NCERT Solutions for Class 12 Science Math Chapter 5 Continuity And Differentiability are provided here with simple step-by-step explanations. These solutions for Continuity And Differentiability are extremely popular among Class 12 Science students for Math Continuity And Differentiability Solutions come handy for quickly completing your homework and preparing for exams. All questions and answers from the NCERT Book of Class 12 Science Math Chapter 5 are provided here for you for free. You will also love the ad-free experience on Meritnationâ€™s NCERT Solutions. All NCERT Solutions for class Class 12 Science Math are prepared by experts and are 100% accurate.

#### Page No 159:

#### Question 1:

Prove that the functionis continuous at

#### Answer:

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

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

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

#### Page No 159:

#### Question 2:

Examine the continuity of the function.

#### Answer:

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

#### Page No 159:

#### Question 3:

Examine the following functions for continuity.

(a) (b)

(c) (d)

#### Answer:

(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.

#### Page No 159:

#### Question 4:

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

#### Answer:

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.

#### Page No 159:

#### Question 5:

Is the function *f*
defined by

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

#### Answer:

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

#### Page No 159:

#### Question 6:

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

#### Answer:

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*.

#### Page No 159:

#### Question 7:

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

#### Answer:

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*.

#### Page No 159:

#### Question 8:

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

#### Answer:

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*.

#### Page No 159:

#### Question 9:

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

#### Answer:

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.

#### Page No 159:

#### Question 10:

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

#### Answer:

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.

#### Page No 159:

#### Question 11:

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

#### Answer:

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.

#### Page No 159:

#### Question 12:

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

#### Answer:

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*.

#### Page No 159:

#### Question 13:

Is the function defined by

a continuous function?

#### Answer:

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*.

#### Page No 160:

#### Question 14:

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

#### Answer:

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

#### Page No 160:

#### Question 15:

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

#### Answer:

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

#### Page No 160:

#### Question 16:

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

#### Answer:

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.

#### Page No 160:

#### Question 17:

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

is
continuous at *x *=
3.

#### Answer:

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,

#### Page No 160:

#### Question 18:

For what value of is the function defined by

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

#### Answer:

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

#### Page No 160:

#### Question 19:

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

#### Answer:

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.

#### Page No 160:

#### Question 20:

Is the function defined by continuous at *x *= $\mathrm{\pi}$?

#### Answer:

The given function is

It is evident that *f* is defined at *x *= $\mathrm{\pi}$.

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

#### Page No 160:

#### Question 21:

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

#### Answer:

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

#### Page No 160:

#### Question 22:

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

#### Answer:

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**

#### Page No 160:

#### Question 23:

Find the points of
discontinuity of *f*,
where

#### Answer:

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.

#### Page No 160:

#### Question 24:

Determine if *f* defined by

is a continuous function?

#### Answer:

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:

$\Rightarrow -{x}^{2}\le {x}^{2}\mathrm{sin}\frac{1}{x}\le {x}^{2}$

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.

#### Page No 161:

#### Question 25:

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

#### Answer:

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.

#### Page No 161:

#### Question 26:

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

#### Answer:

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.

#### Page No 161:

#### Question 27:

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

#### Answer:

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.

#### Page No 161:

#### Question 28:

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

#### Answer:

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

#### Page No 161:

#### Question 29:

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

#### Answer:

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

#### Page No 161:

#### Question 30:

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

is a continuous function.

#### Answer:

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.

#### Page No 161:

#### Question 31:

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

#### Answer:

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.

#### Page No 161:

#### Question 32:

Show that the function defined by is a continuous function.

#### Answer:

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.

#### Page No 161:

#### Question 33:

Examine that is a continuous function.

#### Answer:

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.

#### Page No 161:

#### Question 34:

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

#### Answer:

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.

#### Page No 166:

#### Question 1:

Differentiate the functions with respect to *x*.

#### Answer:

$\mathrm{Let}f\left(x\right)=\mathrm{sin}\left({x}^{2}+5\right),u\left(x\right)={x}^{2}+5,\mathrm{and}v\left(t\right)=\mathrm{sin}t$

$\mathrm{Then},\left(vou\right)=v\left(u\left(x\right)\right)=v\left({x}^{2}+5\right)=\mathrm{tan}\left({x}^{2}+5\right)=f\left(x\right)$

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

**Alternate method**

#### Page No 166:

#### Question 2:

Differentiate the
functions with respect to *x*.

#### Answer:

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

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

By chain rule,

**Alternate method**

#### Page No 166:

#### Question 3:

Differentiate the
functions with respect to *x*.

#### Answer:

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**

#### Page No 166:

#### Question 4:

Differentiate the
functions with respect to *x*.

#### Answer:

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

Hence, by chain rule, we obtain

**Alternate method**

#### Page No 166:

#### Question 5:

Differentiate the
functions with respect to *x*.

#### Answer:

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

#### Page No 166:

#### Question 6:

Differentiate the
functions with respect to *x*.

#### Answer:

The given function is.

#### Page No 166:

#### Question 7:

Differentiate the
functions with respect to *x*.

#### Answer:

#### Page No 166:

#### Question 8:

Differentiate the
functions with respect to *x*.

#### Answer:

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

By using chain rule, we obtain

**Alternate method**

#### Page No 166:

#### Question 9:

Prove that the function
*f *given by

is notdifferentiable at *x* = 1.

#### Answer:

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

#### Page No 166:

#### Question 10:

Prove that the greatest integer function defined byis not

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

#### Answer:

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

#### Page No 169:

#### Question 1:

Find :

#### Answer:

The given relationship is

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

#### Page No 169:

#### Question 2:

Find :

#### Answer:

The given relationship is

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

#### Page No 169:

#### Question 3:

Find :

#### Answer:

The given relationship is

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

Using
chain rule, we obtain** **and

From (1) and (2), we obtain

#### Page No 169:

#### Question 4:

Find :

#### Answer:

The given relationship is

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

#### Page No 169:

#### Question 5:

Find :

#### Answer:

The given relationship is

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

[Derivative of constant function is 0]

#### Page No 169:

#### Question 6:

Find :

#### Answer:

The given relationship is

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

#### Page No 169:

#### Question 7:

Find :

#### Answer:

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

#### Page No 169:

#### Question 8:

Find :

#### Answer:

The given relationship is

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

#### Page No 169:

#### Question 9:

Find :

#### Answer:

$\mathrm{We}\mathrm{have},\phantom{\rule{0ex}{0ex}}y={\mathrm{sin}}^{-1}\left[\frac{2x}{1+{x}^{2}}\right]\phantom{\rule{0ex}{0ex}}\mathrm{put}x=\mathrm{tan}\theta \Rightarrow \theta ={\mathrm{tan}}^{-1}x\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\mathrm{Now},\phantom{\rule{0ex}{0ex}}y={\mathrm{sin}}^{-1}\left[\frac{2\mathrm{tan}\theta}{1+{\mathrm{tan}}^{2}\theta}\right]\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\Rightarrow y={\mathrm{sin}}^{-1}\left(\mathrm{sin}2\theta \right),\left(\mathrm{as}\mathrm{sin}2\theta =\frac{2\mathrm{tan}\theta}{1+{\mathrm{tan}}^{2}\theta}\right)\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\Rightarrow y=2\theta ,\left(\mathrm{as}{\mathrm{sin}}^{-1}\left(\mathrm{sin}x\right)=x\right)\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\Rightarrow y=2{\mathrm{tan}}^{-1}x\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{dy}{dx}=2\times \frac{1}{1+{x}^{2}},\left\{\mathrm{because}\frac{d\left({\mathrm{tan}}^{-1}x\right)}{dx}=\frac{1}{1+{x}^{2}}\right\}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{dy}{dx}=\frac{2}{1+{x}^{2}}$

#### Page No 169:

#### Question 10:

Find :

#### Answer:

The given relationship is

It is known that,

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

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

#### Page No 169:

#### Question 11:

Find :

#### Answer:

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

${\mathrm{sec}}^{2}\left(\frac{\mathrm{y}}{2}\right).\frac{\mathrm{d}}{\mathrm{dx}}\left(\frac{\mathrm{y}}{2}\right)=\frac{\mathrm{d}}{\mathrm{dx}}\left(\mathrm{x}\right)$

$\Rightarrow {\mathrm{sec}}^{2}\frac{y}{2}\times \frac{1}{2}\frac{\mathrm{dy}}{\mathrm{dx}}=1$

$\Rightarrow \frac{dy}{dx}=\frac{2}{se{c}^{2}{\displaystyle \frac{y}{2}}}$

$\Rightarrow \frac{dy}{dx}=\frac{2}{1+{\mathrm{tan}}^{2}{\displaystyle \frac{y}{2}}}$

∴ $\frac{dy}{dx}=\frac{2}{1+{x}^{2}}$

#### Page No 169:

#### Question 12:

Find :

#### Answer:

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

#### Page No 169:

#### Question 13:

Find :

#### Answer:

The given relationship is

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

#### Page No 169:

#### Question 14:

Find :

#### Answer:

The given relationship is

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

#### Page No 169:

#### Question 15:

Find :

#### Answer:

The given relationship is

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

#### Page No 174:

#### Question 1:

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

#### Answer:

Let

By using the quotient rule, we obtain

#### Page No 174:

#### Question 2:

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

#### Answer:

Let

By using the chain rule, we obtain

#### Page No 174:

#### Question 3:

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

#### Answer:

Let

By using the chain rule, we obtain

#### Page No 174:

#### Question 4:

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

#### Answer:

Let

By using the chain rule, we obtain

#### Page No 174:

#### Question 5:

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

#### Answer:

Let

By using the chain rule, we obtain

#### Page No 174:

#### Question 6:

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

#### Answer:

#### Page No 174:

#### Question 7:

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

#### Answer:

Let

Then,

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

#### Page No 174:

#### Question 8:

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

#### Answer:

Let

By using the chain rule, we obtain

,
*x* > 1

#### Page No 174:

#### Question 9:

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

#### Answer:

Let

By using the quotient rule, we obtain

#### Page No 174:

#### Question 10:

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

#### Answer:

Let

By using the chain rule, we obtain

#### Page No 178:

#### Question 1:

Differentiate the
function with respect to *x*.

#### Answer:

Taking logarithm on both the sides, we obtain

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

#### Page No 178:

#### Question 2:

Differentiate the
function with respect to *x*.

#### Answer:

Taking logarithm on both the sides, we obtain

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

#### Page No 178:

#### Question 3:

Differentiate the
function with respect to *x*.

#### Answer:

Taking logarithm on both the sides, we obtain

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

#### Page No 178:

#### Question 4:

Differentiate the
function with respect to *x*.

#### Answer:

*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

#### Page No 178:

#### Question 5:

Differentiate the
function with respect to *x*.

#### Answer:

Taking logarithm on both the sides, we obtain

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

#### Page No 178:

#### Question 6:

Differentiate the
function with respect to *x*.

#### Answer:

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

#### Page No 178:

#### Question 7:

Differentiate the
function with respect to *x*.

#### Answer:

*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

#### Page No 178:

#### Question 8:

Differentiate the
function with respect to *x*.

#### Answer:

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

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

#### Page No 178:

#### Question 9:

Differentiate the
function with respect to *x*.

#### Answer:

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

#### Page No 178:

#### Question 10:

Differentiate the
function with respect to *x*.

#### Answer:

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

#### Page No 178:

#### Question 11:

Differentiate the
function with respect to *x*.

#### Answer:

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

#### Page No 178:

#### Question 12:

Find of function.

#### Answer:

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

#### Page No 178:

#### Question 13:

Find of function.

#### Answer:

The given function is

Taking logarithm on both the sides, we obtain

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

#### Page No 178:

#### Question 14:

Find of function.

#### Answer:

The given function is

Taking logarithm on both the sides, we obtain

Differentiating both sides, we obtain

#### Page No 178:

#### Question 15:

Find of function.

#### Answer:

The given function is

Taking logarithm on both the sides, we obtain

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

#### Page No 178:

#### Question 16:

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

#### Answer:

The given relationship is

Taking logarithm on both the sides, we obtain

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

#### Page No 178:

#### Question 17:

Differentiate in three ways mentioned below

(i) By using product rule.

(ii) By expanding the product to obtain a single polynomial.

(iii By logarithmic differentiation.

Do they all give the same answer?

#### Answer:

(i)

(ii)

(iii)

Taking logarithm on both the sides, we obtain

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

From the above three observations, it can be concluded that all the results of are same.

#### Page No 179:

#### Question 18:

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.

#### Answer:

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

#### Page No 181:

#### Question 1:

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

#### Answer:

The given equations are

#### Page No 181:

#### Question 2:

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

*x* = *a* cos
*θ*, *y* = *b*
cos *θ*

#### Answer:

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

#### Page No 181:

#### Question 3:

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

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

#### Answer:

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

#### Page No 181:

#### Question 4:

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

#### Answer:

The given equations are

#### Page No 181:

#### Question 5:

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

#### Answer:

The given equations are

#### Page No 181:

#### Question 6:

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

#### Answer:

The given equations are

#### Page No 181:

#### Question 7:

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

#### Answer:

The given equations are

#### Page No 181:

#### Question 8:

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

#### Answer:

The given equations are

#### Page No 181:

#### Question 9:

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

#### Answer:

The given equations are

#### Page No 181:

#### Question 10:

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

#### Answer:

The given equations are

#### Page No 181:

#### Question 11:

If

#### Answer:

The given equations are

Hence, proved.

#### Page No 183:

#### Question 1:

Find the second order derivatives of the function.

#### Answer:

Let

Then,

#### Page No 183:

#### Question 2:

Find the second order derivatives of the function.

#### Answer:

Let

Then,

#### Page No 183:

#### Question 3:

Find the second order derivatives of the function.

#### Answer:

Let

Then,

#### Page No 183:

#### Question 4:

Find the second order derivatives of the function.

#### Answer:

Let

Then,

#### Page No 183:

#### Question 5:

Find the second order derivatives of the function.

#### Answer:

Let

Then,

#### Page No 183:

#### Question 6:

Find the second order derivatives of the function.

#### Answer:

Let

Then,

#### Page No 183:

#### Question 7:

Find the second order derivatives of the function.

#### Answer:

Let

Then,

#### Page No 183:

#### Question 8:

Find the second order derivatives of the function.

#### Answer:

Let

Then,

#### Page No 183:

#### Question 9:

Find the second order derivatives of the function.

#### Answer:

Let

Then,

#### Page No 183:

#### Question 10:

Find the second order derivatives of the function.

#### Answer:

Let

Then,

#### Page No 183:

#### Question 11:

If, prove that

#### Answer:

It is given that,

Then,

Hence, proved.

#### Page No 184:

#### Question 12:

If
findin
terms of *y* alone.

#### Answer:

It is given that,

Then,

#### Page No 184:

#### Question 13:

If, show that

#### Answer:

It is given that,

Then,

Hence, proved.

#### Page No 184:

#### Question 14:

Ifshow that

#### Answer:

It is given that,

Then,

Hence, proved.

#### Page No 184:

#### Question 15:

If, show that

#### Answer:

It is given that,

Then,

Hence, proved.

#### Page No 184:

#### Question 16:

If, show that

#### Answer:

The given relationship is

Taking logarithm on both the sides, we obtain

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

Hence, proved.

#### Page No 184:

#### Question 17:

If, show that

#### Answer:

The given relationship is

Then,

Hence, proved.

#### Page No 186:

#### Question 1:

Verify Rolle’s Theorem for the function

#### Answer:

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.

#### Page No 186:

#### Question 2:

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)

#### Answer:

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.

#### Page No 186:

#### Question 3:

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

#### Answer:

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.

#### Page No 186:

#### Question 4:

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

#### Answer:

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.

#### Page No 186:

#### Question 5:

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

#### Answer:

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

#### Page No 186:

#### Question 6:

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

#### Answer:

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.

#### Page No 191:

#### Question 1:

#### Answer:

Using chain rule, we obtain

#### Page No 191:

#### Question 2:

#### Answer:

#### Page No 191:

#### Question 3:

#### Answer:

Taking logarithm on both the sides, we obtain

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

#### Page No 191:

#### Question 4:

#### Answer:

Using chain rule, we obtain

#### Page No 191:

#### Question 5:

#### Answer:

#### Page No 191:

#### Question 6:

#### Answer:

Therefore, equation (1) becomes

#### Page No 191:

#### Question 7:

#### Answer:

Taking logarithm on both the sides, we obtain

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

#### Page No 191:

#### Question 8:

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

#### Answer:

By using chain rule, we obtain

#### Page No 191:

#### Question 9:

#### Answer:

Taking logarithm on both the sides, we obtain

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

#### Page No 191:

#### Question 10:

, for some fixed and

#### Answer:

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

#### Page No 191:

#### Question 11:

, for

#### Answer:

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

#### Page No 191:

#### Question 12:

Find, if

#### Answer:

#### Page No 191:

#### Question 13:

Find, if

#### Answer:

#### Page No 191:

#### Question 14:

If,
for, −1 < *x*
<1, prove that

#### Answer:

_{It is
given that,}

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

_{Hence,
proved.}

#### Page No 191:

#### Question 15:

If, for some prove that

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

#### Answer:

It is given that,

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

Hence, proved.

#### Page No 192:

#### Question 16:

If with _{}prove that

#### Answer:

Then, equation (1) reduces to

$\Rightarrow \mathrm{sin}\left(a+y-y\right)\frac{dy}{dx}={\mathrm{cos}}^{2}\left(a+y\right)\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{dy}{dx}=\frac{{\mathrm{cos}}^{2}\left(a+y\right)}{\mathrm{sin}a}$

Hence, proved.

#### Page No 192:

#### Question 17:

If and, find

#### Answer:

#### Page No 192:

#### Question 18:

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

#### Answer:

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,

#### Page No 192:

#### Question 19:

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

#### Answer:

_{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.}

#### Page No 192:

#### Question 20:

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

#### Answer:

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

#### Page No 192:

#### Question 21:

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

#### Answer:

$y=\left\{\begin{array}{l}\left|x\right|-\infty x\le 1\\ 2-x1\le x\le \infty \end{array}\right.$

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.

#### Page No 192:

#### Question 22:

If, prove that

#### Answer:

Thus,

#### Page No 192:

#### Question 23:

If, show that

#### Answer:

It is given that,

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