respect to *x*, we obtain

Using chain rule, we obtain

Therefore, equation (1) becomes

Taking logarithm on both the sides, we obtain

Differentiating both sides with respect to *x*, we obtain

**, **for some constant *a* and *b*.

By using chain rule, we obtain

Taking logarithm on both the sides, we obtain

Differentiating both sides with respect to *x*, we obtain

, for some fixed and

Differentiating both sides with respect to *x*, we obtain

Differentiating both sides with respect to *x*, we obtain

*s* = *a*^{a}

Since *a* is constant, *a*^{a} is also a constant.

∴

From (1), (2), (3), (4), and (5), we obtain

, for

Differentiating both sides with respect to *x,* we obtain

Differentiating with respect to *x*, we obtain

Also,

Differentiating both sides with respect to *x*, we obtain

Substituting the expressions of in equation (1), we obtain

Find, if

Find, if

If, for, −1 < *x* <1, prove that

_{It is given that,}

_{Differentiating both sides with respect to }_{x}_{, we obtain}

_{Hence, proved.}

If, for some prove that

is a constant independent of *a* and *b*.

It is given that,

Differentiating both sides with respect to *x*, we obtain

Hence, proved.

If with prove that

Then, equation (1) reduces to

⇒sina+y-ydydx=cos2a+y⇒dydx=cos2a+ysina

Hence, proved.

If and, find

If, show that exists for all real *x*, and find it.

It is known that,

Therefore, when *x* ≥ 0,

In this case, and hence,

When *x* < 0,

In this case, and hence,

Thus, for, exists for all real *x* and is given by,

Using mathematical induction prove that for all positive integers *n*.

_{For }_{n}_{ = 1,}

∴P(*n*) is true for *n* = 1

Let P(*k*) is true for some positive integer *k*.

That is,

It has to be proved that P(*k* + 1) is also true.

Thus, P(*k* + 1) is true whenever P (*k*) is true.

Therefore, by the principle of mathematical induction, the statement P(*n*) is true for every positive integer *n*.

_{Hence, proved.}

Using the fact that sin (*A* + *B*) = sin *A* cos *B* + cos *A* sin *B* and the differentiation, obtain the sum formula for cosines.

Differentiating both sides with respect to *x*, we obtain

Does there exist a function which is continuos everywhere but not differentiable at exactly two points? Justify your answer ?

y=x -∞<x≤1 2-x 1≤x≤∞ It can be seen from the above graph that, the given function is continuos everywhere but not differentiable at exactly two points which are 0 and 1.

If, prove that

Thus,

If, show that

It is given that,

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**The topics and sub-topics included in the Determinants chapter are the following:**

Section Name | Topic Name |

4 | Determinants |

4.1 | Introduction |

4.2 | Determinant |

4.3 | Properties of Determinants |

4.4 | Area of a Triangle |

4.5 | Adjoint and Inverse of a Matrix |

4.6 | Applications of Determinants and Matrices |

4.7 | Summary |

Page No 108:

Evaluate the determinants in Exercises 1 and 2.

= 2(−1) − 4(−5) = − 2 + 20 = 18

Evaluate the determinants in Exercises 1 and 2.

(i) (ii)

(i) = (cos *θ*)(cos *θ*) − (−sin *θ*)(sin *θ*) = cos^{2} *θ*+ sin^{2} *θ* = 1

(ii)

= (*x*^{2} − *x* + 1)(*x* + 1) − (*x* − 1)(*x* + 1)

= *x*^{3} − *x*^{2} + *x* + *x*^{2} − *x* + 1 − (*x*^{2} − 1)

= *x*^{3} + 1 − *x*^{2} + 1

= *x*^{3} − *x*^{2} + 2

If, then show that

The given matrix is.

If, then show that

The given matrix is.

It can be observed that in the first column, two entries are zero. Thus, we expand along the first column (C_{1}) for easier calculation.

From equations (i) and (ii), we have:

Hence, the given result is proved.

Evaluate the determinants

(i) (iii)

(ii) (iv)

(i) Let.

It can be observed that in the second row, two entries are zero. Thus, we expand along the second row for easier calculation.

(ii) Let.

By expanding along the first row, we have:

(iii) Let

By expanding along the first row, we have:

(iv) Let

By expanding along the first column, we have:

If, find.

Let

By expanding along the first row, we have:

Find values of *x*, if

(i)

2451=2x46x(ii)

2345=x32x5

(i)

(ii)

If, then *x* is equal to

(A) 6 (B) ±6 (C) −6 (D) 0

**Answer: B**

Hence, the correct answer is B.

Using the property of determinants and without expanding, prove that:

Using the property of determinants and without expanding, prove that:

Here, the two rows R_{1} and R_{3} are identical.

Δ = 0.

Using the property of determinants and without expanding, prove that:

Using the property of determinants and without expanding, prove that:

By applying C_{3 }→ C_{3} + C_{2, }we have:

Here, two columns C_{1} and C_{3 }are proportional.

Δ = 0.

Using the property of determinants and without expanding, prove that:

Applying R_{2} → R_{2} − R_{3}, we have:

Applying R_{1} R_{3} and R_{2} R_{3}, we have:

Applying R_{1 }→ R_{1} − R_{3}, we have:

Applying R_{1} R_{2} and R_{2} R_{3}, we have:

From (1), (2), and (3), we have:

Hence, the given result is proved.

By using properties of determinants, show that:

We have,

Here, the two rows R_{1} and R_{3 }are identical.

∴Δ = 0.

By using properties of determinants, show that:

Applying R_{2 }→ R_{2} + R_{1} and R_{3 }→ R_{3} + R_{1}, we have:

By using properties of determinants, show that:

(i)

(ii)

(i)

Applying R_{1} → R_{1} − R_{3 }and R_{2} → R_{2} − R_{3}, we have:

Applying R_{1} → R_{1} + R_{2}, we have:

Expanding along C_{1}, we have:

Hence, the given result is proved.

(ii) Let.

Applying C_{1} → C_{1} − C_{3 }and C_{2} → C_{2} − C_{3}, we have:

Applying C_{1} → C_{1} + C_{2}, we have:

Expanding along C_{1}, we have:

Hence, the given result is proved.

By using properties of determinants, show that:

Applying R_{2} → R_{2} − R_{1 }and R_{3} → R_{3} − R_{1}, we have:

Applying R_{3} → R_{3} + R_{2}, we have:

Expanding along R_{3}, we have:

Hence, the given result is proved.

By using properties of determinants, show that:

(i)

(ii)

(i)

Applying R_{1} → R_{1} + R_{2 }+ R_{3}, we have:

Applying C_{2} → C_{2} − C_{1}, C_{3} → C_{3} − C_{1}, we have:

Expanding along C_{3}, we have:

Hence, the given result is proved.

(ii)

Applying R_{1} → R_{1} + R_{2 }+ R_{3}, we have:

Applying C_{2} → C_{2} − C_{1 }and C_{3} → C_{3} − C_{1}, we have:

Expanding along C_{3}, we have:

Hence, the given result is proved.

By using properties of determinants, show that:

(i)

(ii)

(i)

Applying R_{1} → R_{1} + R_{2 }+ R_{3}, we have:

Applying C_{2} → C_{2} − C_{1}, C_{3} → C_{3} − C_{1}, we have:

Expanding along C_{3}, we have:

Hence, the given result is proved.

(ii)

Applying C_{1} → C_{1} + C_{2 }+ C_{3}, we have:

Applying R_{2} → R_{2} − R_{1 }and R_{3} → R_{3} − R_{1}, we have:

Expanding along R_{3}, we have:

Hence, the given result is proved.

By using properties of determinants, show that:

Applying R_{1} → R_{1} + R_{2 }+ R_{3}, we have:

Applying C_{2} → C_{2} − C_{1 }and C_{3} → C_{3} − C_{1}, we have:

Expanding along R_{1}, we have:

Hence, the given result is proved.

By using properties of determinants, show that:

Applying R_{1} → R_{1} + *b*R_{3 }and R_{2} → R_{2} − *a*R_{3}, we have:

Expanding along R_{1}, we have:

By using properties of determinants, show that:

Taking out common factors *a*, *b*, and *c* from R_{1}, R_{2}, and R_{3 }respectively, we have:

Applying R_{2} → R_{2} − R_{1 }and R_{3} → R_{3} − R_{1}, we have:

Applying C_{1} → *a*C_{1}, C_{2 }→ *b*C_{2, }and C_{3} → *c*C_{3}, we have:

Expanding along R_{3}, we have:

Hence, the given result is proved.

Choose the correct answer.

Let *A* be a square matrix of order 3 × 3, then is equal to

**A. ** **B. ** **C. ** **D.**

**Answer: C**

*A* is a square matrix of order 3 × 3.

Hence, the correct answer is C.

Which of the following is correct?

**A.** Determinant is a square matrix.

**B.** Determinant is a number associated to a matrix.

**C.** Determinant is a number associated to a square matrix.

**D. **None of these

**Answer: C**

We know that to every square matrix, of order *n*. We can associate a number called the determinant of square matrix *A*, where element of *A*.

Thus, the determinant is a number associated to a square matrix.

Hence, the correct answer is C.

Find area of the triangle with vertices at the point given in each of the following:

(i) (1, 0), (6, 0), (4, 3) (ii) (2, 7), (1, 1), (10, 8)

(iii) (−2, −3), (3, 2), (−1, −8)

(i) The area of the triangle with vertices (1, 0), (6, 0), (4, 3) is given by the relation,

(ii) The area of the triangle with vertices (2, 7), (1, 1), (10, 8) is given by the relation,

(iii) The area of the triangle with vertices (−2, −3), (3, 2), (−1, −8)

is given by the relation,

Hence, the area of the triangle is.

Show that points

are collinear

Area of ΔABC is given by the relation,

Thus, the area of the triangle formed by points A, B, and C is zero.

Hence, the points A, B, and C are collinear.

Find values of *k* if area of triangle is 4 square units and vertices are

(i) (*k*, 0), (4, 0), (0, 2) (ii) (−2, 0), (0, 4), (0, *k*)

We know that the area of a triangle whose vertices are (*x*_{1}, *y*_{1}), (*x*_{2}, *y*_{2}), and

(*x*_{3}, *y*_{3}) is the absolute value of the determinant (Δ), where

It is given that the area of triangle is 4 square units.

∴Δ = ± 4.

(i) The area of the triangle with vertices (*k*, 0), (4, 0), (0, 2) is given by the relation,

Δ =

**∴**−*k* + 4 = ± 4

When −*k* + 4 = − 4, *k* = 8.

When −*k* + 4 = 4, *k* = 0.

Hence, *k* = 0, 8.

(ii) The area of the triangle with vertices (−2, 0), (0, 4), (0, *k*) is given by the relation,

Δ =

∴*k* − 4 = ± 4

When *k* − 4 = − 4, *k* = 0.

When *k* − 4 = 4, *k* = 8.

Hence, *k* = 0, 8.

(i) Find equation of line joining (1, 2) and (3, 6) using determinants

(ii) Find equation of line joining (3, 1) and (9, 3) using determinants

(i) Let P (*x*, *y*) be any point on the line joining points A (1, 2) and B (3, 6). Then, the points A, B, and P are collinear. Therefore, the area of triangle ABP will be zero.

Hence, the equation of the line joining the given points is *y* = 2*x*.

(ii) Let P (*x*, *y*) be any point on the line joining points A (3, 1) and

B (9, 3). Then, the points A, B, and P are collinear. Therefore, the area of triangle ABP will be zero.

Hence, the equation of the line joining the given points is *x* − 3*y* = 0.

If area of triangle is 35 square units with vertices (2, −6), (5, 4), and (*k*, 4). Then *k* is

**A.** 12 **B.** −2 **C.** −12, −2 **D.** 12, −2

**Answer: D**

The area of the triangle with vertices (2, −6), (5, 4), and (*k*, 4) is given by the relation,

It is given that the area of the triangle is ±35.

Therefore, we have:

When 5 − *k* = −7, *k* = 5 + 7 = 12.

When 5 − *k* = 7, *k* = 5 − 7 = −2.

Hence, *k* = 12, −2.

The correct answer is D.

Write Minors and Cofactors of the elements of following determinants:

(i) (ii)

(i) The given determinant is.

Minor of element *a*_{ij} is M_{ij.}

∴M_{11} = minor of element *a*_{11 }= 3

M_{12} = minor of element *a*_{12 }= 0

M_{21} = minor of element *a*_{21 }= −4

M_{22} = minor of element *a*_{22 }= 2

Cofactor of *a*_{ij} is A_{ij} = (−1)^{i + j} M_{ij}.

∴A_{11} = (−1)^{1+1} M_{11} = (−1)^{2} (3) = 3

A_{12} = (−1)^{1+2} M_{12} = (−1)^{3} (0) = 0

A_{21} = (−1)^{2+1} M_{21} = (−1)^{3} (−4) = 4

A_{22} = (−1)^{2+2} M_{22} = (−1)^{4} (2) = 2

(ii) The given determinant is.

Minor of element *a*_{ij} is M_{ij}.

∴M_{11} = minor of element *a*_{11 }= *d*

M_{12} = minor of element *a*_{12 }= *b*

M_{21} = minor of element *a*_{21 }= *c*

M_{22} = minor of element *a*_{22 }= *a*

Cofactor of *a*_{ij} is A_{ij} = (−1)^{i + j} M_{ij.}

∴A_{11} = (−1)^{1+1} M_{11} = (−1)^{2} (*d*) = *d*

A_{12} = (−1)^{1+2} M_{12} = (−1)^{3} (*b*) = −*b*

A_{21} = (−1)^{2+1} M_{21} = (−1)^{3} (*c*) = −*c*

A_{22} = (−1)^{2+2} M_{22} = (−1)^{4} (*a*) = *a*

(i) (ii)

(i) The given determinant is.

By the definition of minors and cofactors, we have:

M_{11 }= minor of *a*_{11}=

M_{12 }= minor of *a*_{12}=

M_{13 }= minor of *a*_{13 }=

M_{21 }= minor of *a*_{21 }=

M_{22 }= minor of *a*_{22 }=

M_{23 }= minor of *a*_{23 }=

M_{31 }= minor of *a*_{31}=

M_{32 }= minor of *a*_{32 }=

M_{33 }= minor of *a*_{33 }=

A_{11 }= cofactor of *a*_{11}= (−1)^{1+1} M_{11} = 1

A_{12 }= cofactor of *a*_{12 }= (−1)^{1+2} M_{12} = 0

A_{13 }= cofactor of *a*_{13 }= (−1)^{1+3} M_{13} = 0

A_{21 }= cofactor of *a*_{21 }= (−1)^{2+1} M_{21} = 0

A_{22 }= cofactor of *a*_{22 }= (−1)^{2+2} M_{22} = 1

A_{23 }= cofactor of *a*_{23 }= (−1)^{2+3} M_{23} = 0

A_{31 }= cofactor of *a*_{31 }= (−1)^{3+1} M_{31} = 0

A_{32 }= cofactor of *a*_{32 }= (−1)^{3+2} M_{32} = 0

A_{33 }= cofactor of *a*_{33 }= (−1)^{3+3} M_{33} = 1

(ii) The given determinant is.

By definition of minors and cofactors, we have:

M_{11 }= minor of *a*_{11}=

M_{12 }= minor of *a*_{12}=

M_{13 }= minor of *a*_{13 }=

M_{21 }= minor of *a*_{21 }=

M_{22 }= minor of *a*_{22 }=

M_{23 }= minor of *a*_{23 }=

M_{31 }= minor of *a*_{31}=

M_{32 }= minor of *a*_{32 }=

M_{33 }= minor of *a*_{33 }=

A_{11 }= cofactor of *a*_{11}= (−1)^{1+1} M_{11} = 11

A_{12 }= cofactor of *a*_{12 }= (−1)^{1+2} M_{12} = −6

A_{13 }= cofactor of *a*_{13 }= (−1)^{1+3} M_{13} = 3

A_{21 }= cofactor of *a*_{21 }= (−1)^{2+1} M_{21} = 4

A_{22 }= cofactor of *a*_{22 }= (−1)^{2+2} M_{22} = 2

A_{23 }= cofactor of *a*_{23 }= (−1)^{2+3} M_{23} = −1

A_{31 }= cofactor of *a*_{31 }= (−1)^{3+1} M_{31} = −20

A_{32 }= cofactor of *a*_{32 }= (−1)^{3+2} M_{32} = 13

A_{33 }= cofactor of *a*_{33 }= (−1)^{3+3} M_{33} = 5

Using Cofactors of elements of second row, evaluate.

The given determinant is.

We have:

M_{21 }=

∴A_{21 }= cofactor of *a*_{21 }= (−1)^{2+1} M_{21} = 7

M_{22 }=

∴A_{22 }= cofactor of *a*_{22 }= (−1)^{2+2} M_{22} = 7

M_{23 }=

∴A_{23 }= cofactor of *a*_{23 }= (−1)^{2+3} M_{23} = −7

We know that Δ is equal to the sum of the product of the elements of the second row with their corresponding cofactors.

∴Δ = *a*_{21}A_{21} + *a*_{22}A_{22} + *a*_{23}A_{23} = 2(7) + 0(7) + 1(−7) = 14 − 7 = 7

Using Cofactors of elements of third column, evaluate

The given determinant is.

We have:

M_{13 }=

M_{23 }=

M_{33 }=

∴A_{13 }= cofactor of *a*_{13 }= (−1)^{1+3} M_{13} = (*z − y*)

A_{23 }= cofactor of *a*_{23 }= (−1)^{2+3} M_{23} = − (*z − x*) = (*x − z*)

A_{33 }= cofactor of *a*_{33 }= (−1)^{3+3} M_{33} = (*y* − *x*)

We know that Δ is equal to the sum of the product of the elements of the second row with their corresponding cofactors.

Hence,

If and A_{ij} is Cofactors of *a*_{ij}, then value of Δ is given by

**Answer: D**

We know that:

Δ = Sum of the product of the elements of a column (or a row) with their corresponding cofactors

∴Δ = *a*_{11}A_{11} + *a*_{21}A_{21} + *a*_{31}A_{31}

Hence, the value of Δ is given by the expression given in alternative **D**.

The correct answer is D.

Find adjoint of each of the matrices.

Find adjoint of each of the matrices.

Verify *A* (*adj A*) = (*adj A*) *A* = *I* .

Verify *A* (*adj A*) = (*adj A*) *A* = *I* .

Find the inverse of each of the matrices (if it exists).

Find the inverse of each of the matrices (if it exists).

Find the inverse of each of the matrices (if it exists).

Find the inverse of each of the matrices (if it exists).

Find the inverse of each of the matrices (if it exists).

Find the inverse of each of the matrices (if it exists).

.

Find the inverse of each of the matrices (if it exists).

Let and. Verify that

From (1) and (2), we have:

(*AB*)^{−1} = *B*^{−1}*A*^{−1}

Hence, the given result is proved.

If, show that. Hence find.

For the matrix, find the numbers *a* and *b* such that *A*^{2} + *aA* + *bI *= *O*.

We have:

Comparing the corresponding elements of the two matrices, we have:

Hence, −4 and 1 are the required values of *a* and *b* respectively.

For the matrixshow that *A*^{3} − 6*A*^{2} + 5*A* + 11 *I* = O. Hence, find *A*^{−1.}

From equation (1), we have:

If verify that *A*^{3} − 6*A*^{2} + 9*A* − 4*I* = *O* and hence find *A*^{−1}

From equation (1), we have:

Let *A* be a nonsingular square matrix of order 3 × 3. Then is equal to

**A.** **B.** **C.** **D. **

**Answer: B**

We know that,

Hence, the correct answer is B.

If *A* is an invertible matrix of order 2, then det (*A*^{−1}) is equal to

**A.** det (*A*) **B.** **C.** 1 **D. **0

Since *A* is an invertible matrix,

Hence, the correct answer is B.

Examine the consistency of the system of equations.

*x *+ 2*y *= 2

2*x* + 3*y *= 3

The given system of equations is:

*x *+ 2*y *= 2

2*x* + 3*y *= 3

The given system of equations can be written in the form of *AX* = *B*, where

∴ *A* is non-singular.

Therefore, *A*^{−1} exists.

Hence, the given system of equations is consistent.

Examine the consistency of the system of equations.

2*x *− *y* = 5

*x* + *y *= 4

The given system of equations is:

2*x *− *y* = 5

*x* + *y *= 4

The given system of equations can be written in the form of *AX* = *B*, where

∴ *A* is non-singular.

Therefore, *A*^{−1} exists.

Hence, the given system of equations is consistent.

Examine the consistency of the system of equations.

*x* + 3*y* = 5

2*x* + 6*y* = 8

The given system of equations is:

*x* + 3*y* = 5

2*x* + 6*y* = 8

The given system of equations can be written in the form of *AX* = *B*, where

∴ *A* is a singular matrix.

Thus, the solution of the given system of equations does not exist. Hence, the system of equations is inconsistent.

Examine the consistency of the system of equations.

*x* +* y *+ *z* = 1

2*x* + 3*y* + 2*z* = 2

*ax* + *ay* + 2*az* = 4

The given system of equations is:

*x* +* y *+ *z* = 1

2*x* + 3*y* + 2*z* = 2

*ax* + *ay* + 2*az* = 4

This system of equations can be written in the form *AX* = *B*, where

∴ *A* is non-singular.

Therefore, *A*^{−1} exists.

Hence, the given system of equations is consistent.

Examine the consistency of the system of equations.

3*x* −* y *− 2z = 2

2*y* − *z* = −1

3*x* − 5*y* = 3

The given system of equations is:

3*x* −* y *− 2z = 2

2*y* − *z* = −1

3*x* − 5*y* = 3

This system of equations can be written in the form of *AX* = *B*, where

∴ *A* is a singular matrix.

Thus, the solution of the given system of equations does not exist. Hence, the system of equations is inconsistent.

Examine the consistency of the system of equations.

5*x* −* y *+ 4*z* = 5

2*x* + 3*y* + 5*z* = 2

5*x* − 2*y* + 6*z* = −1

The given system of equations is:

5*x* −* y *+ 4*z* = 5

2*x* + 3*y* + 5*z* = 2

5*x* − 2*y* + 6*z* = −1

This system of equations can be written in the form of *AX* = *B*, where

∴ *A* is non-singular.

Therefore, *A*^{−1} exists.

Hence, the given system of equations is consistent.

Solve system of linear equations, using matrix method.

The given system of equations can be written in the form of *AX* = *B*, where

Thus, *A* is non-singular. Therefore, its inverse exists.

Solve system of linear equations, using matrix method.

The given system of equations can be written in the form of *AX* = *B*, where

Thus, *A* is non-singular. Therefore, its inverse exists.

Solve system of linear equations, using matrix method.

The given system of equations can be written in the form of *AX* = *B*, where

Thus, *A* is non-singular. Therefore, its inverse exists.

Solve system of linear equations, using matrix method.

5*x* + 2*y* = 3

3*x* + 2*y* = 5

The given system of equations can be written in the form of *AX* = *B*, where

Thus, *A* is non-singular. Therefore, its inverse exists.

Solve system of linear equations, using matrix method.

The given system of equations can be written in the form of *AX* = *B*, where

Thus, *A* is non-singular. Therefore, its inverse exists.

Solve system of linear equations, using matrix method.

*x* − *y* + *z* = 4

2*x* + *y* − 3*z* = 0

*x* + *y* + *z* = 2

The given system of equations can be written in the form of *AX* = *B*, where

Thus, *A* is non-singular. Therefore, its inverse exists.

Solve system of linear equations, using matrix method.

2*x* + 3*y* + 3*z* = 5

*x* − 2*y* + *z* = −4

3*x* − *y* − 2*z* = 3

The given system of equations can be written in the form *AX* = *B*, where

Thus, *A* is non-singular. Therefore, its inverse exists.

Solve system of linear equations, using matrix method.

*x* − *y* + 2*z* = 7

3*x* + 4*y* − 5*z* = −5

2*x* −* y* + 3*z* = 12

The given system of equations can be written in the form of *AX* = *B*, where

Thus, *A* is non-singular. Therefore, its inverse exists.

If, find *A*^{−1}. Using A^{−1} solve the system of equations

Now, the given system of equations can be written in the form of *AX* = *B*, where

The cost of 4 kg onion, 3 kg wheat and 2 kg rice is Rs 60. The cost of 2 kg onion, 4 kg

wheat and 6 kg rice is Rs 90. The cost of 6 kg onion 2 kg wheat and 3 kg rice is Rs 70.

Find cost of each item per kg by matrix method.

Let the cost of onions, wheat, and rice per kg be Rs *x*, Rs *y*,and Rs *z* respectively.

Then, the given situation can be represented by a system of equations as:

This system of equations can be written in the form of *AX* = *B*, where

Now,

*X* = *A*^{−1} *B*

Hence, the cost of onions is Rs 5 per kg, the cost of wheat is Rs 8 per kg, and the cost of rice is Rs 8 per kg.

Prove that the determinant is independent of *θ*.

Hence, Î” is independent of *Î¸*.

Without expanding the determinant, prove that

Hence, the given result is proved.

Evaluate

Expanding along C_{3}, we have:

If *a*, *b* and *c *are real numbers, and,

Show that either *a* + *b* + *c* = 0 or *a* = *b* = *c*.

Expanding along R_{1}, we have:

Hence, if Δ = 0, then either *a* + *b* + *c* = 0 or *a* = *b* = *c*.

Solve the equations

Prove that

Expanding along R_{3}, we have:

Hence, the given result is proved.

If

We know that.

Let verify that

(i)

(ii)

(i)

We have,

(ii)

Evaluate

Expanding along R_{1}, we have:

Evaluate

Expanding along C_{1}, we have:

Using properties of determinants, prove that:

Expanding along R_{3}, we have:

Hence, the given result is proved.

Using properties of determinants, prove that:

Expanding along R_{3}, we have:

Hence, the given result is proved.

Using properties of determinants, prove that:

Expanding along C_{1}, we have:

Hence, the given result is proved.

Using properties of determinants, prove that:

Expanding along C_{1}, we have:

Hence, the given result is proved.

Using properties of determinants, prove that:

Hence, the given result is proved.

Solve the system of the following equations

Let

Then the given system of equations is as follows:

This system can be written in the form of *AX *= *B*, where

A

Thus, *A* is non-singular. Therefore, its inverse exists.

Now,

*A*_{11} = 75, *A*_{12} = 110, *A*_{13} = 72

*A*_{21} = 150, *A*_{22} = −100, *A*_{23} = 0

*A*_{31} = 75, *A*_{32} = 30, *A*_{33} = − 24

Choose the correct answer.

If *a*, *b*, *c*, are in A.P., then the determinant

**A.** 0 **B.** 1 **C.** *x ***D. **2*x*

**Answer:** **A**

Here, all the elements of the first row (R_{1}) are zero.

Hence, we have Δ = 0.

The correct answer is A.

Choose the correct answer.

If *x*, *y*, *z* are nonzero real numbers, then the inverse of matrix is

**A.** **B.**

**C.** **D. **

**Answer: A**

The correct answer is A.

Choose the correct answer.

Let, where 0 ≤ *θ*≤ 2π, then

**A.** Det (A) = 0

**B.** Det (A) ∈ (2, ∞)

**C.** Det (A) ∈ (2, 4)

**D. **Det (A)∈ [2, 4]

Answer: D

Now,

0≤θ≤2π

⇒-1≤sinθ≤1 The correct answer is D.

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]]>If you have not checked solution of Chapter 2 i.e Inverse Trigonometric Functions, then you can check it from the link given above.

**Topics and Sub Topics in Class 11 Maths Chapter 3 Matrices:**

Section Name | Topic Name |

3 | Matrices |

3.1 | Introduction |

3.2 | Matrix |

3.3 | Types of Matrices |

3.4 | Operations on Matrices |

3.5 | Transpose of a Matrix |

3.6 | Symmetric and Skew Symmetric Matrices |

3.7 | Elementary Operation (Transformation) of a Matrix |

3.8 | Invertible Matrices |

In the matrix, write:

(i) The order of the matrix (ii) The number of elements,

(iii) Write the elements *a*_{13}, *a*_{21}, *a*_{33}, *a*_{24}, *a*_{23}

**(i)** In the given matrix, the number of rows is 3 and the number of columns is 4. Therefore, the order of the matrix is 3 × 4.

**(ii)** Since the order of the matrix is 3 × 4, there are 3 × 4 = 12 elements in it.

**(iii)** *a*_{13} = 19, *a*_{21} = 35, *a*_{33} = −5, *a*_{24} = 12, *a*_{23} =

If a matrix has 24 elements, what are the possible order it can have? What, if it has 13 elements?

We know that if a matrix is of the order *m* × *n*, it has *mn *elements. Thus, to find all the possible orders of a matrix having 24 elements, we have to find all the ordered pairs of natural numbers whose product is 24.

The ordered pairs are: (1, 24), (24, 1), (2, 12), (12, 2), (3, 8), (8, 3), (4, 6), and (6, 4)

Hence, the possible orders of a matrix having 24 elements are:

1 × 24, 24 × 1, 2 × 12, 12 × 2, 3 × 8, 8 × 3, 4 × 6, and 6 × 4

(1, 13) and (13, 1) are the ordered pairs of natural numbers whose product is 13.

Hence, the possible orders of a matrix having 13 elements are 1 × 13 and 13 × 1.

If a matrix has 18 elements, what are the possible orders it can have? What, if it has 5 elements?

We know that if a matrix is of the order *m* × *n*, it has *mn* elements. Thus, to find all the possible orders of a matrix having 18 elements, we have to find all the ordered pairs of natural numbers whose product is 18.

The ordered pairs are: (1, 18), (18, 1), (2, 9), (9, 2), (3, 6,), and (6, 3)

Hence, the possible orders of a matrix having 18 elements are:

1 × 18, 18 × 1, 2 × 9, 9 × 2, 3 × 6, and 6 × 3

(1, 5) and (5, 1) are the ordered pairs of natural numbers whose product is 5.

Hence, the possible orders of a matrix having 5 elements are 1 × 5 and 5 × 1.

Construct a 2 × 2 matrix,, whose elements are given by:

(i)

(ii)

(iii)

In general, a 2 × 2 matrix is given by

**(i)**

Therefore, the required matrix is

**(ii)**

Therefore, the required matrix is

**(iii)**

Therefore, the required matrix is

Construct a 3 × 4 matrix, whose elements are given by

(i) (ii)

In general, a 3 × 4 matrix is given by

**(i)**

Therefore, the required matrix is

**(ii)**

Therefore, the required matrix is

Find the value of *x*, *y*, and *z* from the following equation:

(i) (ii)

(iii)

**(i)**

As the given matrices are equal, their corresponding elements are also equal.

Comparing the corresponding elements, we get:

*x* = 1, *y* = 4, and *z* = 3

**(ii)**

As the given matrices are equal, their corresponding elements are also equal.

Comparing the corresponding elements, we get:

*x* + *y* = 6, *xy* = 8, 5 + *z *= 5

Now, 5 + *z* = 5 ⇒ *z* = 0

We know that:

(*x* − *y*)^{2} = (*x* + *y*)^{2} − 4*xy*

⇒ (*x* − *y*)^{2} = 36 − 32 = 4

⇒ *x* − *y* = ±2

Now, when *x* − *y* = 2 and *x* +* y *= 6, we get *x* = 4 and *y* = 2

When *x* − *y *= − 2 and *x* + *y* = 6, we get *x* = 2 and *y *= 4

∴*x* = 4, *y* = 2, and *z* = 0 or *x* = 2, *y* = 4, and *z* = 0

**(iii)**

As the two matrices are equal, their corresponding elements are also equal.

Comparing the corresponding elements, we get:

*x* + *y* + *z* = 9 … (1)

*x* + *z* = 5 … (2)

*y* + *z* = 7 … (3)

From (1) and (2), we have:

*y *+ 5 = 9

⇒ *y* = 4

Then, from (3), we have:

4 + *z* = 7

⇒ *z* = 3

∴ *x* +* z *= 5

⇒ *x* = 2

∴ *x* = 2, *y* = 4, and *z* = 3

Find the value of *a*, *b*, *c*, and *d* from the equation:

As the two matrices are equal, their corresponding elements are also equal.

Comparing the corresponding elements, we get:

*a* − *b* = −1 … (1)

2*a* − *b* = 0 … (2)

2*a* + *c* = 5 … (3)

3*c* + *d* = 13 … (4)

From (2), we have:

*b* = 2*a*

Then, from (1), we have:

*a *− 2*a* = −1

⇒ *a* = 1

⇒ *b* = 2

Now, from (3), we have:

2 ×1 +* c* = 5

⇒ *c* = 3

From (4) we have:

3 ×3 + *d* = 13

⇒ 9 + *d *= 13 ⇒ *d* = 4

∴*a* = 1, *b* = 2, *c* = 3, and *d* = 4

is a square matrix, if

**(A)** *m* < *n*

**(B)** *m* > *n*

**(C)** *m* = *n*

**(D)** None of these

The correct answer is C.

It is known that a given matrix is said to be a square matrix if the number of rows is equal to the number of columns.

Therefore, is a square matrix, if *m* = *n*.

Which of the given values of *x* and *y* make the following pair of matrices equal

**(A)**

**(B)** Not possible to find

**(C)**

**(D)**

The correct answer is B.

It is given that

Equating the corresponding elements, we get:

We find that on comparing the corresponding elements of the two matrices, we get two different values of *x*, which is not possible.

Hence, it is not possible to find the values of *x* and *y* for which the given matrices are equal.

The number of all possible matrices of order 3 × 3 with each entry 0 or 1 is:

**(A)** 27

**(B) **18

**(C)** 81

**(D)** 512

The correct answer is D.

The given matrix of the order 3 × 3 has 9 elements and each of these elements can be either 0 or 1.

Now, each of the 9 elements can be filled in two possible ways.

Therefore, by the multiplication principle, the required number of possible matrices is 2^{9} = 512

Let

Find each of the following

(i) (ii) (iii)

(iv) (v)

**(i)**

**(ii)**

**(iii)**

**(iv)** Matrix *A* has 2 columns. This number is equal to the number of rows in matrix *B*. Therefore, *AB* is defined as:

**(v)** Matrix* B *has 2 columns. This number is equal to the number of rows in matrix *A*. Therefore, *BA* is defined as:

Compute the following:

(i) (ii)

(iii)

(v)

**(i)**

**(ii)**

**(iii)**

**(iv)**

Compute the indicated products

(i)

(ii)

(iii)

(iv)

(v)

(vi)

**(i)**

**(ii)**

**(iii)**

**(iv)**

**(v)**

**(****vi)**

If, and, then compute and. Also, verify that

If andthen compute.

Simplify

Find *X *and *Y*, if

(i) and

(ii) and

**(i)**

Adding equations (1) and (2), we get:

**(ii)**

Multiplying equation (3) with (2), we get:

Multiplying equation (4) with (3), we get:

From (5) and (6), we have:

Now,

Find *X*, if and

Find *x* and *y*, if

Comparing the corresponding elements of these two matrices, we have:

∴*x* = 3 and *y* = 3

Solve the equation for *x*, *y*, *z* and *t* if

Comparing the corresponding elements of these two matrices, we get:

If, find values of *x* and *y*.

Comparing the corresponding elements of these two matrices, we get:

2*x* − *y* = 10 and 3*x* + *y* = 5

Adding these two equations, we have:

5*x* = 15

⇒ *x* = 3

Now, 3*x* + *y* = 5

⇒ *y* = 5 − 3*x*

⇒ *y* = 5 − 9 = −4

∴*x* = 3 and *y* = −4

Given, find the values of *x*, *y*, *z* and

*w*.

Comparing the corresponding elements of these two matrices, we get:

If, show that.

Show that

(i)

(ii)

**(i)**

**(ii)**

Find if

We have *A*^{2} = *A* × *A*

If, prove that

If and, find *k* so that

Comparing the corresponding elements, we have:

Thus, the value of *k* is 1.

Ifand *I* is the identity matrix of order 2, show that

A trust fund has Rs 30,000 that must be invested in two different types of bonds. The first bond pays 5% interest per year, and the second bond pays 7% interest per year. Using matrix multiplication, determine how to divide Rs 30,000 among the two types of bonds. If the trust fund must obtain an annual total interest of:

(a) Rs 1,800 (b) Rs 2,000

**(a)** Let Rs *x* be invested in the first bond. Then, the sum of money invested in the second bond will be Rs (30000 − *x*).

It is given that the first bond pays 5% interest per year and the second bond pays 7% interest per year.

Therefore, in order to obtain an annual total interest of Rs 1800, we have:

Thus, in order to obtain an annual total interest of Rs 1800, the trust fund should invest Rs 15000 in the first bond and the remaining Rs 15000 in the second bond.

**(b)** Let Rs *x* be invested in the first bond. Then, the sum of money invested in the second bond will be Rs (30000 − *x*).

Therefore, in order to obtain an annual total interest of Rs 2000, we have:

Thus, in order to obtain an annual total interest of Rs 2000, the trust fund should invest Rs 5000 in the first bond and the remaining Rs 25000 in the second bond.

The bookshop of a particular school has 10 dozen chemistry books, 8 dozen physics books, 10 dozen economics books. Their selling prices are Rs 80, Rs 60 and Rs 40 each respectively. Find the total amount the bookshop will receive from selling all the books using matrix algebra.

The bookshop has 10 dozen chemistry books, 8 dozen physics books, and 10 dozen economics books.

The selling prices of a chemistry book, a physics book, and an economics book are respectively given as Rs 80, Rs 60, and Rs 40.

The total amount of money that will be received from the sale of all these books can be represented in the form of a matrix as:

Thus, the bookshop will receive Rs 20160 from the sale of all these books.

Assume *X*, *Y*, *Z*, *W* and *P* are matrices of order, and respectively. The restriction on *n*, *k* and *p* so that will be defined are:

**A.** *k* = 3, *p* = *n*

**B.** *k* is arbitrary, *p* = 2

**C.** *p* is arbitrary, *k* = 3

**D.** *k* = 2, *p* = 3

Matrices *P* and *Y* are of the orders *p* × *k* and 3 × *k* respectively.

Therefore, matrix *PY *will be defined if *k* = 3. Consequently, *PY* will be of the order *p* × *k*.

Matrices *W* and *Y* are of the orders *n *× 3 and 3 × *k* respectively.

Since the number of columns in *W* is equal to the number of rows in *Y*, matrix *WY* is well-defined and is of the order *n *× *k*.

Matrices *PY* and *WY* can be added only when their orders are the same.

However, *PY* is of the order *p* × *k* and *WY* is of the order *n* × *k*. Therefore, we must have *p* = *n*.

Thus, *k *= 3 and *p* = *n* are the restrictions on *n*, *k*, and *p* so that will be defined.

Assume *X*, *Y*, *Z*, *W* and *P* are matrices of order, and respectively. If *n* = *p*, then the order of the matrix is

**A** *p* × 2 **B** 2 × *n* **C** *n* × 3 **D** *p* × *n*

The correct answer is B.

Matrix *X* is of the order 2 × *n*.

Therefore, matrix 7*X* is also of the same order.

Matrix *Z* is of the order 2 × *p*, i.e., 2 × *n* [Since *n* = *p*]

Therefore, matrix 5*Z* is also of the same order.

Now, both the matrices 7*X* and 5*Z* are of the order 2 × *n*.

Thus, matrix 7*X* − 5*Z* is well-defined and is of the order 2 × *n*.

Find the transpose of each of the following matrices:

(i) (ii) (iii)

**(i)**

**(ii)**

**(iii)**

If

and, then verify that

(i)

(ii)

We have:

**(i)**

**(ii)**

If

and, then verify that

(i)

(ii)

**(i)** It is known that

Therefore, we have:

**(ii)**

If

and, then find

We know that

For the matrices *A* and *B*, verify that (*AB*)′ = where

(i)

(ii)

**(i)**

**(ii)**

If (i) , then verify that

(ii) , then verify that

**(i)**

**(ii)**

(i) Show that the matrix is a symmetric matrix

(ii) Show that the matrix is a skew symmetric matrix

**(i)** We have:

Hence, *A* is a symmetric matrix.

**(ii)** We have:

Hence, *A* is a skew-symmetric matrix.

For the matrix, verify that

(i) is a symmetric matrix

(ii) is a skew symmetric matrix

**(i)**

Hence,

is a symmetric matrix.

**(ii)**

Hence,

is a skew-symmetric matrix.

Find

and, when

The given matrix is

Express the following matrices as the sum of a symmetric and a skew symmetric matrix:

(i)

(ii)

(iii)

(iv)

**(i)**

Thus, is a symmetric matrix.

Thus, is a skew-symmetric matrix.

Representing *A* as the sum of *P* and *Q*:

**(ii)**

Thus, is a symmetric matrix.

Thus, is a skew-symmetric matrix.

Representing *A* as the sum of *P* and *Q*:

**(iii)**

Thus, is a symmetric matrix.

Thus, is a skew-symmetric matrix.

Representing *A* as the sum of *P* and *Q*:

**(iv)**

Thus, is a symmetric matrix.

Thus, is a skew-symmetric matrix.

Representing *A* as the sum of *P* and *Q*:

If *A*, *B* are symmetric matrices of same order, then *AB* − *BA* is a

**A.** Skew symmetric matrix **B.** Symmetric matrix

**C.** Zero matrix **D. **Identity matrix

The correct answer is A.

*A* and *B* are symmetric matrices, therefore, we have:

Thus, (*AB* − *BA*) is a skew-symmetric matrix.

If, then, if the value of α is

**A.** **B.**

**C. **π **D. **

The correct answer is B.

Comparing the corresponding elements of the two matrices, we have:

Find the inverse of each of the matrices, if it exists.

We know that *A* = *IA*

Find the inverse of each of the matrices, if it exists.

We know that *A* = *IA*

Find the inverse of each of the matrices, if it exists.

We know that *A* = *IA*

Find the inverse of each of the matrices, if it exists.

We know that *A* = *IA*

Find the inverse of each of the matrices, if it exists.

We know that *A* = *IA*

Find the inverse of each of the matrices, if it exists.

We know that *A* = *IA*

Find the inverse of each of the matrices, if it exists.

We know that *A* = *AI*

Find the inverse of each of the matrices, if it exists.

We know that *A* = *IA*

Find the inverse of each of the matrices, if it exists.

We know that *A* = *IA*

Find the inverse of each of the matrices, if it exists.

We know that *A* = *AI*

Find the inverse of each of the matrices, if it exists.

We know that *A* = *AI*

Find the inverse of each of the matrices, if it exists.

We know that *A* = *IA*

Now, in the above equation, we can see all the zeros in the second row of the matrix on the L.H.S.

Therefore, *A*^{−1} does not exist.

Find the inverse of each of the matrices, if it exists.

We know that *A* = *IA*

Find the inverse of each of the matrices, if it exists.

We know that *A* = *IA*

Applying, we have:

Now, in the above equation, we can see all the zeros in the first row of the matrix on the L.H.S.

Therefore, *A*^{−1} does not exist.

Find the inverse of each of the matrices, if it exists.

We know that *A* = *IA*

Find the inverse of each of the matrices, if it exists.

We know that *A* = *IA*

Applying R_{2} → R_{2} + 3R_{1} and R_{3} → R_{3} − 2R_{1}, we have:

Find the inverse of each of the matrices, if it exists.

We know that *A* = *IA*

Applying, we have:

Matrices *A* and *B* will be inverse of each other only if

**A.** *AB* = *BA*

**C.** *AB* = 0, *BA* = *I*

**B.** *AB* = *BA* = 0

**D. ***AB* = *BA* = *I*

**Answer: D**

We know that if *A* is a square matrix of order *m*, and if there exists another square matrix *B* of the same order *m*, such that *AB* = *BA* = *I*, then *B* is said to be the inverse of *A*. In this case, it is clear that *A* is the inverse of *B*.

Thus, matrices *A* and *B* will be inverses of each other only if *AB* = *BA* = *I*.

Let, show that, where *I* is the identity matrix of order 2 and *n* ∈ **N**

It is given that

We shall prove the result by using the principle of mathematical induction.

For *n* = 1, we have:

Therefore, the result is true for *n *= 1.

Let the result be true for *n* = *k*.

That is,

Now, we prove that the result is true for *n* = *k* + 1.

Consider

From (1), we have:

Therefore, the result is true for *n* = *k* + 1.

Thus, by the principle of mathematical induction, we have:

If, prove that

It is given that

We shall prove the result by using the principle of mathematical induction.

For *n* = 1, we have:

Therefore, the result is true for *n* = 1.

Let the result be true for *n* = *k*.

That is

Now, we prove that the result is true for *n* = *k* + 1.

Therefore, the result is true for *n* = *k* + 1.

Thus by the principle of mathematical induction, we have:

If, then prove where *n* is any positive integer

It is given that

We shall prove the result by using the principle of mathematical induction.

For *n* = 1, we have:

Therefore, the result is true for *n* = 1.

Let the result be true for *n* = *k*.

That is,

Now, we prove that the result is true for *n* = *k* + 1.

Therefore, the result is true for *n* = *k* + 1.

Thus, by the principle of mathematical induction, we have:

If *A* and *B* are symmetric matrices, prove that *AB* − *BA* is a skew symmetric matrix.

It is given that *A* and *B* are symmetric matrices. Therefore, we have:

Thus, (*AB* − *BA*) is a skew-symmetric matrix.

Show that the matrix is symmetric or skew symmetric according as *A* is symmetric or skew symmetric.

We suppose that *A* is a symmetric matrix, then … (1)

Consider

Thus, if *A* is a symmetric matrix, thenis a symmetric matrix.

Now, we suppose that *A* is a skew-symmetric matrix.

Then,

Thus, if *A* is a skew-symmetric matrix, thenis a skew-symmetric matrix.

Hence, if A is a symmetric or skew-symmetric matrix, thenis a symmetric or skew-symmetric matrix accordingly.

Find the values of *x*, *y*, *z* if the matrix satisfy the equation

Now,

On comparing the corresponding elements, we have:

For what values of?

We have:

∴4 + 4*x* = 0

⇒ *x* = −1

Thus, the required value of *x* is −1.

If, show that

Find *x*, if

We have:

A manufacturer produces three products *x*, *y*, *z* which he sells in two markets.

Annual sales are indicated below:

Market | Products | ||

I | 10000 | 2000 | 18000 |

II | 6000 | 20000 | 8000 |

(a) If unit sale prices of *x*, *y* and *z *are Rs 2.50, Rs 1.50 and Rs 1.00, respectively, find the total revenue in each market with the help of matrix algebra.

(b) If the unit costs of the above three commodities are Rs 2.00, Rs 1.00 and 50 paise respectively. Find the gross profit.

**(a)** The unit sale prices of *x*, *y*, and *z *are respectively given as Rs 2.50, Rs 1.50, and Rs 1.00.

Consequently, the total revenue in market **I** can be represented in the form of a matrix as:

The total revenue in market **II** can be represented in the form of a matrix as:

Therefore, the total revenue in market **I **isRs 46000 and the same in market **II **isRs 53000.

**(b)** The unit cost prices of *x*, *y*, and *z *are respectively given as Rs 2.00, Rs 1.00, and 50 paise.

Consequently, the total cost prices of all the products in market **I** can be represented in the form of a matrix as:

Since the total revenue in market **I **isRs 46000, the gross profit in this marketis (Rs 46000 − Rs 31000) Rs 15000.

The total cost prices of all the products in market **II** can be represented in the form of a matrix as:

Since the total revenue in market **II **isRs 53000, the gross profit in this market is (Rs 53000 − Rs 36000) Rs 17000.

Find the matrix *X* so that

It is given that:

The matrix given on the R.H.S. of the equation is a 2 × 3 matrix and the one given on the L.H.S. of the equation is a 2 × 3 matrix. Therefore, *X* has to be a 2 × 2 matrix.

Now, let

Therefore, we have:

Equating the corresponding elements of the two matrices, we have:

Thus, *a* = 1, *b* = 2, *c* = −2, *d* = 0

Hence, the required matrix *X* is

If *A* and *B* are square matrices of the same order such that *AB* = *BA*, then prove by induction that. Further, prove that for all *n* ∈ **N**

*A* and *B* are square matrices of the same order such that *AB* = *BA*.

For *n* = 1, we have:

Therefore, the result is true for *n* = 1.

Let the result be true for *n* = *k*.

Now, we prove that the result is true for *n* = *k* + 1.

Therefore, the result is true for *n* = *k* + 1.

Thus, by the principle of mathematical induction, we have

Now, we prove that for all *n* ∈ **N**

For *n* = 1, we have:

Therefore, the result is true for *n* = 1.

Let the result be true for *n* = *k*.

Now, we prove that the result is true for *n* = *k* + 1.

Therefore, the result is true for *n* = *k* + 1.

Thus, by the principle of mathematical induction, we have, for all natural numbers.

Choose the correct answer in the following questions:

If is such that then

**A.**

**B.**

**C.**

**D. **

**Answer: C**

On comparing the corresponding elements, we have:

If the matrix *A* is both symmetric and skew symmetric, then

**A.** *A* is a diagonal matrix

**B.** *A* is a zero matrix

**C.** *A* is a square matrix

**D. **None of these

**Answer: B**

If *A* is both symmetric and skew-symmetric matrix, then we should have

Therefore, *A* is a zero matrix.

If *A* is square matrix such that then is equal to

**A.** *A* **B.** *I* − *A* **C.** *I* **D. **3*A*

**Answer: C**

**Conclusion:** If you like all the questions about **NCERT Class 12 Maths Solution Chapter 3 on Matrices**, then share it with others. You can also leave a comment in the comment box.

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]]>The post Class 12 Maths NCERT Solutions Chapter 2 – Inverse Trigonometric Functions appeared first on Babaji Academy.

]]>**Topics and Sub Topics in Class 11 Maths Chapter 2 Inverse Trigonometric Functions:**

Section Name | Topic Name |

2 | Inverse Trigonometric Functions |

2.1 | Introduction |

2.2 | Basic Concepts |

2.3 | Properties of Inverse Trigonometric Functions |

Find the principal value of

Let sin^{-1 } Then sin *y* =

We know that the range of the principal value branch of sin^{−1} is

and sin

Therefore, the principal value of

Find the principal value of

We know that the range of the principal value branch of cos^{−1} is

.

Therefore, the principal value of.

Find the principal value of cosec^{−1} (2)

Let cosec^{−1} (2) = *y*. Then,

We know that the range of the principal value branch of cosec^{−1} is

Therefore, the principal value of

Find the principal value of

We know that the range of the principal value branch of tan^{−1} is

Therefore, the principal value of

Find the principal value of

We know that the range of the principal value branch of cos^{−1} is

Therefore, the principal value of

Find the principal value of tan^{−1} (−1)

Let tan^{−1} (−1) = *y*. Then,

We know that the range of the principal value branch of tan^{−1} is

Therefore, the principal value of

Find the principal value of

We know that the range of the principal value branch of sec^{−1} is

Therefore, the principal value of

Find the principal value of

We know that the range of the principal value branch of cot^{−1} is (0,π) and

Therefore, the principal value of

Find the principal value of

We know that the range of the principal value branch of cos^{−1} is [0,π] and

.

Therefore, the principal value of

Find the principal value of

We know that the range of the principal value branch of cosec^{−1} is

Therefore, the principal value of

Find the value of

Find the value of

Find the value of if sin^{−1} *x *= *y*, then

**(****A)** **(B)**

**(****C)** **(D) **

It is given that sin^{−1} *x *= *y*.

We know that the range of the principal value branch of sin^{−1} is

Therefore,.

Find the value of is equal to

**(****A)** π (**B)** (**C)** (**D) **

Prove

To prove:

Let *x* = sin*θ*. Then,

We have,

R.H.S. =

= 3*θ*

= L.H.S.

Prove

To prove:

Let *x* = cos*θ*. Then, cos^{−1} *x* =*θ*.

We have,

Prove

To prove:

Prove

To prove:

Write the function in the simplest form:

Write the function in the simplest form:

Put *x* = cosec *θ* ⇒ *θ* = cosec^{−1} *x*

Write the function in the simplest form:

Write the function in the simplest form:

tan-1cosx-sinxcosx+sinx=tan-11-sinxcosx1+sinxcosx=tan-11-tanx1+tanx=tan-11-tan-1tanx tan-1x-y1+xy=tan-1x-tan-1y=π4-x

Write the function in the simplest form:

Write the function in the simplest form:

Find the value of

Let. Then,

Find the value of

Find the value of

Let *x* = tan *θ*. Then, *θ* = tan^{−1} *x*.

Let *y* = tan *Φ*. Then, *Φ* = tan^{−1} *y*.

If, then find the value of *x*.

On squaring both sides, we get:

Hence, the value of *x* is

If, then find the value of *x*.

Hence, the value of *x* is

Find the values of

We know that sin^{−1} (sin *x*) =* x* if, which is the principal value branch of sin^{−1}*x*.

Here,

Now, can be written as:

Find the values of

We know that tan^{−1} (tan *x*) =* x* if, which is the principal value branch of tan^{−1}*x*.

Here,

Now, can be written as:

Find the values of

Let. Then,

Find the values of is equal to

**(A)** **(B)** **(C)** **(D)**

We know that cos^{−1} (cos *x*) =* x* if, which is the principal value branch of cos ^{−1}*x*.

Here,

Now, can be written as:

cos-1cos7π6 = cos-1cosπ+π6cos-1cos7π6 = cos-1- cosπ6 as, cosπ+θ = – cos θcos-1cos7π6 = cos-1- cosπ-5π6cos-1cos7π6 = cos-1– cos 5π6 as, cosπ-θ = – cos θ

The correct answer is B.

Find the values of is equal to

**(A)** **(B)** **(C)** **(D)** 1

Let. Then,

We know that the range of the principal value branch of.

∴

The correct answer is D.

Find the values of is equal to

**(A)** π **(B)** **(C)** 0 **(D)**

Let. Then,

We know that the range of the principal value branch of

Let.

The range of the principal value branch of

The correct answer is B.

Find the value of

We know that cos^{−1} (cos *x*) =* x* if, which is the principal value branch of cos ^{−1}*x*.

Here,

Now, can be written as:

Find the value of

We know that tan^{−1} (tan *x*) =* x* if, which is the principal value branch of tan ^{−1}*x*.

Here,

Now,

can be written as:

Prove

Now, we have:

Prove

Now, we have:

Prove

Now, we will prove that:

Prove

Now, we have:

Prove

Using (1) and (2), we have

Prove

Prove

Prove

Prove [**Hint: **put*x* = cos 2*θ*]

Prove

Solve

Solve

Solveis equal to

**(A) ** (**B) ** (**C) ** (**D) **

Let tan^{−1} *x* = *y*. Then,

The correct answer is D.

Solve**, **then *x* is equal to

**(****A) ** (**B) ** (**C)** 0 (**D) **

Therefore, from equation (1), we have

Put *x* = sin *y*. Then, we have:

But, when, it can be observed that:

is not the solution of the given equation.

Thus, *x* = 0.

Hence, the correct answer is **C**.

Solveis equal to

**(A)** **(B).** **(C)** **(D) **

Hence, the correct answer is **C**.

**Conclusion:** If you liked Class 12 Maths Solution chapter 2- Inverse Trigonometric Functions, then share this post with your friends. You can also leave a comment below.

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]]>The post NCERT Solutions for Class 12 Maths Chapter 1 Relations and functions appeared first on Babaji Academy.

]]>This will clear your doubts in regards to any question and improve your application skills. In this article, we will provide you with NCERT Solutions for Class 12 Maths – Chapter 1 – Relations and Functions Ex 1.1. Read on to find out.

**Topics and Sub Topics in Class 11 Maths Chapter 1 Relations and Functions Ex 1.1:**

- 1 – Relations and Functions
- 1.1 Introduction
- 1.2 Types of Relations
- 1.3 Types of Functions
- 1.4 Composition of Functions and Invertible Function
- 1.5 Binary Operations

Determine whether each of the following relations are reflexive, symmetric and transitive:

(i)Relation R in the set *A* = {1, 2, 3…13, 14} defined as

R = {(*x*, *y*): 3*x* − *y* = 0}

(ii) Relation R in the set **N** of natural numbers defined as

R = {(*x*, *y*): *y* = *x* + 5 and *x* < 4}

(iii) Relation R in the set *A* = {1, 2, 3, 4, 5, 6} as

R = {(*x*, *y*): *y* is divisible by *x*}

(iv) Relation R in the set **Z** of all integers defined as

R = {(*x*, *y*): *x* − *y* is as integer}

(v) Relation R in the set *A* of human beings in a town at a particular time given by

(a) R = {(*x*, *y*): *x *and *y* work at the same place}

(b) R = {(*x*, *y*): *x* and *y* live in the same locality}

(c) R = {(*x*, *y*): *x *is exactly 7 cm taller than *y*}

(d) R = {(*x*, *y*): *x* is wife of *y*}

(e) R = {(*x*, *y*): *x* is father of* y*}

(i) *A* = {1, 2, 3 … 13, 14}

R = {(*x*, *y*): 3*x* − *y* = 0}

∴R = {(1, 3), (2, 6), (3, 9), (4, 12)}

R is not reflexive since (1, 1), (2, 2) … (14, 14) ∉ R.

Also, R is not symmetric as (1, 3) ∈R, but (3, 1) ∉ R. [3(3) − 1 ≠ 0]

Also, R is not transitive as (1, 3), (3, 9) ∈R, but (1, 9) ∉ R.[3(1) − 9 ≠ 0]

Hence, R is neither reflexive, nor symmetric, nor transitive.

(ii) R = {(*x*, *y*): *y* = *x* + 5 and *x* < 4} = {(1, 6), (2, 7), (3, 8)}

It is seen that (1, 1) ∉ R.

∴R is not reflexive.

(1, 6) ∈R

But,

(6, 1) ∉ R.

∴R is not symmetric.

Now, since there is no pair in R such that (*x*, *y*) and (*y*, *z*) ∈R, then (*x*, *z*) cannot belong to R.

∴ R is not transitive.

Hence, R is neither reflexive, nor symmetric, nor transitive.

(iii) *A* = {1, 2, 3, 4, 5, 6}

R = {(*x*, *y*): *y* is divisible by *x*}

We know that any number (*x)* is divisible by itself.

(*x*, *x*) ∈R

∴R is reflexive.

Now,

(2, 4) ∈R [as 4 is divisible by 2]

But,

(4, 2) ∉ R. [as 2 is not divisible by 4]

∴R is not symmetric.

Let (*x*, *y*), (*y*, *z*) ∈ R. Then, *y* is divisible by *x* and *z* is divisible by *y*.

∴*z* is divisible by *x*.

⇒ (*x*, *z*) ∈R

∴R is transitive.

Hence, R is reflexive and transitive but not symmetric.

(iv) R = {(*x*, *y*): *x* − *y* is an integer}

Now, for every *x* ∈ **Z**, (*x*, *x*) ∈R as *x* − *x* = 0 is an integer.

∴R is reflexive.

Now, for every *x*, *y* ∈ **Z** if (*x*, *y*) ∈ R, then *x* − *y* is an integer.

⇒ −(*x* − *y*) is also an integer.

⇒ (*y* − *x*) is an integer.

∴ (*y*, *x*) ∈ R

∴R is symmetric.

Now,

Let (*x*, *y*) and (*y*, *z*) ∈R, where *x*, *y*, *z* ∈ **Z**.

⇒ (*x* − *y*) and (*y* − *z*) are integers.

⇒ *x *− *z* = (*x* − *y*) + (*y* − *z*) is an integer.

∴ (*x*, *z*) ∈R

∴R is transitive.

Hence, R is reflexive, symmetric, and transitive.

(v) (a) R = {(*x*, *y*): *x* and *y* work at the same place}

(*x*, *x*) ∈ R

∴ R is reflexive.

If (*x*, *y*) ∈ R, then *x* and *y* work at the same place.

⇒ *y* and *x* work at the same place.

⇒ (*y*, *x*) ∈ R.

∴R is symmetric.

Now, let (*x*, *y*), (*y*, *z*) ∈ R

⇒ *x* and *y* work at the same place and *y* and *z* work at the same place.

⇒ *x* and *z* work at the same place.

⇒ (*x*, *z*) ∈R

∴ R is transitive.

Hence, R is reflexive, symmetric, and transitive.

(b) R = {(*x*, *y*): *x* and *y* live in the same locality}

Clearly (*x*, *x*) ∈ R as *x* and *x* is the same human being.

∴ R is reflexive.

If (*x*, *y*) ∈R, then *x* and *y* live in the same locality.

⇒ *y* and *x* live in the same locality.

⇒ (*y*, *x*) ∈ R

∴R is symmetric.

Now, let (*x*, *y*) ∈ R and (*y*, *z*) ∈ R.

⇒ *x* and *y* live in the same locality and *y* and *z* live in the same locality.

⇒ *x* and *z* live in the same locality.

⇒ (*x,* *z*) ∈ R

∴ R is transitive.

Hence, R is reflexive, symmetric, and transitive.

(c) R = {(*x*, *y*): *x* is exactly 7 cm taller than *y*}

Now,

(*x*, *x*) ∉ R

Since human being *x *cannot be taller than himself.

∴R is not reflexive.

Now, let (*x*, *y*) ∈R.

⇒ *x* is exactly 7 cm taller than *y*.

Then, *y* is not taller than *x*.

∴ (*y*, *x*) ∉R

Indeed if *x* is exactly 7 cm taller than *y*, then *y* is exactly 7 cm shorter than *x*.

∴R is not symmetric.

Now,

Let (*x*, *y*), (*y*, *z*) ∈ R.

⇒ *x* is exactly 7 cm taller than*y *and *y* is exactly 7 cm taller than z.

⇒ *x* is exactly 14 cm taller than *z *.

∴(*x*, *z*) ∉R

∴ R is not transitive.

Hence, R is neither reflexive, nor symmetric, nor transitive.

(d) R = {(*x*, *y*): *x* is the wife of *y*}

Now,

(*x*, *x*) ∉ R

Since *x* cannot be the wife of herself.

∴R is not reflexive.

Now, let (*x*, *y*) ∈ R

⇒ *x* is the wife of *y.*

Clearly *y* is not the wife of *x*.

∴(*y*, *x*) ∉ R

Indeed if *x* is the wife of *y*, then *y* is the husband of *x*.

∴ R is not symmetric.

Let (*x*, *y*), (*y*, *z*) ∈ R

⇒ *x* is the wife of *y* and *y* is the wife of *z*.

This case is not possible. Also, this does not imply that *x* is the wife of *z*.

∴(*x*, *z*) ∉ R

∴R is not transitive.

Hence, R is neither reflexive, nor symmetric, nor transitive.

(e) R = {(*x*, *y*): *x* is the father of *y*}

(*x*, *x*) ∉ R

As *x* cannot be the father of himself.

∴R is not reflexive.

Now, let (*x*, *y*) ∈R.

⇒ *x* is the father of *y.*

⇒ *y* cannot be the father of *y.*

Indeed, *y* is the son or the daughter of *y.*

∴(*y*, *x*) ∉ R

∴ R is not symmetric.

Now, let (*x*, *y*) ∈ R and (*y*, *z*) ∈ R.

⇒ *x* is the father of *y* and *y* is the father of *z*.

⇒ *x* is not the father of *z*.

Indeed *x* is the grandfather of *z*.

∴ (*x*, *z*) ∉ R

∴R is not transitive.

Hence, R is neither reflexive, nor symmetric, nor transitive.

Show that the relation R in the set **R **of real numbers, defined as

R = {(*a*, *b*): *a* ≤ *b*^{2}} is neither reflexive nor symmetric nor transitive.

R = {(*a*, *b*): *a* ≤ *b*^{2}}

It can be observed that

∴R is not reflexive.

Now, (1, 4) ∈ R as 1 < 4^{2}

But, 4 is not less than 1^{2}.

∴(4, 1) ∉ R

∴R is not symmetric.

Now,

(3, 2), (2, 1.5) ∈ R

(as 3 < 2^{2} = 4 and 2 < (1.5)^{2} = 2.25)

But, 3 > (1.5)^{2} = 2.25

∴(3, 1.5) ∉ R

∴ R is not transitive.

Hence, R is neither reflexive, nor symmetric, nor transitive.

Check whether the relation R defined in the set {1, 2, 3, 4, 5, 6} as

R = {(*a*, *b*): *b* = *a* + 1} is reflexive, symmetric or transitive.

Let *A* = {1, 2, 3, 4, 5, 6}.

A relation R is defined on set *A* as:

R = {(*a*, *b*): *b* = *a* + 1}

∴R = {(1, 2), (2, 3), (3, 4), (4, 5), (5, 6)}

We can find (*a*, *a*) ∉ R, where *a *∈ A.

For instance,

(1, 1), (2, 2), (3, 3), (4, 4), (5, 5), (6, 6) ∉ R

∴R is not reflexive.

It can be observed that (1, 2) ∈ R, but (2, 1) ∉ R.

∴R is not symmetric.

Now, (1, 2), (2, 3) ∈ **R**

But,

(1, 3) ∉ R

∴R is not transitive

Hence, R is neither reflexive, nor symmetric, nor transitive.

Show that the relation R in **R** defined as R = {(*a*, *b*): *a* ≤ *b*}, is reflexive and transitive but not symmetric.

R = {(*a*, *b*); *a* ≤ *b*}

Clearly (*a*, *a*) ∈ R as *a *= *a*.

∴R is reflexive.

Now,

(2, 4) ∈ R (as 2 < 4)

But, (4, 2) ∉ R as 4 is greater than 2.

∴ R is not symmetric.

Now, let (*a*, *b*), (*b*, *c*) ∈ R.

Then,

*a* ≤ *b* and *b* ≤ *c*

⇒ *a* ≤ *c*

⇒ (*a*, *c*) ∈ R

∴R is transitive.

Hence,R is reflexive and transitive but not symmetric.

Check whether the relation R in **R** defined as R = {(*a*, *b*): *a* ≤ *b*^{3}} is reflexive, symmetric or transitive.

R = {(*a*, *b*): *a *â‰¤ *b*^{3}}

It is observed that

âˆ´ R is not reflexive.

Now,

(1, 2) âˆˆ R (as 1 < 2^{3} = 8)

But,

(2, 1) âˆ‰ R (as 2 > 1^{3} = 1)

âˆ´ R is not symmetric.

We have

But

âˆ´ R is not transitive.

Hence, R is neither reflexive, nor symmetric, nor transitive.

Show that the relation R in the set {1, 2, 3} given by R = {(1, 2), (2, 1)} is symmetric but neither reflexive nor transitive.

Let *A* = {1, 2, 3}.

A relation R on *A* is defined as R = {(1, 2), (2, 1)}.

It is seen that (1, 1), (2, 2), (3, 3) ∉R.

∴ R is not reflexive.

Now, as (1, 2) ∈ R and (2, 1) ∈ R, then R is symmetric.

Now, (1, 2) and (2, 1) ∈ R

However,

(1, 1) ∉ R

∴ R is not transitive.

Hence, R is symmetric but neither reflexive nor transitive.

Show that the relation R in the set *A* of all the books in a library of a college, given by R = {(*x*, *y*): *x* and *y* have same number of pages} is an equivalence relation.

Set *A* is the set of all books in the library of a college.

R = {*x*, *y*): *x* and *y* have the same number of pages}

Now, R is reflexive since (*x*, *x*) ∈ R as *x* and *x* has the same number of pages.

Let (*x*, *y*) ∈ R ⇒ *x* and *y* have the same number of pages.

⇒ *y* and *x* have the same number of pages.

⇒ (*y*, *x*) ∈ R

∴R is symmetric.

Now, let (*x*, *y*) ∈R and (*y*, *z*) ∈ R.

⇒ *x* and *y* and have the same number of pages and *y* and *z* have the same number of pages.

⇒ *x* and *z* have the same number of pages.

⇒ (*x*, *z*) ∈ R

∴R is transitive.

Hence, R is an equivalence relation.

Show that the relation R in the set *A* = {1, 2, 3, 4, 5} given by

, is an equivalence relation. Show that all the elements of {1, 3, 5} are related to each other and all the elements of {2, 4} are related to each other. But no element of {1, 3, 5} is related to any element of {2, 4}.

*A* = {1, 2, 3, 4, 5}

It is clear that for any element *a* ∈*A*, we have (which is even).

∴R is reflexive.

Let (*a*, *b*) ∈ R.

∴R is symmetric.

Now, let (*a*, *b*) ∈ R and (*b*, *c*) ∈ R.

⇒ (*a*, *c*) ∈ R

∴R is transitive.

Hence, R is an equivalence relation.

Now, all elements of the set {1, 3, 5} are related to each other as all the elements of this subset are odd. Thus, the modulus of the difference between any two elements will be even.

Similarly, all elements of the set {2, 4} are related to each other as all the elements of this subset are even.

Also, no element of the subset {1, 3, 5} can be related to any element of {2, 4} as all elements of {1, 3, 5} are odd and all elements of {2, 4} are even. Thus, the modulus of the difference between the two elements (from each of these two subsets) will not be even.

Show that each of the relation R in the set, given by

(i)

(ii)

is an equivalence relation. Find the set of all elements related to 1 in each case.

(i)

For any element *a* ∈A, we have (*a*, *a*) ∈ R as is a multiple of 4.

∴R is reflexive.

Now, let (*a*, *b*) ∈ R ⇒ is a multiple of 4.

⇒ (*b*, *a*) ∈ R

∴R is symmetric.

Now, let (*a*, *b*), (*b*, *c*) ∈ R.

⇒ (*a*, *c*) ∈R

∴ R is transitive.

Hence, R is an equivalence relation.

The set of elements related to 1 is {1, 5, 9} since

(ii) R = {(*a*, *b*): *a* = *b*}

For any element *a* ∈A, we have (*a*, *a*) ∈ R, since *a* = *a*.

∴R is reflexive.

Now, let (*a*, *b*) ∈ R.

⇒ *a* = *b*

⇒ *b* = *a*

⇒ (*b*, *a*) ∈ R

∴R is symmetric.

Now, let (*a*, *b*) ∈ R and (*b*, *c*) ∈ R.

⇒ *a* = *b* and *b* = *c*

⇒ *a* = *c*

⇒ (*a*, *c*) ∈ R

∴ R is transitive.

Hence, R is an equivalence relation.

The elements in R that are related to 1 will be those elements from set A which are equal to 1.

Hence, the set of elements related to 1 is {1}.

Given an example of a relation. Which is

(i) Symmetric but neither reflexive nor transitive.

(ii) Transitive but neither reflexive nor symmetric.

(iii) Reflexive and symmetric but not transitive.

(iv) Reflexive and transitive but not symmetric.

(v) Symmetric and transitive but not reflexive.

(i) Let *A* = {5, 6, 7}.

Define a relation R on* A* as R = {(5, 6), (6, 5)}.

Relation R is not reflexive as (5, 5), (6, 6), (7, 7) ∉ R.

Now, as (5, 6) ∈ R and also (6, 5) ∈ R, R is symmetric.

(5, 6), (6, 5) ∈ R, but (5, 5) ∉ R

∴R is not transitive.

Hence, relation R is symmetric but not reflexive or transitive.

(ii)Consider a relation R in **R **defined as:

R = {(*a*, *b*): *a* < *b*}

For any *a *∈ R, we have (*a*, *a*) ∉ R since *a* cannot be strictly less than *a* itself. In fact, *a* = *a*.

∴ R is not reflexive.

Now,

(1, 2) ∈ R (as 1 < 2)

But, 2 is not less than 1.

∴ (2, 1) ∉ R

∴ R is not symmetric.

Now, let (*a*, *b*), (*b*, *c*) ∈ R.

⇒ *a* < *b* and *b* < *c*

⇒ *a* < *c*

⇒ (*a*, *c*) ∈ R

∴R is transitive.

Hence, relation R is transitive but not reflexive and symmetric.

(iii)Let *A* = {4, 6, 8}.

Define a relation R on A as:

*A* = {(4, 4), (6, 6), (8, 8), (4, 6), (6, 4), (6, 8), (8, 6)}

Relation R is reflexive since for every *a* ∈ *A*, (*a*, *a*) ∈R i.e., (4, 4), (6, 6), (8, 8)} ∈ R.

Relation R is symmetric since (*a*, *b*) ∈ R ⇒ (*b*, *a*) ∈ R for all *a*, *b* ∈ R.

Relation R is not transitive since (4, 6), (6, 8) ∈ R, but (4, 8) ∉ R.

Hence, relation R is reflexive and symmetric but not transitive.

(iv) Define a relation R in **R** as:

R = {*a*, *b*): *a*^{3} ≥ *b*^{3}}

Clearly (*a*, *a*) ∈ R as *a*^{3} = *a*^{3}.

∴R is reflexive.

Now,

(2, 1) ∈ R (as 2^{3} ≥ 1^{3})

But,

(1, 2) ∉ R (as 1^{3} < 2^{3})

∴ R is not symmetric.

Now,

Let (*a*, *b*), (*b*, *c*) ∈ R.

⇒ *a*^{3} ≥ *b*^{3} and *b*^{3} ≥ *c*^{3}

⇒ *a*^{3} ≥ *c*^{3}

⇒ (*a*, *c*) ∈ R

∴R is transitive.

Hence, relation R is reflexive and transitive but not symmetric.

(v) Let *A* = {−5, −6}.

Define a relation R on *A* as:

R = {(−5, −6), (−6, −5), (−5, −5)}

Relation R is not reflexive as (−6, −6) ∉ R.

Relation R is symmetric as (−5, −6) ∈ R and (−6, −5}∈R.

It is seen that (−5, −6), (−6, −5) ∈ R. Also, (−5, −5) ∈ R.

∴The relation R is transitive.

Hence, relation R is symmetric and transitive but not reflexive.

Show that the relation R in the set *A* of points in a plane given by R = {(P, Q): distance of the point P from the origin is same as the distance of the point Q from the origin}, is an equivalence relation. Further, show that the set of all point related to a point P ≠ (0, 0) is the circle passing through P with origin as centre.

R = {(P, Q): distance of point P from the origin is the same as the distance of point Q from the origin}

Clearly, (P, P) ∈ R since the distance of point P from the origin is always the same as the distance of the same point P from the origin.

∴R is reflexive.

Now,

Let (P, Q) ∈ R.

⇒ The distance of point P from the origin is the same as the distance of point Q from the origin.

⇒ The distance of point Q from the origin is the same as the distance of point P from the origin.

⇒ (Q, P) ∈ R

∴R is symmetric.

Now,

Let (P, Q), (Q, S) ∈ R.

⇒ The distance of points P and Q from the origin is the same and also, the distance of points Q and S from the origin is the same.

⇒ The distance of points P and S from the origin is the same.

⇒ (P, S) ∈ R

∴R is transitive.

Therefore, R is an equivalence relation.

The set of all points related to P ≠ (0, 0) will be those points whose distance from the origin is the same as the distance of point P from the origin.

In other words, if O (0, 0) is the origin and OP = *k*, then the set of all points related to P is at a distance of *k* from the origin.

Hence, this set of points forms a circle with the centre as the origin and this circle passes through point P.

Show that the relation R defined in the set *A* of all triangles as R = {(*T*_{1}, *T*_{2}): *T*_{1} is similar to *T*_{2}}, is equivalence relation. Consider three right angle triangles *T*_{1} with sides 3, 4, 5, *T*_{2} with sides 5, 12, 13 and *T*_{3} with sides 6, 8, 10. Which triangles among *T*_{1}, *T*_{2} and *T*_{3} are related?

R = {(*T*_{1}, *T*_{2}): *T*_{1} is similar to *T*_{2}}

R is reflexive since every triangle is similar to itself.

Further, if (*T*_{1}, *T*_{2}) ∈ R, then *T*_{1} is similar to *T*_{2.}

⇒ *T*_{2} is similar to *T*_{1.}

⇒ (*T*_{2}, *T*_{1}) ∈R

∴R is symmetric.

Now,

Let (*T*_{1}, *T*_{2}), (*T*_{2}, *T*_{3}) ∈ R.

⇒ *T*_{1} is similar to *T*_{2} and *T*_{2} is similar to *T*_{3.}

⇒ *T*_{1} is similar to *T*_{3.}

⇒ (*T*_{1}, *T*_{3}) ∈ R

∴ R is transitive.

Thus, R is an equivalence relation.

Now, we can observe that:

∴The corresponding sides of triangles *T*_{1} and *T*_{3} are in the same ratio.

Then, triangle *T*_{1} is similar to triangle *T*_{3.}

Hence, *T*_{1} is related to *T*_{3}.

Show that the relation R defined in the set *A* of all polygons as R = {(*P*_{1}, *P*_{2}): *P*_{1} and *P*_{2} have same number of sides}, is an equivalence relation. What is the set of all elements in *A* related to the right angle triangle *T* with sides 3, 4 and 5?

R = {(*P*_{1}, *P*_{2}): *P*_{1} and *P*_{2} have same the number of sides}

R is reflexive since (*P*_{1}, *P*_{1}) ∈ R as the same polygon has the same number of sides with itself.

Let (*P*_{1}, *P*_{2}) ∈ R.

⇒ *P*_{1} and *P*_{2} have the same number of sides.

⇒ *P*_{2} and *P*_{1} have the same number of sides.

⇒ (*P*_{2}, *P*_{1}) ∈ R

∴R is symmetric.

Now,

Let (*P*_{1}, *P*_{2}), (*P*_{2}, *P*_{3}) ∈ R.

⇒ *P*_{1} and *P*_{2} have the same number of sides. Also, *P*_{2} and *P*_{3} have the same number of sides.

⇒ *P*_{1} and *P*_{3} have the same number of sides.

⇒ (*P*_{1}, *P*_{3}) ∈ R

∴R is transitive.

Hence, R is an equivalence relation.

The elements in *A* related to the right-angled triangle (*T)* with sides 3, 4, and 5 are those polygons which have 3 sides (since *T* is a polygon with 3 sides).

Hence, the set of all elements in *A* related to triangle *T* is the set of all triangles.

Let *L* be the set of all lines in XY plane and R be the relation in *L* defined as R = {(*L*_{1}, *L*_{2}): *L*_{1} is parallel to *L*_{2}}. Show that R is an equivalence relation. Find the set of all lines related to the line *y* = 2*x* + 4.

R = {(*L*_{1}, *L*_{2}): L_{1} is parallel to *L*_{2}}

R is reflexive as any line *L*_{1} is parallel to itself i.e., (*L*_{1}, *L*_{1}) ∈ R.

Now,

Let (*L*_{1}, *L*_{2}) ∈ R.

⇒ *L*_{1} is parallel to *L*_{2.}

⇒ *L*_{2} is parallel to *L*_{1.}

⇒ (*L*_{2}, *L*_{1}) ∈ R

∴ R is symmetric.

Now,

Let (*L*_{1}, *L*_{2}), (*L*_{2}, *L*_{3}) ∈R.

⇒ *L*_{1} is parallel to *L*_{2}. Also, *L*_{2} is parallel to *L*_{3.}

⇒ *L*_{1} is parallel to *L*_{3.}

∴R is transitive.

Hence, R is an equivalence relation.

The set of all lines related to the line *y* = 2*x* + 4 is the set of all lines that are parallel to the line *y* = 2*x* + 4.

Slope of line *y* = 2*x* + 4 is *m* = 2

It is known that parallel lines have the same slopes.

The line parallel to the given line is of the form *y* = 2*x* + *c*, where *c* ∈**R**.

Hence, the set of all lines related to the given line is given by *y* = 2*x* + *c*, where *c* ∈ **R**.

Let R be the relation in the set {1, 2, 3, 4} given by R = {(1, 2), (2, 2), (1, 1), (4, 4), (1, 3), (3, 3), (3, 2)}. Choose the correct answer.

(A) R is reflexive and symmetric but not transitive.

(B) R is reflexive and transitive but not symmetric.

(C) R is symmetric and transitive but not reflexive.

(D) R is an equivalence relation.

R = {(1, 2), (2, 2), (1, 1), (4, 4), (1, 3), (3, 3), (3, 2)}

It is seen that (*a*, *a*) ∈ **R,** for every *a* ∈{1, 2, 3, 4}.

∴ R is reflexive.

It is seen that (1, 2) ∈ R, but (2, 1) ∉ R.

∴R is not symmetric.

Also, it is observed that (*a*, *b*), (*b*, *c*) ∈ R ⇒ (*a*, *c*) ∈ R for all *a*, *b*, *c* ∈ {1, 2, 3, 4}.

∴ R is transitive.

Hence, R is reflexive and transitive but not symmetric.

The correct answer is B.

Let R be the relation in the set **N **given by R = {(*a*, *b*): *a *= *b* − 2, *b *> 6}. Choose the correct answer.

(A) (2, 4) ∈ R (B) (3, 8) ∈R (C) (6, 8) ∈R (D) (8, 7) ∈ R

R = {(*a*, *b*): *a *= *b* − 2, *b* > 6}

Now, since *b* > 6, (2, 4) ∉ R

Also, as 3 ≠ 8 − 2, (3, 8) ∉ R

And, as 8 ≠ 7 − 2

(8, 7) ∉ R

Now, consider (6, 8).

We have 8 > 6 and also, 6 = 8 − 2.

∴(6, 8) ∈ R

The correct answer is C.

Show that the function *f*: **R**_{*} → **R**_{*} defined byis one-one and onto, where **R**_{*} is the set of all non-zero real numbers. Is the result true, if the domain **R**_{*} is replaced by **N** with co-domain being same as **R**_{*}?

It is given that* f*: **R**_{*} → **R**_{*} is defined by

One-one:

∴*f* is one-one.

Onto:

It is clear that for *y*∈** R**_{*}, there existssuch that

∴*f* is onto.

Thus, the given function (*f)* is one-one and onto.

Now, consider function *g*:** N **→ **R**_{*}defined by

We have,

∴*g* is one-one.

Further, it is clear that *g* is not onto as for 1.2 ∈**R**_{*} there does not exit any *x* in **N** such that *g*(*x*) =.

Hence, function *g* is one-one but not onto.

Check the injectivity and surjectivity of the following functions:

(i) *f*: **N** → **N** given by *f*(*x*) = *x*^{2}

(ii) *f*: **Z** → **Z** given by *f*(*x*) = *x*^{2}

(iii) *f*: **R** → **R** given by *f*(*x*) = *x*^{2}

(iv) *f*: **N **→ **N** given by *f*(*x*) = *x*^{3}

(v) *f*: **Z** → **Z** given by *f*(*x*) = *x*^{3}

(i) *f*: **N** → **N** is given by,

*f*(*x*) = *x*^{2}

It is seen that for *x*, *y* ∈**N**, *f*(*x*) = *f*(*y*) ⇒ *x*^{2} = *y*^{2} ⇒ *x* = *y*.

∴*f* is injective.

Now, 2 ∈ **N**. But, there does not exist any *x* in **N** such that *f*(*x*) = *x*^{2} = 2.

∴ *f* is not surjective.

Hence, function *f* is injective but not surjective.

(ii) *f*: **Z** → **Z** is given by,

*f*(*x*) = *x*^{2}

It is seen that *f*(−1) = *f*(1) = 1, but −1 ≠ 1.

∴ *f *is not injective.

Now,−2 ∈ **Z**. But, there does not exist any element *x *∈**Z** such that *f*(*x*) = *x*^{2} = −2.

∴ *f* is not surjective.

Hence, function *f* is neither injective nor surjective.

(iii)* f*: **R** → **R** is given by,

*f*(*x*) = *x*^{2}

It is seen that *f*(−1) = *f*(1) = 1, but −1 ≠ 1.

∴ *f *is not injective.

Now,−2 ∈ **R**. But, there does not exist any element *x *∈ **R** such that *f*(*x*) = *x*^{2} = −2.

∴ *f* is not surjective.

Hence, function *f* is neither injective nor surjective.

(iv)* f*: **N** → **N** given by,

*f*(*x*) = *x*^{3}

It is seen that for *x*, *y* ∈**N**, *f*(*x*) = *f*(*y*) ⇒ *x*^{3} = *y*^{3} ⇒ *x* = *y*.

∴*f* is injective.

Now, 2 ∈ **N**. But, there does not exist any element *x* in domain **N** such that *f*(*x*) = *x*^{3} = 2.

∴ *f* is not surjective

Hence, function *f* is injective but not surjective.

(v) *f*: **Z **→ **Z** is given by,

*f*(*x*) = *x*^{3}

It is seen that for *x*, *y* ∈ **Z**, *f*(*x*) = *f*(*y*) ⇒ *x*^{3} = *y*^{3} ⇒ *x* = *y*.

∴ *f* is injective.

Now, 2 ∈ **Z**. But, there does not exist any element *x* in domain **Z** such that *f*(*x*) = *x*^{3} = 2.

∴ *f* is not surjective.

Hence, function *f* is injective but not surjective.

Prove that the Greatest Integer Function *f*: **R **→ **R** given by *f*(*x*) = [*x*], is neither one-once nor onto, where [*x*] denotes the greatest integer less than or equal to *x*.

*f*: **R** → **R** is given by,

*f*(*x*) = [*x*]

It is seen that *f*(1.2) = [1.2] = 1, *f*(1.9) = [1.9] = 1.

∴ *f*(1.2) = *f*(1.9), but 1.2 ≠ 1.9.

∴ *f* is not one-one.

Now, consider 0.7 ∈ **R**.

It is known that *f*(*x*) = [*x*] is always an integer. Thus, there does not exist any element *x* ∈ **R** such that *f*(*x*) = 0.7.

∴ *f* is not onto.

Hence, the greatest integer function is neither one-one nor onto.

Show that the Modulus Function *f*: **R **→ **R** given by, is neither one-one nor onto, where is *x*, if *x* is positive or 0 andis − *x*, if *x* is negative.

*f*: **R** → **R** is given by,

It is seen that.

∴*f*(−1) = *f*(1), but −1 ≠ 1.

∴ *f* is not one-one.

Now, consider −1 ∈ **R**.

It is known that *f*(*x*) = is always non-negative. Thus, there does not exist any element *x* in domain **R** such that *f*(*x*) = = −1.

∴ *f* is not onto.

Hence, the modulus function is neither one-one nor onto.

Show that the Signum Function* f*: **R** → **R**, given by

is neither one-one nor onto.

*f*: **R** → **R** is given by,

It is seen that *f*(1) = *f*(2) = 1, but 1 ≠ 2.

∴*f* is not one-one.

Now, as *f*(*x*) takes only 3 values (1, 0, or −1) for the element −2 in co-domain **R**, there does not exist any* x* in domain **R** such that *f*(*x*) = −2.

∴ *f* is not onto.

Hence, the signum function is neither one-one nor onto.

Let *A* = {1, 2, 3}, *B *= {4, 5, 6, 7} and let *f* = {(1, 4), (2, 5), (3, 6)} be a function from *A* to *B*. Show that *f* is one-one.

It is given that *A* = {1, 2, 3}, *B *= {4, 5, 6, 7}.

*f*: *A* → *B* is defined as *f* = {(1, 4), (2, 5), (3, 6)}.

∴ *f* (1) = 4, *f* (2) = 5, *f* (3) = 6

It is seen that the images of distinct elements of *A* under *f* are distinct.

Hence, function *f* is one-one.

In each of the following cases, state whether the function is one-one, onto or bijective. Justify your answer.

(i) *f*: **R **→ **R** defined by *f*(*x*) = 3 − 4*x*

(ii) *f*: **R **→ **R** defined by *f*(*x*) = 1 + *x*^{2}

(i) *f*: **R **→ **R** is defined as *f*(*x*) = 3 − 4*x*.

.

∴ *f* is one-one.

For any real number (*y)* in **R**, there existsin **R** such that

∴*f *is onto.

Hence, *f *is bijective.

(ii) *f*: **R** →** R** is defined as

.

.

∴does not imply that

For instance,

∴ *f* is not one-one.

Consider an element −2 in co-domain **R**.

It is seen thatis positive for all *x* ∈ **R**.

Thus, there does not exist any* x* in domain **R** such that *f*(*x*) = −2.

∴ *f* is not onto.

Hence, *f* is neither one-one nor onto.

Let *A* and *B* be sets. Show that *f*: *A* × *B* → *B *× *A* such that (*a*, *b*) = (*b*, *a*) is bijective function.

*f*: *A* × *B* → *B* × *A* is defined as *f*(*a*, *b*) = (*b*, *a*).

.

∴ *f* is one-one.

Now, let (*b*, *a*) ∈ *B* × *A* be any element.

Then, there exists (*a*, *b*) ∈*A* × *B* such that *f*(*a*, *b*) = (*b*, *a*). [By definition of *f*]

∴ *f* is onto.

Hence,* f* is bijective.

Let *f*: **N** → **N** be defined by

State whether the function f is bijective. Justify your answer.

*f*: **N** → **N** is defined as

It can be observed that:

∴ *f* is not one-one.

Consider a natural number (*n)* in co-domain **N**.

Case **I: ***n* is odd

∴*n* = 2*r* + 1 for some *r* ∈ **N. **Then, there exists 4*r *+ 1∈**N** such that

.

Case **II: ***n* is even

∴*n* = 2*r* for some *r* ∈ **N. **Then,there exists 4*r* ∈**N** such that.

∴ *f* is onto.

Hence, *f* is not a bijective function.

Let A = **R** − {3} and B = **R** − {1}. Consider the function *f*: A → B defined by

. Is *f* one-one and onto? Justify your answer.

A = **R** − {3}, B = **R** − {1}

*f*: A → B is defined as.

.

∴ *f* is one-one.

Let *y* ∈B = **R** − {1}. Then, *y* ≠ 1.

The function *f *is onto if there exists *x* ∈A such that *f*(*x*) = *y.*

Now,

Thus, for any *y *∈ B, there existssuch that

Hence, function* f* is one-one and onto.

Let *f*: **R** → **R** be defined as* f*(*x*) = *x*^{4}. Choose the correct answer.

(A) *f* is one-one onto (B) *f* is many-one onto

(C) *f* is one-one but not onto (D) *f* is neither one-one nor onto

*f*: **R** → **R** is defined as

Let *x*, *y *∈ **R** such that *f*(*x*) = *f*(*y*).

∴does not imply that.

For instance,

∴ *f* is not one-one.

Consider an element 2 in co-domain **R**. It is clear that there does not exist any *x* in domain **R** such that* f*(*x*) = 2.

∴ *f* is not onto.

Hence, function *f* is neither one-one nor onto.

The correct answer is D.

Let *f*: **R** → **R** be defined as *f*(*x*) = 3*x*. Choose the correct answer.

(A) *f* is one-one onto (B) *f* is many-one onto

(C) *f* is one-one but not onto (D) *f* is neither one-one nor onto

*f*: **R** → **R** is defined as *f*(*x*) = 3*x*.

Let *x*, *y *∈ **R** such that *f*(*x*) = *f*(*y*).

⇒ 3*x* = 3*y*

⇒ *x* = *y*

∴*f *is one-one.

Also, for any real number (*y)* in co-domain **R**, there exists in **R** such that.

∴*f *is onto.

Hence, function *f* is one-one and onto.

The correct answer is A.

Let *f*: {1, 3, 4} → {1, 2, 5} and *g*: {1, 2, 5} → {1, 3} be given by *f *= {(1, 2), (3, 5), (4, 1)} and *g *= {(1, 3), (2, 3), (5, 1)}. Write down *g*o*f*.

The functions *f*: {1, 3, 4} → {1, 2, 5} and *g*: {1, 2, 5} → {1, 3} are defined as

*f *= {(1, 2), (3, 5), (4, 1)} and *g *= {(1, 3), (2, 3), (5, 1)}.

Let *f*, *g* and *h* be functions from **R **to **R**. Show that

To prove:

Find *g*o*f *and *f*o*g*, if

(i)

(ii)

(i)

(ii)

If, show that* f *o *f*(*x*) = *x*, for all. What is the inverse of *f*?

It is given that.

Hence, the given function *f* is invertible and the inverse of *f* is *f* itself.

State with reason whether following functions have inverse

(i) *f*: {1, 2, 3, 4} → {10} with

*f* = {(1, 10), (2, 10), (3, 10), (4, 10)}

(ii) *g*: {5, 6, 7, 8} → {1, 2, 3, 4} with

*g* = {(5, 4), (6, 3), (7, 4), (8, 2)}

(iii) *h*: {2, 3, 4, 5} → {7, 9, 11, 13} with

*h* = {(2, 7), (3, 9), (4, 11), (5, 13)}

(i) *f*: {1, 2, 3, 4} → {10}defined as:

*f* = {(1, 10), (2, 10), (3, 10), (4, 10)}

From the given definition of *f*, we can see that *f* is a many one function as: *f*(1) = *f*(2) = *f*(3) = *f*(4) = 10

∴*f *is not one-one.

Hence, function *f *does not have an inverse.

(ii) *g*: {5, 6, 7, 8} → {1, 2, 3, 4} defined as:

*g* = {(5, 4), (6, 3), (7, 4), (8, 2)}

From the given definition of *g*, it is seen that *g* is a many one function as: *g*(5) = *g*(7) = 4.

∴*g *is not one-one,

Hence, function g does not have an inverse.

(iii) *h*: {2, 3, 4, 5} → {7, 9, 11, 13} defined as:

*h* = {(2, 7), (3, 9), (4, 11), (5, 13)}

It is seen that all distinct elements of the set {2, 3, 4, 5} have distinct images under *h*.

∴Function *h* is one-one.

Also, *h* is onto since for every element *y* of the set {7, 9, 11, 13}, there exists an element *x* in the set {2, 3, 4, 5}such that *h*(*x*) = *y*.

Thus, *h* is a one-one and onto function. Hence, *h* has an inverse.

Show that *f*: [−1, 1] → **R**, given byis one-one. Find the inverse of the function *f*: [−1, 1] → Range *f*.

(Hint: For *y* ∈Range *f*, *y* =, for some *x* in [−1, 1], i.e.,)

*f*: [−1, 1] → R is given as

Let *f*(*x*) = *f*(*y*).

∴ *f* is a one-one function.

It is clear that *f*: [−1, 1] → Range *f* is onto.

∴ *f*: [−1, 1] → Range *f* is one-one and onto and therefore, the inverse of the function:

*f*: [−1, 1] → Range *f *exists.

Let *g*: Range *f* → [−1, 1] be the inverse of *f*.

Let* y* be an arbitrary element of range *f*.

Since *f*: [−1, 1] → Range *f* is onto, we have:

Now, let us define *g*: Range* f* → [−1, 1] as

∴*g*o*f* = and *f*o*g *=

*f*^{−1} = *g*

⇒

Consider *f*: **R** → **R** given by *f*(*x*) = 4*x* + 3. Show that *f* is invertible. Find the inverse of *f*.

*f*: **R** → **R** is given by,

*f*(*x*) = 4*x* + 3

One-one:

Let *f*(*x*) = *f*(*y*).

∴ *f* is a one-one function.

Onto:

For *y* ∈ **R**, let *y* = 4*x* + 3.

Therefore, for any *y *∈ **R**, there exists such that

∴ *f* is onto.

Thus, *f* is one-one and onto and therefore, *f*^{−1} exists.

Let us define *g*: **R**→ **R** by.

∴

Hence, *f* is invertible and the inverse of *f* is given by

Consider *f*: **R**_{+ }→ [4, ∞) given by *f*(*x*) =* x*^{2} + 4. Show that *f* is invertible with the inverse *f*^{−1} of given *f *by, where **R**_{+} is the set of all non-negative real numbers.

*f*: **R**_{+} → [4, ∞) is given as* f*(*x*) = *x*^{2} + 4.

One-one:

Let *f*(*x*) = *f*(*y*).

∴ *f* is a one-one function.

Onto:

For *y* ∈ [4, ∞), let *y* = *x*^{2} + 4.

Therefore, for any *y *∈ **R**, there exists such that

.

∴ *f* is onto.

Thus, *f* is one-one and onto and therefore, *f*^{−1} exists.

Let us define *g*: [4, ∞) → **R**_{+ }by,

∴

Hence, *f* is invertible and the inverse of *f* is given by

Consider *f*: **R**_{+} → [−5, ∞) given by *f*(*x*) = 9*x*^{2} + 6*x* − 5. Show that *f* is invertible with.

*f*: **R**_{+} → [−5, ∞) is given as *f*(*x*) = 9*x*^{2} + 6*x* − 5.

Let *y* be an arbitrary element of [−5, ∞).

Let *y* = 9*x*^{2} + 6*x *− 5.

∴*f* is onto, thereby range *f* = [−5, ∞).

Let us define *g*: [−5, ∞) → **R**_{+} as

We now have:

∴and

Hence, *f* is invertible and the inverse of *f* is given by

Let *f*: *X* → *Y* be an invertible function. Show that* f *has unique inverse.

(Hint: suppose *g*_{1} and *g*_{2} are two inverses of *f*. Then for all *y* ∈ *Y*,

*f*o*g*_{1}(*y*) = I_{Y}(*y*) = *f*o*g*_{2}(*y*). Use one-one ness of *f*).

Let *f*: *X* → *Y* be an invertible function.

Also, suppose *f* has two inverses (say_{).}

Then, for all *y* ∈*Y*, we have:

Hence,* f* has a unique inverse.

Consider *f*: {1, 2, 3} → {*a*, *b*, *c*} given by *f*(1) = *a*, *f*(2) = *b* and *f*(3) = *c*. Find *f*^{−1} and show that (*f*^{−1})^{−1} =* f*.

Function *f*: {1, 2, 3} → {*a*, *b*, *c*} is given by,

*f*(1) = *a*, *f*(2) = *b,* and *f*(3) = *c*

If we define *g*: {*a*, *b*, *c*} → {1, 2, 3} as *g*(*a*) = 1, *g*(*b*) = 2, *g*(*c*) = 3, then we have:

∴and, where *X* = {1, 2, 3} and *Y*= {*a*, *b*, *c*}.

Thus, the inverse of *f *exists and *f*^{−1} **=** *g*.

∴*f*^{−1}: {*a*, *b*, *c*} → {1, 2, 3} is given by,

*f*^{−1}(*a*) = 1, *f*^{−1}(*b*) = 2, *f*^{-1}(*c*) = 3

Let us now find the inverse of *f*^{−1} i.e., find the inverse of *g*.

If we define *h*: {1, 2, 3} → {*a*, *b*, *c*} as

*h*(1) = *a*, *h*(2) = *b*, *h*(3) = *c*, then we have:

∴, where *X* = {1, 2, 3} and *Y* = {*a*, *b*, *c*}.

Thus, the inverse of *g *exists and *g*^{−1} = *h* ⇒ (*f*^{−1})^{−1} = *h*.

It can be noted that *h* = *f*.

Hence, (*f*^{−1})^{−1} = *f*.

Let *f*: *X* → *Y* be an invertible function. Show that the inverse of *f*^{−1} is *f*, i.e.,

(*f*^{−1})^{−1} = *f*.

Let *f*: *X* → *Y* be an invertible function.

Then, there exists a function *g*: *Y* → *X* such that *g*o*f* = I_{X}and *f*o*g *= I_{Y}.

Here, *f*^{−1} = *g*.

Now, *g*o*f* = I_{X}and *f*o*g *= I_{Y}

⇒ *f*^{−1}o*f* = I_{X}and *f*o*f*^{−1}= I_{Y}

Hence,* f*^{−1}: *Y* → *X* is invertible and *f* is the inverse of *f*^{−1}

i.e., (*f*^{−1})^{−1} = *f*.

If* f*: **R **→ **R **be given by, then *f*o*f*(*x*) is

(A) (B) *x*^{3} (C) *x* (D) (3 − *x*^{3})

*f*: **R** → **R** is given as.

The correct answer is C.

Letbe a function defined as. The inverse of *f* is map *g*: Range

(A) (B)

(C) (D)

It is given that

Let *y* be an arbitrary element of Range *f*.

Then, there exists *x* ∈such that

Let us define *g*: Rangeas

Now,

∴

Thus, *g* is the inverse of *f* i.e.,* f*^{−1} = *g*.

Hence, the inverse of *f* is the map *g*: Range, which is given by

The correct answer is B.

Determine whether or not each of the definition of given below gives a binary operation. In the event that * is not a binary operation, give justification for this.

(i) On **Z**^{+}, define * by *a ** *b *= *a *− *b*

(ii) On **Z**^{+}, define * by *a ** *b *= *ab*

(iii) On **R**, define * by *a ** *b *= *ab*^{2}

(iv) On **Z**^{+}, define * by *a ** *b *= |*a *− *b*|

(v) On **Z**^{+}, define * by *a ** *b *= *a*

(i) On **Z**^{+}, * is defined by *a *** b = a − b*.

It is not a binary operation as the image of (1, 2) under * is 1 * 2 = 1 − 2 = −1 ∉ **Z**^{+}.

(ii) On **Z**^{+}, * is defined by* a *** b* = *ab*.

It is seen that for each *a*, *b* ∈ **Z**^{+}, there is a unique element *ab* in **Z**^{+}.

This means that * carries each pair (*a*, *b*) to a unique element *a *** b *= *ab *in **Z**^{+}.

Therefore, * is a binary operation.

(iii) On **R**, * is defined by* a *** b* = *ab*^{2}.

It is seen that for each *a*, *b* ∈ **R**, there is a unique element *ab*^{2} in **R**.

This means that * carries each pair (*a*, *b*) to a unique element *a *** b *= *ab*^{2 }in **R**.

Therefore, * is a binary operation.

(iv) On **Z**^{+}, * is defined by *a *** b* = |*a − b*|.

It is seen that for each *a*, *b* ∈ **Z**^{+}, there is a unique element |*a − b*| in **Z**^{+}.

This means that * carries each pair (*a*, *b*) to a unique element *a *** b *= |*a − b*| in **Z**^{+}.

Therefore, * is a binary operation.

(v) On **Z**^{+}, * is defined by* a *** b* = *a*.

* carries each pair (*a*, *b*) to a unique element *a *** b *= *a* in **Z**^{+}.

Therefore, * is a binary operation.

For each binary operation * defined below, determine whether * is commutative or associative.

(i) On **Z**, define *a ** *b *= *a *− *b*

(ii) On **Q**, define *a ** *b *= *ab *+ 1

(iii) On **Q**, define *a ** *b *

(iv) On **Z**^{+}, define *a ** *b *= 2^{ab}

(v) On **Z**^{+}, define *a ** *b *= *a*^{b}

(vi) On **R **− {−1}, define

(i) On **Z**, * is defined by* a *** b* = *a − b*.

It can be observed that 1 * 2 = 1 − 2 = 1 and 2 * 1 = 2 − 1 = 1.

∴1 * 2 ≠ 2 * 1; where 1, 2 ∈ **Z**

Hence, the operation * is not commutative.

Also we have:

(1 * 2) * 3 = (1 − 2) * 3 = −1 * 3 = −1 − 3 = −4

1 * (2 * 3) = 1 * (2 − 3) = 1 * −1 = 1 − (−1) = 2

∴(1 * 2) * 3 ≠ 1 * (2 * 3) ; where 1, 2, 3 ∈ **Z**

Hence, the operation * is not associative.

(ii) On **Q**, * is defined by *a *** b* = *ab* + 1.

It is known that:

*ab* = *ba* &mnForE; *a, b* ∈ **Q**

⇒ *ab* + 1 = *ba *+ 1 &mnForE; *a, b* ∈ **Q**

⇒ *a *** b* = *a *** b* &mnForE; *a, b* ∈ **Q**

Therefore, the operation * is commutative.

It can be observed that:

(1 * 2) * 3 = (1 × 2 + 1) * 3 = 3 * 3 = 3 × 3 + 1 = 10

1 * (2 * 3) = 1 * (2 × 3 + 1) = 1 * 7 = 1 × 7 + 1 = 8

∴(1 * 2) * 3 ≠ 1 * (2 * 3) ; where 1, 2, 3 ∈ **Q**

Therefore, the operation * is not associative.

(iii) On **Q**, * is defined by *a *** b*

It is known that:

*ab* = *ba* &mnForE; *a, b* ∈ **Q**

⇒ &mnForE; *a, b* ∈ **Q**

⇒ *a *** b* = *b *** a* &mnForE; *a, b* ∈ **Q**

Therefore, the operation * is commutative.

For all *a, b*,* c* ∈ **Q**, we have:

∴

Therefore, the operation * is associative.

(iv) On **Z**^{+}, * is defined by *a *** b* = 2^{ab}.

It is known that:

*ab* = *ba* &mnForE; *a, b* ∈ **Z**^{+}

⇒ 2^{ab} = 2^{ba} &mnForE; *a, b* ∈ **Z**^{+}

⇒ *a *** b* = *b *** a* &mnForE; *a, b* ∈ **Z**^{+}

Therefore, the operation * is commutative.

It can be observed that:

∴(1 * 2) * 3 ≠ 1 * (2 * 3) ; where 1, 2, 3 ∈ **Z**^{+}

Therefore, the operation * is not associative.

(v) On **Z**^{+}, * is defined by *a *** b* = *a*^{b}.

It can be observed that:

and

∴ 1 * 2 ≠ 2 * 1 ; where 1, 2 ∈ **Z**^{+}

Therefore, the operation * is not commutative.

It can also be observed that:

∴(2 * 3) * 4 ≠ 2 * (3 * 4) ; where 2, 3, 4 ∈ **Z**^{+}

Therefore, the operation * is not associative.

(vi) On **R**, * − {−1} is defined by

It can be observed that and

∴1 * 2 ≠ 2 * 1 ; where 1, 2 ∈ **R **− {−1}

Therefore, the operation * is not commutative.

It can also be observed that:

∴ (1 * 2) * 3 ≠ 1 * (2 * 3) ; where 1, 2, 3 ∈ **R **− {−1}

Therefore, the operation * is not associative.

Consider the binary operation ∨ on the set {1, 2, 3, 4, 5} defined by *a *∨*b *= min {*a*, *b*}. Write the operation table of the operation∨.

The binary operation ∨ on the set {1, 2, 3, 4, 5} is defined as *a *∨*b* = min {*a*, *b*}

&mnForE; *a*, *b* ∈ {1, 2, 3, 4, 5}.

Thus, the operation table for the given operation ∨ can be given as:

∨ | 1 | 2 | 3 | 4 | 5 |

1 | 1 | 1 | 1 | 1 | 1 |

2 | 1 | 2 | 2 | 2 | 2 |

3 | 1 | 2 | 3 | 3 | 3 |

4 | 1 | 2 | 3 | 4 | 4 |

5 | 1 | 2 | 3 | 4 | 5 |

Consider a binary operation * on the set {1, 2, 3, 4, 5} given by the following multiplication table.

(i) Compute (2 * 3) * 4 and 2 * (3 * 4)

(ii) Is * commutative?

(iii) Compute (2 * 3) * (4 * 5).

(Hint: use the following table)

* | 1 | 2 | 3 | 4 | 5 |

1 | 1 | 1 | 1 | 1 | 1 |

2 | 1 | 2 | 1 | 2 | 1 |

3 | 1 | 1 | 3 | 1 | 1 |

4 | 1 | 2 | 1 | 4 | 1 |

5 | 1 | 1 | 1 | 1 | 5 |

(i) (2 * 3) * 4 = 1 * 4 = 1

2 * (3 * 4) = 2 * 1 = 1

(ii) For every *a, b* ∈{1, 2, 3, 4, 5}, we have *a *** b = b *** a*. Therefore, the operation * is commutative.

(iii) (2 * 3) = 1 and (4 * 5) = 1

∴(2 * 3) * (4 * 5) = 1 * 1 = 1

Let*′ be the binary operation on the set {1, 2, 3, 4, 5} defined by *a **′ *b *= H.C.F. of *a *and *b*. Is the operation *′ same as the operation * defined in Exercise 4 above? Justify your answer.

The binary operation *′ on the set {1, 2, 3 4, 5} is defined as *a **′ *b* = H.C.F of *a* and *b*.

The operation table for the operation *′ can be given as:

*′ | 1 | 2 | 3 | 4 | 5 |

1 | 1 | 1 | 1 | 1 | 1 |

2 | 1 | 2 | 1 | 2 | 1 |

3 | 1 | 1 | 3 | 1 | 1 |

4 | 1 | 2 | 1 | 4 | 1 |

5 | 1 | 1 | 1 | 1 | 5 |

We observe that the operation tables for the operations * and *′ are the same.

Thus, the operation *′ is same as the operation*.

Let * be the binary operation on **N **given by *a* * *b *= L.C.M. of *a *and *b*. Find

(i) 5 * 7, 20 * 16 (ii) Is * commutative?

(iii) Is * associative? (iv) Find the identity of * in **N**

(v) Which elements of **N **are invertible for the operation *?

The binary operation * on **N** is defined as *a *** b* = L.C.M. of *a* and *b*.

(i) 5 * 7 = L.C.M. of 5 and 7 = 35

20 * 16 = L.C.M of 20 and 16 = 80

(ii) It is known that:

L.C.M of *a* and *b* = L.C.M of *b* and *a* &mnForE; *a, b* ∈ **N**.

∴*a* * *b* = *b *** a*

Thus, the operation * is commutative.

(iii) For *a, b*, *c *∈ **N**, we have:

(*a *** b*) ** c *= (L.C.M of *a* and *b*) * *c* = LCM of *a*, *b*, and *c*

*a* * (*b* * *c*) = *a* * (LCM of *b* and *c*) = L.C.M of *a*, *b*, and *c*

∴(*a *** b*) * *c* = *a* * (*b *** c*)

Thus, the operation * is associative.

(iv) It is known that:

L.C.M. of *a* and 1 *= a* = L.C.M. 1 and *a* &mnForE; *a* ∈ **N**

⇒ *a* * 1 = *a* = 1 * *a* &mnForE; *a* ∈ **N**

Thus, 1 is the identity of * in **N**.

(v) An element *a* in **N** is invertible with respect to the operation * if there exists an element *b* in **N**, such that *a *** b = e = b *** a*.

Here, *e* = 1

This means that:

L.C.M of *a* and *b* = 1 = L.C.M of *b* and *a*

This case is possible only when *a* and *b* are equal to 1.

Thus, 1 is the only invertible element of **N** with respect to the operation *.

Is * defined on the set {1, 2, 3, 4, 5} by *a ** *b *= L.C.M. of *a *and *b *a binary operation? Justify your answer.

The operation * on the set A = {1, 2, 3, 4, 5} is defined as

*a* * *b* = L.C.M. of *a* and *b*.

Then, the operation table for the given operation * can be given as:

* | 1 | 2 | 3 | 4 | 5 |

1 | 1 | 2 | 3 | 4 | 5 |

2 | 2 | 2 | 6 | 4 | 10 |

3 | 3 | 6 | 3 | 12 | 15 |

4 | 4 | 4 | 12 | 4 | 20 |

5 | 5 | 10 | 15 | 20 | 5 |

It can be observed from the obtained table that:

3 * 2 = 2 * 3 = 6 ∉ A, 5 * 2 = 2 * 5 = 10 ∉ A, 3 * 4 = 4 * 3 = 12 ∉ A

3 * 5 = 5 * 3 = 15 ∉ A, 4 * 5 = 5 * 4 = 20 ∉ A

Hence, the given operation * is not a binary operation.

Let * be the binary operation on **N **defined by *a ** *b *= H.C.F. of *a *and *b*. Is * commutative? Is * associative? Does there exist identity for this binary operation on **N**?

The binary operation * on **N** is defined as:

*a *** b* = H.C.F. of *a* and *b*

It is known that:

H.C.F. of *a* and *b* = H.C.F. of *b* and *a* &mnForE; *a, b* ∈ **N**.

∴*a* * *b* = *b *** a*

Thus, the operation * is commutative.

For *a*, *b*, *c *∈ **N**, we have:

(*a *** b*)** c* = (H.C.F. of *a* and *b*) * *c* = H.C.F. of *a*, *b*, and *c*

*a **(*b *** c*)= *a **(H.C.F. of *b* and *c*) = H.C.F. of *a*, *b*, and *c*

∴(*a *** b*) * *c = a* * (*b *** c*)

Thus, the operation * is associative.

Now, an element *e *∈ **N** will be the identity for the operation * if *a *** e* = *a* = e** a* *a *∈ **N**.

But this relation is not true for any *a *∈ **N**.

Thus, the operation * does not have any identity in **N**.

Let * be a binary operation on the set **Q **of rational numbers as follows:

(i) *a ** *b *= *a *− *b *(ii) *a ** *b *= *a*^{2} + *b*^{2}

(iii) *a ** *b *= *a *+ *ab *(iv) *a ** *b *= (*a *− *b*)^{2}

(v) (vi) *a ** *b *= *ab*^{2}

Find which of the binary operations are commutative and which are associative.

(i) On **Q**, the operation * is defined as *a *** b = a − b*.

It can be observed that:

and

∴ ; where

Thus, the operation * is not commutative.

It can also be observed that:

Thus, the operation * is not associative.

(ii) On **Q**, the operation * is defined as *a *** b = **a*^{2}* + **b*^{2}.

For *a, b* ∈ **Q**, we have:

∴*a*** b = b *** a*

Thus, the operation * is commutative.

It can be observed that:

(1*2)*3 =(12 + 22)*3 = (1+4)*3 = 5*3 = 52+32= 25 + 9 = 34

1*(2*3)=1*(22+32) = 1*(4+9) = 1*13 = 12+ 132 = 1 + 169 = 170∴ (1*2)*3 ≠ 1*(2*3) , where 1, 2, 3 ∈ Q

Thus, the operation * is not associative.

(iii) On **Q**, the operation * is defined as *a *** b = a + ab*.

It can be observed that:

Thus, the operation * is not commutative.

It can also be observed that:

Thus, the operation * is not associative.

(iv) On **Q**, the operation * is defined by* a *** b* = (*a − b*)^{2}.

For *a*, *b* ∈ **Q**, we have:

*a *** b* = (*a − b*)^{2}

*b *** a* = (*b − a*)^{2} = [− (*a − b*)]^{2} = (*a − b*)^{2}

∴ *a *** b = b *** a*

Thus, the operation * is commutative.

It can be observed that:

Thus, the operation * is not associative.

(v) On **Q**, the operation * is defined as

For *a*, *b* ∈ **Q**, we have:

∴ *a *** b* = *b *** a*

Thus, the operation * is commutative.

For *a, b, c* ∈ **Q**, we have:

∴(*a *** b*) * *c* = *a* * (*b *** c*)

Thus, the operation * is associative.

(vi) On **Q**, the operation * is defined as *a *** b = **ab*^{2}

It can be observed that:

Thus, the operation * is not commutative.

It can also be observed that:

Thus, the operation * is not associative.

Hence, the operations defined in (ii), (iv), (v) are commutative and the operation defined in (v) is associative.

Find which of the operations given above has identity.

An element *e *∈ **Q** will be the identity element for the operation * if

*a *** e = a = e *** a*, *a* ∈ **Q**.

We are given

*a *** b =*

ab4

⇒ a*e = a⇒ae4=a⇒ e=4Similarly, it can be checked for e*a=a, we get e=4Thus, e = 4 is the identity.

Let A = **N **× **N **and * be the binary operation on A defined by

(*a*, *b*) * (*c*, *d*) = (*a *+ *c*, *b *+ *d*)

Show that * is commutative and associative. Find the identity element for * on A, if any.

A = **N** × **N**

* is a binary operation on A and is defined by:

(*a, b*) * (*c, d*) = (*a + c, b + d*)

Let (*a, b*), (*c, d*) ∈ A

Then, *a, b, c, d* ∈ **N**

We have:

(*a, b*) * (*c, d*) = (*a + c, b + d*)

(*c, d*) * (*a, b*) = (*c + a, d + b*) = (*a + c, b + d*)[Addition is commutative in the set of natural numbers]

∴(*a, b*) * (*c, d*) = (*c, d*) * (*a, b*)

Therefore, the operation * is commutative.

Now, let (*a, b*), (*c, d*), (*e, f*) ∈A

Then, *a, b, c, d, e*, *f *∈ **N**

We have:

Therefore, the operation * is associative.

An element will be an identity element for the operation * if

, i.e., which is not true for any element in A.

Therefore, the operation * does not have any identity element.

State whether the following statements are true or false. Justify.

(i) For an arbitrary binary operation * on a set **N**, *a ** *a *= *a * *a ** **N**.

(ii) If * is a commutative binary operation on **N**, then *a ** (*b ** *c*) = (*c ** *b*) * *a*

(i) Define an operation * on **N** as:

*a *** b* = *a + b* *a, b *∈ **N**

Then, in particular, for *b* = *a* = 3, we have:

3 * 3 = 3 + 3 = 6 ≠ 3

Therefore, statement (i) is false.

(ii) R.H.S. = (*c *** b*) * *a*

= (*b *** c*) * *a *[* is commutative]

= *a* * (*b *** c*) [Again, as * is commutative]

= L.H.S.

∴ *a* * (*b *** c*) = (*c *** b*) * *a*

Therefore, statement (ii) is true.

Consider a binary operation * on **N **defined as *a ** *b *= *a*^{3} + *b*^{3}. Choose the correct answer.

(A) Is * both associative and commutative?

(B) Is * commutative but not associative?

(C) Is * associative but not commutative?

(D) Is * neither commutative nor associative?

On **N**, the operation * is defined as *a *** b* = *a*^{3} + *b*^{3}.

For, *a*, *b*, ∈ **N**, we have:

*a *** b* = *a*^{3} + *b*^{3} = *b*^{3} + *a*^{3} = *b *** a* [Addition is commutative in **N**]

Therefore, the operation * is commutative.

It can be observed that:

∴(1 * 2) * 3 ≠ 1 * (2 * 3) ; where 1, 2, 3 ∈ **N**

Therefore, the operation * is not associative.

Hence, the operation * is commutative, but not associative. Thus, the correct answer is B.

Let *f*: **R** → **R **be defined as *f*(*x*) = 10*x* + 7. Find the function *g*: **R** → **R** such that *g* o *f* = *f* o *g *= 1_{R}.

It is given that* f*: **R** → **R** is defined as *f*(*x*) = 10*x* + 7.

One-one:

Let *f*(*x*) = *f*(*y*), where *x*, *y* ∈**R**.

⇒ 10*x* + 7 = 10*y* + 7

⇒ *x* = *y*

∴ *f *is a one-one function.

Onto:

For *y *∈ **R**, let *y* = 10*x* + 7.

Therefore, for any *y *∈ **R**, there exists such that

∴ *f *is onto.

Therefore, *f *is one-one and onto.

Thus, *f* is an invertible function.

Let us define *g*: **R** → **R** as

Now, we have:

Hence, the required function *g*: **R** → **R** is defined as.

Let *f*: W → W be defined as *f*(*n*) = *n* − 1, if is odd and *f*(*n*) = *n* + 1, if *n* is even. Show that* f* is invertible. Find the inverse of *f*. Here, W is the set of all whole numbers.

It is given that:

*f*: W → W is defined as

One-one:

Let *f*(*n*) = *f*(*m*).

It can be observed that if *n* is odd and *m* is even, then we will have *n* − 1 = *m* + 1.

⇒ *n* − *m* = 2

However, this is impossible.

Similarly, the possibility of *n* being even and *m* being odd can also be ignored under a similar argument.

∴Both *n* and *m* must be either odd or even.

Now, if both *n* and *m* are odd, then we have:

*f*(*n*) = *f*(*m*) ⇒ *n* − 1 = *m* − 1 ⇒ *n* = *m*

Again, if both *n* and *m* are even, then we have:

*f*(*n*) = *f*(*m*) ⇒ *n* + 1 = *m* + 1 ⇒ *n* = *m*

∴*f* is one-one.

It is clear that any odd number 2*r *+ 1 in co-domain **N **is the image of 2*r *in domain **N **and any even number 2*r *in co-domain **N **is the image of 2*r *+ 1 in domain **N**.

∴*f* is onto.

Hence, *f *is an invertible function.

Let us define *g*: W → W as:

Now, when *n* is odd:

And, when *n* is even:

Similarly, when *m *is odd:

When *m *is even:

∴

Thus, *f* is invertible and the inverse of *f *is given by *f*^{—1} = *g, *which is the same as *f*.

Hence, the inverse of *f* is *f* itself.

If *f*: **R **→ **R** is defined by *f*(*x*) = *x*^{2} − 3*x *+ 2, find *f*(*f*(*x*)).

It is given that* f*: **R** → **R** is defined as *f*(*x*) = *x*^{2} − 3*x* + 2.

Show that function *f*: **R** → {*x* ∈ **R**: −1 < *x* < 1} defined by *f*(*x*) =, *x *∈**R** is one-one and onto function.

It is given that *f*: **R** → {*x* ∈ **R**: −1 < *x* < 1} is defined as *f*(*x*) =, *x *∈**R**.

Suppose *f*(*x*) = *f*(*y*), where *x*, *y *∈ **R**.

It can be observed that if *x* is positive and *y* is negative, then we have:

Since *x *is positive and *y* is negative:

*x* > *y* ⇒ *x* − *y* > 0

But, 2*xy* is negative.

Then, .

Thus, the case of *x* being positive and *y* being negative can be ruled out.

Under a similar argument, *x* being negative and *y* being positive can also be ruled out

*x* and *y* have to be either positive or negative.

When *x* and *y* are both positive, we have:

When *x* and *y* are both negative, we have:

∴ *f* is one-one.

Now, let *y* ∈ **R** such that −1 < *y *< 1.

If *x* is negative, then there existssuch that

If *x* is positive, then there existssuch that

∴ *f* is onto.

Hence, *f* is one-one and onto.

Show that the function *f*: **R** → **R** given by *f*(*x*) = *x*^{3} is injective.

*f*: **R** → **R** is given as *f*(*x*) = *x*^{3}.

Suppose *f*(*x*) = *f*(*y*), where *x*, *y* ∈ **R**.

⇒ *x*^{3} = *y*^{3} … (1)

Now, we need to show that *x* = *y*.

Suppose *x* ≠ *y*, their cubes will also not be equal.

*x*^{3} ≠ *y*^{3}

However, this will be a contradiction to (1).

∴ *x* = *y*

Hence, *f* is injective.

Give examples of two functions *f*: **N** → **Z** and *g*: **Z** → **Z** such that *g* o *f* is injective but *g* is not injective.

(Hint: Consider *f*(*x*) = *x* and *g*(*x*) =)

Define *f*: **N** → **Z** as *f*(*x*) = *x* and *g*: **Z** → **Z** as *g*(*x*) =.

We first show that *g* is not injective.

It can be observed that:

*g*(−1) =

*g*(1) =

∴ *g*(−1) = *g*(1), but −1 ≠ 1.

∴ *g* is not injective.

Now, *g*o*f*: **N** → **Z** is defined as.

Let *x*, *y* ∈ **N** such that go*f*(*x*) = *g*o*f*(*y*).

⇒

Since *x* and *y* ∈ **N,** both are positive.

Hence, *g*o*f* is injective

Given examples of two functions *f*: **N** → **N** and *g*: **N** → **N** such that *g*o*f* is onto but *f *is not onto.

(Hint: Consider *f*(*x*) = *x* + 1 and

Define *f*: **N** → **N** by,

*f*(*x*) =* x *+ 1

And, *g*: **N** → **N** by,

We first show that *g* is not onto.

For this, consider element 1 in co-domain **N**. It is clear that this element is not an image of any of the elements in domain **N**.

∴ *f* is not onto.

Now, *g*o*f*: **N** → **N** is defined by,

Then, it is clear that for *y* ∈ **N**, there exists *x *= *y *∈ **N** such that *g*o*f*(*x*) = *y*.

Hence, *g*o*f* is onto.

Given a non empty set *X*, consider P(*X*) which is the set of all subsets of *X*.

Define the relation R in P(*X*) as follows:

For subsets *A*, *B* in P(*X*), *A*R*B* if and only if *A* ⊂ *B*. Is R an equivalence relation on P(*X*)? Justify you answer:

Since every set is a subset of itself, *A*R*A* for all *A* ∈ P(*X*).

∴R is reflexive.

Let *A*R*B* ⇒ *A* ⊂ *B.*

This cannot be implied to *B* ⊂ *A.*

For instance, if *A *= {1, 2} and *B* = {1, 2, 3}, then it cannot be implied that *B* is related to *A*.

∴ R is not symmetric.

Further, if *A*R*B *and *B*R*C*, then *A* ⊂ *B* and *B *⊂ *C*.

⇒ *A* ⊂ *C*

⇒ *A*R*C*

∴ R is transitive.

Hence, R is not an equivalence relation since it is not symmetric.

Given a non-empty set *X*, consider the binary operation *: P(*X*) × P(*X*) → P(*X*) given by *A* * *B* = *A* ∩ *B* &mnForE; *A*, *B* in P(*X*) is the power set of *X*. Show that *X *is the identity element for this operation and *X *is the only invertible element in P(*X*) with respect to the operation*.

It is given that.

We know that.

Thus, *X* is the identity element for the given binary operation *.

Now, an elementis invertible if there existssuch that

This case is possible only when *A* = *X* = *B*.

Thus, *X* is the only invertible element in P(*X*) with respect to the given operation*.

Hence, the given result is proved.

Find the number of all onto functions from the set {1, 2, 3, â€¦ , *n*) to itself.

Onto functions from the set {1, 2, 3, â€¦ ,*n*} to itself is simply a permutation on *n* symbols 1, 2, â€¦, *n*.

Thus, the total number of onto maps from {1, 2, â€¦ , *n*} to itself is the same as the total number of permutations on *n* symbols 1, 2, â€¦, *n*, which is *n*!.

Let *S* = {*a*, *b*, *c*} and *T* = {1, 2, 3}. Find F^{−1} of the following functions F from *S* to *T*, if it exists.

(i) *F* = {(*a*, 3), (*b*, 2), (*c*, 1)} (ii) *F* = {(*a*, 2), (*b*, 1), (*c*, 1)}

*S* = {*a*, *b*, *c*}, *T* = {1, 2, 3}

(i) F: *S* → *T* is defined as:

F = {(*a*, 3), (*b*, 2), (*c*, 1)}

⇒ F (*a*) = 3, F (*b*) = 2, F(*c*) = 1

Therefore, F^{−1}: *T* → *S* is given by

F^{−1} = {(3, *a*), (2, *b*), (1, *c*)}.

(ii) F: *S* → *T* is defined as:

F = {(*a*, 2), (*b*, 1), (*c*, 1)}

Since F (*b*) = F (*c*) = 1, F is not one-one.

Hence, F is not invertible i.e., F^{−1} does not exist.

Consider the binary operations*: **R **×**R **→ and o: **R** × **R** → **R **defined as and *a* o *b* = *a*, &mnForE;*a*, *b* ∈ **R**. Show that * is commutative but not associative, o is associative but not commutative. Further, show that &mnForE;*a*, *b*, *c* ∈ **R**, *a**(*b* o *c*) = (*a* * *b*) o (*a* * *c*). [If it is so, we say that the operation * distributes over the operation o]. Does o distribute over *? Justify your answer.

It is given that *: **R **×**R **→ and o: **R** × **R** → **R **isdefined as

and *a* o *b* = *a*, &mnForE;*a*, *b* ∈ **R**.

For *a*, *b* ∈ **R**, we have:

∴*a* * *b* = *b* * *a*

∴ The operation * is commutative.

It can be observed that,

∴The operation * is not associative.

Now, consider the operation o:

It can be observed that 1 o 2 = 1 and 2 o 1 = 2.

∴1 o 2 ≠ 2 o 1 (where 1, 2 ∈ **R**)

∴The operation o is not commutative.

Let *a*, *b*, *c *∈ **R**. Then, we have:

(*a *o *b*) o *c* = *a* o* c *= *a*

*a* o (*b* o *c*) = *a* o *b* = *a*

⇒ *a *o *b*) o c = *a* o (*b* o *c*)

∴ The operation o is associative.

Now, let *a*, *b*, *c *∈ **R**, then we have:

*a* * (*b* o *c*) = a * *b* =

(*a ** *b*) o (*a* * *c*) =

Hence, *a ** (*b *o *c*) = (*a ** *b*) o (*a ** *c*).

Now,

1 o (2 * 3) =

(1 o 2) * (1 o 3) = 1 * 1 =

∴1 o (2 * 3) ≠ (1 o 2) * (1 o 3) (where 1, 2, 3 ∈ **R**)

The operation o does not distribute over *.

Given a non-empty set *X*, let *: P(*X*) × P(*X*) → P(*X*) be defined as *A* * *B* = (*A* − *B*) ∪ (*B* − *A*), &mnForE; *A*, *B* ∈ P(*X*). Show that the empty set *Φ* is the identity for the operation * and all the elements *A* of P(*X*) are invertible with *A*^{−1} = *A*. (Hint: (*A* − *Φ*) ∪ (*Φ* − *A*) = *A* and (*A* − *A*) ∪ (*A* − *A*) = *A* * *A* = *Φ*).

It is given that *: P(*X*) × P(*X*) → P(*X*) is defined as

*A* * *B* = (*A* − *B*) ∪ (*B* − *A*) &mnForE; *A*, *B* ∈ P(*X*).

Let *A *∈ P(*X*). Then, we have:

*A* * *Φ* = (*A* − *Φ*) ∪ (*Φ* − *A*) = *A* ∪ *Φ* = *A*

Φ * *A* = (*Φ* − *A*) ∪ (*A* − *Φ*) = *Φ* ∪ *A* = *A*

∴*A* * *Φ* = *A* = *Φ* * *A*. &mnForE; *A* ∈ P(*X*)

Thus, *Φ* is the identity element for the given operation*.

Now, an element *A* ∈ P(*X*) will be invertible if there exists *B* ∈ P(*X*) such that

*A* * *B* = *Φ* = *B* * *A*. (As Φ is the identity element)

Now, we observed that.

Hence, all the elements *A* of P(*X*) are invertible with *A*^{−1} = *A*.

Define a binary operation *on the set {0, 1, 2, 3, 4, 5} as

Show that zero is the identity for this operation and each element *a* ≠ 0 of the set is invertible with 6 − *a* being the inverse of *a*.

Let *X* = {0, 1, 2, 3, 4, 5}.

The operation * on X is defined as:

An element *e* ∈ *X* is the identity element for the operation *, if

Thus, 0 is the identity element for the given operation *.

An element *a* ∈ *X* is invertible if there exists *b*∈ *X* such that *a* * *b* = 0 = *b* * *a*.

i.e.,

*a* = −*b* or *b* = 6 − *a*

But, *X* = {0, 1, 2, 3, 4, 5} and *a*, *b* ∈ *X.* Then, *a* ≠ −*b*.

∴*b* = 6 − *a* is the inverse of *a* &mnForE; *a* ∈ *X*.

Hence, the inverse of an element *a* ∈*X*, *a* ≠ 0 is 6 − *a* i.e., *a*^{−1} = 6 − *a*.

Let *A* = {−1, 0, 1, 2}, *B* = {−4, −2, 0, 2} and *f*, *g*: *A* → *B* be functions defined by *f*(*x*) = *x*^{2} − *x*, *x* ∈ A and. Are *f* and *g* equal?

Justify your answer. (Hint: One may note that two function *f*: *A* → *B* and g: *A* → B such that *f*(*a*) = g(*a*) &mnForE;*a* ∈*A*, are called equal functions).

It is given that *A* = {−1, 0, 1, 2}, *B* = {−4, −2, 0, 2}.

Also, it is given that* f*, *g*: *A* → *B* are defined by *f*(*x*) = *x*^{2} − *x*, *x* ∈ *A* and.

It is observed that:

Hence, the functions *f *and *g* are equal.

Let *A* = {1, 2, 3}. Then number of relations containing (1, 2) and (1, 3) which are reflexive and symmetric but not transitive is

(A) 1 (B) 2 (C) 3 (D) 4

The given set is A = {1, 2, 3}.

The smallest relation containing (1, 2) and (1, 3) which is reflexive and symmetric, but not transitive is given by:

R = {(1, 1), (2, 2), (3, 3), (1, 2), (1, 3), (2, 1), (3, 1)}

This is because relation R is reflexive as (1, 1), (2, 2), (3, 3) ∈ R.

Relation R is symmetric since (1, 2), (2, 1) ∈R and (1, 3), (3, 1) ∈R.

But relation R is not transitive as (3, 1), (1, 2) ∈ R, but (3, 2) ∉ R.

Now, if we add any two pairs (3, 2) and (2, 3) (or both) to relation R, then relation R will become transitive.

Hence, the total number of desired relations is one.

The correct answer is A.

Let *A* = {1, 2, 3}. Then number of equivalence relations containing (1, 2) is

(A) 1 (B) 2 (C) 3 (D) 4

It is given that* A* = {1, 2, 3}.

The smallest equivalence relation containing (1, 2) is given by,

R_{1} = {(1, 1), (2, 2), (3, 3), (1, 2), (2, 1)}

Now, we are left with only four pairs i.e., (2, 3), (3, 2), (1, 3), and (3, 1).

If we odd any one pair [say (2, 3)] to R_{1}, then for symmetry we must add (3, 2). Also, for transitivity we are required to add (1, 3) and (3, 1).

Hence, the only equivalence relation (bigger than R_{1}) is the universal relation.

This shows that the total number of equivalence relations containing (1, 2) is two.

The correct answer is B.

Let *f*: **R** → **R** be the Signum Function defined as

and *g*: **R** → **R **be the Greatest Integer Function given by *g*(*x*) = [*x*], where [*x*] is greatest integer less than or equal to *x*. Then does *f*o*g* and *g*o*f* coincide in (0, 1]?

It is given that,

*f*: **R** → **R** is defined as

Also,* g*: **R** → **R** is defined as *g*(*x*) = [*x*], where [*x*] is the greatest integer less than or equal to *x*.

Now, let *x* ∈ (0, 1].

Then, we have:[*x*] = 1 if *x* = 1 and [*x*] = 0 if 0 < *x* < 1.

Thus, when *x* ∈ (0, 1), we have *f*o*g*(*x*) = 0and *g*o*f* (*x*) = 1.

Hence, *f*o*g* and *g*o*f* do not coincide in (0, 1].

Number of binary operations on the set {*a*, *b*} are

(A) 10 (B) 16 (C) 20 (D) 8

A binary operation * on {*a*, *b*} is a function from {*a*, *b*} × {*a*, *b*} → {*a*, *b*}

i.e., * is a function from {(*a*, *a*), (*a*, *b*), (*b*, *a*), (*b*, *b*)} → {*a*, *b*}.

Hence, the total number of binary operations on the set {*a*, *b*} is 2^{4} i.e., 16.

The correct answer is B.

**Conclusion**: If you liked the solution of Relations and functions, then share this post with your friends. You can also leave a comment in the comment box.

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]]>The post Things You Must Do While Preparing For UPPCS Exam appeared first on Babaji Academy.

]]>It comes up with a total number of three stages called Preliminary, Mains, and Interview. Moreover, each stage has its specific syllabus indeed. Therefore, you need to put your best efforts to clear this exam. Today, we are going to share some excellent tips and tricks to crack this exam. Let’s check it out in a detailed manner –

**Do Adhere To Your Time Table –**

Yes, it is quite important to stay adhered to your timetable. You should make a timetable adding deadlines. Some candidates do make their timetable when they receive **UPPCS Admit Card****. **Moreover, it is not ideal since you are not left with so much time and it will start creating more pressure on you.

Do not miss those deadlines but adhere to them if you want to have excellent results. Do not waste your time but start following your timetable. Time Table makes you know how much time you would need to accomplish the entire syllabus, how much time you need to give on each section, and how much time you are left with to finish your syllabus.

**Ideal Ways To Make Smart Notes –**

You need to understand that coaching centers do not care if you are hired or not. You need to do everything possible to clear your exam. So, make sure that you do not make excuses to yourself that institutes would be providing you notes.

- Start preparing notes on every topic and that should be short notes.
- Candidates should also emphasize reading the state board so that you could accurate pertinent information regarding a variety of topics. Moreover, it makes you confident too.
- Do not skip any topic while preparing the UPPSC syllabus at all since each one is important and will be in your exam.

- Do write them in your language so that you can get them easily. Moreover, it builds up an emotional connection when you read your stuff written in your language which helps to learn quickly indeed.
- These quick and short notes will also help you during revision saving you precious time too. Make sure that you are depending on your institute only to get notes.
- Whether it is History, Polity or Geography, you should read it understanding its state’s aspects indeed.
- Go with the previous years’ question papers too.

**Consider Only Best Study Material –**

Having quality study material is important since it helps you during exam time as well as revision time. Go with only good quality study material. It will help to get good results indeed. Make sure your study material is available in simple language so that you remain comfortable while preparing it.

To clear this exam, you need to go with a sophisticated preparation strategy as well as a comprehensive booklist regarding the ongoing PCS Main and Prelims indeed. You may also search online for what new best books are available in the market. Make sure that you go with the excellent books to get high scores in your Prelims and Mains both at the same time.

**Self-Study Is Important –**

Coaching is important but it does not mean that Self-study could be ignored. You can rely on coaching centers only for your preparation. Though some coaching centers are quite good and give you incredible value. Self-study is important since it gives you an ideal insight regarding your preparation. You get to understand what you are doing currently and how much you need to cover. Every day you should learn and practice what you were taught in the class. Moreover, it is quite helpful if you read before you get-go to take your class to the coaching center. It will help you to find out doubts, which you can raise in your class in front of your teacher.

**Stay Confident –**

You need to start your preparation right from the first day. This way helps you to take your confidence to the next level. It is time to create a bright future indeed. You should not ever lose hope. Once you are done with a chapter, do not forget to take your test. It will help to realize what things you are good or wrong at. Taking a test means getting feedback on your preparation.

**In The Last – **

This exam is quite important for the candidate who wants to make their career in this field. Since experts design the entire exam, you also need to go with excellent preparation. If you prove yourself, this wonderful job post is going yours.

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]]>The post Basic Computer Knowledge Book Pdf free download in Hindi appeared first on Babaji Academy.

]]>The name of this pdf is **Basic Computer Notes In Hindi. **You can easily download its PDF free of cost. If you are making the Computer Notes in Hindi as well as learning the Computer information in Hindi, then you are at the right place.

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यहाँ हम आपको बता देते हैं की आपको इन नोट्स मे क्या क्या पढ़ने को मिलने वाला है। इन Computer Fundamental Notes मे आपको संपूर्ण ** Basic of Computer in Hindi** की जानकारी आसानी से प्राप्त हो जाएगी तथा यह आपके लिए बहुत ही उपयोगी है। इसमें आपको Introduction of Computer in Hindi दिया रहेगा तथा अन्य नीचे दी हुई महत्वपूर्ण जानकारी भी दी गई हैं।

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Name of the Book | Basic Computer Notes in Hindi |

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Pages in the Book | 52 |

Language of Book | Hindi language |

You can check all the important details about Basic Computer Book Pdf have been given above. If you are facing any issue in downloading this Hindi Book Pdf, then email us on our email id.

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]]>The post Best Noise Cancelling Headphones In India appeared first on Babaji Academy.

]]>इस तरह के हैडफोन बाहर होने वाले शोर को काफी हद तक कम कर देते हैं। इससे म्यूजिक सुनने या फिल्म देखने के दौरान शानदार एक्सपीरियंस मिलता है। Noise Cancellations Headphones in India are available on various online stores. अगर आप भी नॉइज कैंसलिंग हेडफोन Purchase करना चाहते हैं तो यहां दी गई सारी डीटेल आपको इस तरह के Headphone purchase करने में मदद मिलेगी।

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Sennheiser Momentum 3 Wireless भी काफी पॉप्युलर फ्लैगशिप हेडफोन है. इसमें शानदार ऑडियो परफॉर्मेंस के साथ कई फीचर्स दिए गए हैं। Sennheiser के इस हेडफोन में ऐक्टिव नॉइज कैंसलेशन के साथ ऑटो ऑन/ऑफ, स्मार्ट पॉज, तीन ANC मोड्स, ऐमजॉन अलेक्सा, सीरी व गूगल असिस्टेंट इंटीग्रेशन और सिंक विथ Sennheiser Smart Control ऐप जैसे फीचर मिलेंगे। ऐमजॉन पर यह हेडफोन 34,990 रुपये में बेचा जा रहा है।

**04. Bose Noise Cancelling Headphones 700**

चौथा नंबर का बेस्ट हैडफोन है Bose कंपनी का जिसकी डिजाइन काफी शानदार है। स्लीक हेड बैंड और स्टाइलिश ईयर कप डिजाइन इसे भारतीय बाजार में उपलब्ध बेस्ट नॉइज कैंसलिंग हेडफोन में से एक बनाता है। इसमें भी आपको अच्छी साउंड क्वॉलिटी और बेहतरीन नॉइज कैंसलेशन मिलेगा. इसकी कीमत ऐमजॉन पर इस समय 34,500 रुपये है।

**05. Sony WH-XB900N**

पांचबां नंबर का बेस्ट Headphone है Sony कंपनी का जिसका नंबर है Sony WH-XB900N कम दाम वाले बेस्ट नॉइज कैंसलिंग हेडफोन्स में से एक है। इसकी डिजाइन कंपनी के फ्लैगशिप हेडफोन्स से ली गई है। सोनी के इस हेडफोन में एम्बिएंट नॉइज एडजस्टमेंट, क्विक चार्जिंग और टच कंट्रोल्स जैसे कई शानदार फीचर दिए गए हैं। अगर कम दाम में नॉइज कैंसलिंग हेडफोन खरीदना चाह रहे हैं, तो Sony WH-XB900N बेस्ट ऑप्शन रहेगा. ऐमजॉन पर यह हेडफोन 16,990 रुपये में उपलब्ध है।

**निष्कर्ष**: अगर आपको ऊपर दी गई संपूर्ण जानकारी पसंद आई तो इसे ज्यादा से ज्यादा लोगों तक शेयर करें ताकि अन्य लोगों को भी इसका फायदा मिलें।

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]]>These Chandra Institute Notes are very important to crack any competitive exams in India. You can easily get all following subject notes provided by Chandra Institute Coaching Allahabad.

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