Chapter 5 Continuity and Differentiability

8/13/2019 Chapter 5 Continuity and Differentiability

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Class XII Chapter 5 Continuity and Differentiability Maths

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Exercise 5.1

Question 1:

Prove that the function is continuous at

Answer

Therefore, fis continuous atx= 0

Therefore, f is continuous atx= 3


Question 2:

Examine the continuity of the function .

Answer

Thus, fis continuous atx= 3


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Question 3:

Examine the following functions for continuity.

(a) (b)

(c) (d)

Answer

(a) The given function is

It is evident that fis defined at every real number kand its value at kis 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, fis continuous at every point in the domain of fand therefore, it is a continuous

function.

(c) The given function is

For any real number c 5, we obtain


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Hence, fis continuous at every point in the domain of fand therefore, it is a continuous

function.

(d) The given function is

This function fis defined at all points of the real line.

Let cbe 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, fis continuous at all real numbers less than 5.

Case II : c= 5

Then,

Therefore, f is continuous atx= 5

Case III: c> 5

Therefore, fis continuous at all real numbers greater than 5.

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


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Question 4:

Prove that the function is continuous atx= n, where nis a positive integer.

Answer

The given function is f(x)=xn

It is evident that fis defined at all positive integers, n, and its value at nis nn.

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

Question 5:

Is the function fdefined by

continuous atx= 0? Atx= 1? Atx= 2?

Answer

The given function fis

Atx= 0,

It is evident thatf is defined at 0 and its value at 0 is 0.


Atx= 1,

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

The left hand limit off atx= 1 is,

The right hand limit of f atx= 1 is,


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Therefore,f is not continuous atx= 1

Atx = 2,

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

Therefore, fis continuous atx = 2

Question 6:

Find all points of discontinuity of f, where fis defined by

Answer


It is evident that the given function fis defined at all the points of the real line.Let cbe a point on the real line. Then, three cases arise.

(i) c< 2

(ii) c> 2

(iii) c= 2

Case (i) c< 2

Therefore, fis continuous at all pointsx, such thatx< 2

Case (ii) c> 2


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Therefore, fis continuous at all pointsx, such thatx> 2

Case (iii) c= 2

Then, the left hand limit of f atx = 2 is,

The right hand limit of fatx= 2 is,

It is observed that the left and right hand limit of fatx= 2 do not coincide.

Therefore, fis not continuous atx= 2

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

Question 7:


Answer


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

Let cbe a point on the real line.

Case I:


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Case II:


Case III:

Therefore, fis continuous in (3, 3).

Case IV:

If c= 3, then the left hand limit of f atx = 3 is,

The right hand limit of f atx = 3 is,



Case V:



Question 8:



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Answer


It is known that,

Therefore, the given function can be rewritten as



Case I:

Therefore, fis continuous at all pointsx< 0

Case II:

If c= 0, then the left hand limit of f atx = 0 is,




Case III:


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Question 9:


Answer


It is known that,

Therefore, the given function can be rewritten as

Let cbe any real number. Then,

Also,

Therefore, the given function is a continuous function.

Hence, the given function has no point of discontinuity.

Question 10:



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Answer




Case I:


Case II:

The left hand limit of f atx = 1 is,



Case III:

Therefore, fis continuous at all pointsx, such thatx> 1Hence, the given function f has no point of discontinuity.

Question 11:



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Answer




Case I:


Case II:


Case III:


Thus, the given function fis continuous at every point on the real line.

Hence, f has no point of discontinuity.

Question 12:



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Answer




Case I:


Case II:

If c= 1, then the left hand limit of fatx= 1 is,

The right hand limit of fatx = 1 is,



Case III:


Thus, from the above observation, it can be concluded thatx= 1 is the only point ofdiscontinuity of f.

Question 13:

Is the function defined by


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a continuous function?

Answer

The given function is



Case I:


Case II:

The left hand limit of f atx= 1 is,




Case III:

Therefore, fis continuous at all pointsx, such thatx> 1Thus, from the above observation, it can be concluded thatx= 1 is the only point of

discontinuity of f.

Question 14:


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Discuss the continuity of the function f, where fis defined by

Answer


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

Let cbe a point in the interval [0, 10].Case I:

Therefore, fis continuous in the interval [0, 1).

Case II:



It is observed that the left and right hand limits of fatx= 1 do not coincide.


Case III:

Therefore, fis continuous at all points of the interval (1, 3).

Case IV:


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Case V:

Therefore, fis continuous at all points of the interval (3, 10].

Hence, f is not continuous atx = 1 andx = 3

Question 15:


Answer


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


Case I:

Therefore, fis continuous at all pointsx, such thatx < 0


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Case II:




Case III:


Case IV:





Case V:

Therefore, fis continuous at all pointsx, such thatx> 1Hence, f is not continuous only atx = 1

Question 16:



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Answer


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


Case I:


Case II:




Case III:


Case IV:


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Case V:


Thus, from the above observations, it can be concluded that fis continuous at all points

of the real line.

Question 17:

Find the relationship between aand bso that the function fdefined by

is continuous atx = 3.

Answer


If fis continuous atx= 3, then


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Therefore, from (1), we obtain

Therefore, the required relationship is given by,

Question 18:

For what value of is the function defined by

continuous atx= 0? What about continuity atx= 1?

Answer


If fis continuous atx= 0, then

Therefore, there is no value of for which fis continuous atx= 0


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Atx= 1,

f(1) = 4x+ 1 = 4 1 + 1 = 5

Therefore, for any values of , fis continuous atx= 1

Question 19:

Show that the function defined by is discontinuous at all integral point.

Here denotes the greatest integer less than or equal tox.Answer


It is evident that gis defined at all integral points.

Let nbe an integer.

Then,

The left hand limit of f atx= nis,

The right hand limit of fatx = nis,

It is observed that the left and right hand limits of fatx= ndo not coincide.

Therefore, fis not continuous atx=n

Hence, gis discontinuous at all integral points.

Question 20:

Is the function defined by continuous atx = p?

Answer



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It is evident that fis defined atx = p

Therefore, the given function fis continuous atx =

Question 21:

Discuss the continuity of the following functions.

(a) f(x) = sinx+ cosx

(b) f(x) = sinx cosx

(c) f(x) = sinx 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) = sinx and h(x) = cosxare continuous functions.Let g (x) = sinx

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

Let c be a real number. Putx= c+ h

Ifxc, then h0


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Therefore, gis a continuous function.

Let h(x) = cosx

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


Ifxc, then h0

h (c) = cos c

Therefore, his a continuous function.

Therefore, it can be concluded that

(a) f(x) = g(x) + h(x) = sinx+ cosxis a continuous function

(b) f(x) = g(x) h(x) = sinx cosxis a continuous function

(c) f(x) = g(x) h(x) = sinx cosxis a continuous function


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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) = sinx and h(x) = cosxare continuous functions.Let g (x) = sinx

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


Ifx c, then h 0

Therefore, gis a continuous function.

Let h(x) = cosx

It is evident that h(x) = cosxis defined for every real number.Let c be a real number. Putx= c+ h

Ifx c, then h 0

h (c) = cos c


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Therefore, h(x) = cosxis continuous function.

It can be concluded that,

Therefore, cosecant is continuous except atx = np, n Z

Therefore, secant is continuous except at

Therefore, cotangent is continuous except atx = np, n Z

Question 23:

Find the points of discontinuity of f, where


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Answer


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

Let cbe a real number.

Case I:


Case II:


Case III:

The left hand limit of fatx= 0 is,

The right hand limit of fatx= 0 is,


From the above observations, it can be concluded that fis continuous at all points of the

real line.Thus, fhas no point of discontinuity.


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Question 24:

Determine if fdefined by

is a continuous function?

Answer




Case I:

Therefore, fis continuous at all pointsx 0

Case II:


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From the above observations, it can be concluded thatfis continuous at every point of

the real line.

Thus, fis a continuous function.

Question 25:

Examine the continuity of f, where fis defined by

Answer




Case I:


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Therefore, fis continuous at all pointsx, such thatx 0

Case II:


From the above observations, it can be concluded thatfis continuous at every point of

the real line.

Thus, fis a continuous function.

Question 26:

Find the values of k so that the function fis continuous at the indicated point.

Answer


The given function fis continuous at , if fis defined at and if the value of the f

at equals the limit of fat .


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It is evident that f is defined at and

Therefore, the required value of kis 6.

Question 27:


Answer


The given function fis continuous atx= 2, if fis defined atx= 2 and if the value of fat

x= 2 equals the limit of fatx= 2

It is evident that f is defined atx= 2 and


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Therefore, the required value of .

Question 28:


Answer


The given function fis continuous atx= p, if fis defined atx= p and if the value of fat

x= p equals the limit of fatx= p

It is evident that f is defined atx= p and

Therefore, the required value of


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Question 29:


Answer

The given function f is

The given function fis continuous atx= 5, if fis defined atx= 5 and if the value of fat

x= 5 equals the limit of fatx= 5

It is evident that f is defined atx= 5 and

Therefore, the required value of

Question 30:

Find the values of aand bsuch that the function defined by

is a continuous function.


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Answer

The given function f is

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

If fis a continuous function, then fis continuous at all real numbers.

In particular, fis continuous atx = 2 andx = 10

Since fis continuous atx = 2, we obtain

Since fis continuous atx = 10, we obtain

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

8a= 16

a= 2

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

2 2 + b= 5

4 + b= 5


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b= 1

Therefore, the values of aand bfor whichfis a continuous function are 2 and 1

respectively.

Question 31:

Show that the function defined byf (x) = cos (x2) is a continuous function.

Answer

The given function is f (x) = cos (x2)

This function fis defined for every real number and fcan be written as the composition

of two functions as,

f= g o h, where g(x) = cosxand h(x) =x2

It has to be first proved that g (x) = cosxand h(x) =x2are continuous functions.

It is evident that gis defined for every real number.


Then, g(c) = cos c

Therefore, g(x) = cosxis continuous function.


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h(x) =x2

Clearly, his defined for every real number.

Let kbe a real number, thenh (k) = k2


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.

Question 32:

Show that the function defined by is a continuous function.

Answer




f= g o h, where

It has to be first proved that are continuous functions.

Clearly, gis defined for all real numbers.

Let cbe a real number.Case I:


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Therefore, gis continuous at all pointsx, such thatx < 0

Case II:

Therefore, gis continuous at all pointsx, such thatx> 0

Case III:

Therefore, gis continuous atx= 0

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

h (x) = cosx

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


Ifxc, then h0

h (c) = cos c

Therefore, h(x) = cosxis a continuous function.




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Question 33:

Examine that is a continuous function.

Answer



f= g o h, where

It has to be proved first that are continuous functions.


Let cbe a real number.Case I:


Case II:


Case III:


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h (x) = sinx

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

Let c be a real number. Putx= c+ k

Ifxc, then k0

h (c) = sin c





Question 34:

Find all the points of discontinuity of f defined by .

Answer



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The two functions, gand h, are defined as

Then, f= g h

The continuity of gand h is examined first.



Case I:


Case II:


Case III:



Clearly, his defined for every real number.


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Let c be a real number.

Case I:

Therefore, his continuous at all pointsx, such thatx < 1

Case II:

Therefore, his continuous at all pointsx, such thatx> 1

Case III:

Therefore, his continuous atx= 1

From the above three observations, it can be concluded thathis continuous at all points

of the real line.

gand hare continuous functions. Therefore, f = g h is also a continuous function.

Therefore, f has no point of discontinuity.


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Exercise 5.2

Question 1:

Differentiate the functions with respect tox.

Answer

Thus, fis a composite of two functions.

Alternate method

Question 2:



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Answer

Thus, f is a composite function of two functions.

Put t= u(x) = sinx

By chain rule,

Alternate method

Question 3:


Answer

Thus, f is a composite function of two functions, uand v.

Put t= u(x) = ax+ b

Hence, by chain rule, we obtain


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Alternate method

Question 4:


Answer

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

Hence, by chain rule, we obtain


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Alternate method

Question 5:


Answer

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

h(x) = cos (cx + d)


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g is a composite function of two functions, uand v.

Therefore, by chain rule, we obtain

his a composite function of two functions,pand q.

Put y=p(x) = cx + d

Therefore, by chain rule, we obtain


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Question 6:


Answer

The given function is .

Question 7:


Answer


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Question 8:


Answer

Clearly, f is a composite function of two functions, u andv, such that


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By using chain rule, we obtain

Alternate method

Question 9:

Prove that the function f given by

is notdifferentiable atx= 1.

Answer



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It is known that a function fis differentiable at a pointx= cin its domain if both

are finite and equal.

To check the differentiability of the given function atx= 1,

consider the left hand limit of fatx= 1

Since the left and right hand limits of fatx= 1 are not equal, fis not differentiable atx

= 1

Question 10:

Prove that the greatest integer function defined by is not

differentiable atx= 1 andx= 2.

Answer


It is known that a function fis differentiable at a pointx= cin its domain if both

are finite and equal.

To check the differentiability of the given function atx= 1, consider the left hand limit of

fatx= 1


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Since the left and right hand limits of fatx= 1 are not equal, fis not differentiable at

x= 1

To check the differentiability of the given function atx= 2, consider the left hand limit

of fatx= 2

Since the left and right hand limits of fatx= 2 are not equal, fis not differentiable atx

= 2


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Exercise 5.3

Question 1:

Find :

Answer

The given relationship is

Differentiating this relationship with respect tox, we obtain

Question 2:

Find :

Answer




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Question 3:

Find :

Answer



Using chain rule, we obtain and

From (1) and (2), we obtain

Question 4:

Find :


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Answer

The given relationship isDifferentiating this relationship with respect tox, we obtain

Question 5:

Find :

Answer



[Derivative of constant function is 0]


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Question 6:

Find :

Answer



Question 7:

Find :


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Answer

The given relationship isDifferentiating this relationship with respect tox, we obtain

Using chain rule, we obtain

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

Question 8:

Find :

Answer




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Question 9:

Find :

Answer



The function, , is of the form of .

Therefore, by quotient rule, we obtain


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Also,


Question 10:

Find :

Answer



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It is known that,

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


Question 11:

Find :

x Answer

The given relationship is,


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


Question 12:

Find :

Answer



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Alternate method



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Question 13:

Find :

Answer




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Question 14:

Find :

Answer

lationship is



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Question 15:

Find :

Answer



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Exercise 5.4

Question 1:

Differentiate the following w.r.t.x:

Answer

Let

By using the quotient rule, we obtain

Question 2:


Answer

Let

By using the chain rule, we obtain


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Question 2:

Show that the function given by f(x) = e2xis strictly increasing on R.

Answer

Let be any two numbers in R.

Then, we have:

Hence, fis strictly increasing on R.

Question 3:


Answer

Let


Question 4:



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Answer

Let


Question 5:


Answer

Let



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Question 6:


Answer

Question 7:


Answer

Let

Then,

By differentiating this relationship with respect tox, we obtain


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Question 8:


Answer

Let


,x> 1

Question 9:



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Answer

Let

By using the quotient rule, we obtain

Question 10:


Answer

Let



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Exercise 5.5

Question 1:

Differentiate the function with respect tox.

Answer

Taking logarithm on both the sides, we obtain

Differentiating both sides with respect tox, we obtain

Question 2:


Answer



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Question 3:


Answer




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Question 4:


Answer

u =xx



v= 2sinx

Taking logarithm on both the sides with respect tox, we obtain


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Question 5:


Answer




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Question 6:


Answer



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

Question 7:


Answer

u = (logx)x



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Question 8:



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Answer




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Question 9:


Answer




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Question 10:


Answer



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Question 11:



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Answer



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Question 12:

Find of function.


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Answer


Letxy= uand yx= v

Then, the function becomesu + v= 1





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Question 13:

Find of function.

Answer




Question 14:

Find of function.


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Answer



Differentiating both sides, we obtain

Question 15:

Find of function.

Answer




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Question 16:

Find the derivative of the function given by and hence

find .

Answer





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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)


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(ii)

(iii)




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



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Exercise 5.6

Question 1:

Ifxand yare connected parametrically by the equation, without eliminating the

parameter, find .

Answer

The given equations are

Question 2:


parameter, find .

x= acos , y= bcos

Answer

The given equations arex= acos and y= bcos


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Question 3:


parameter, find .

x= sin t, y= cos 2t

Answer

The given equations arex= sin tand y= cos 2t

Question 4:


parameter, find .

Answer



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Question 5:


parameter, find .

Answer


Question 6:


parameter, find .


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Answer


Question 7:


parameter, find .

Answer



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Question 8:


parameter, find .

Answer



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Question 9:


parameter, find .


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Answer


Question 10:


parameter, find .

Answer



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Question 11:

If

Answer



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Hence, proved.


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Exercise 5.7

Question 1:

Find the second order derivatives of the function.

Answer

Let

Then,

Question 2:


Answer

LetThen,

Question 3:


Answer

Let

Then,


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Question 4:


Answer

Let

Then,

Question 5:


Answer

Let

Then,


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Question 6:


Answer

Let

Then,


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Question 7:


Answer

Let

Then,

Question 8:


Answer

Let

Then,


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Question 9:


Answer

Let

Then,

Question 10:


Answer

Let

Then,


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Question 11:

If , prove that

Answer

It is given that,

Then,


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Hence, proved.

Question 12:

If find in terms of yalone.

Answer

It is given that,

Then,


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Question 13:

If , show that

Answer

It is given that,

Then,


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Hence, proved.

Question 14:

If show that


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Answer

It is given that,Then,

Hence, proved.

Question 15:

If , show that

Answer

It is given that,

Then,


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Hence, proved.

Question 16:

If , show that

Answer





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Hence, proved.

Question 17:

If , show that

Answer


Then,

Hence, proved.


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Exercise 5.8

Question 1:

Verify Rolles 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.

Rolles Theorem states that there is a point c(4, 2) such that

Hence, Rolles Theorem is verified for the given function.


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Question 2:

Examine if Rolles Theorem is applicable to any of the following functions. Can you say

some thing about the converse of Rolles Theorem from these examples?

(i)

(ii)

(iii)

Answer

By Rolles Theorem, for a function , if

(a) fis continuous on [a, b](b) fis differentiable on (a, b)

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

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

Therefore, Rolles 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 atx = 5 andx = 9

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

The differentiability of fin (5, 9) is checked as follows.


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

Since the left and right hand limits of fatx= nare not equal, fis not differentiable atx

= n

f is not differentiable in (5, 9).

It is observed that fdoes not satisfy all the conditions of the hypothesis of Rolles

Theorem.

Hence, Rolles Theorem is not applicable for .

(ii)






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= n


It is observed that fdoes not satisfy all the conditions of the hypothesis of Rolles

Theorem.


(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 fdoes not satisfy a condition of the hypothesis of Rolles Theorem.


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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) fis 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.

Question 4:

Verify Mean Value Theorem, if in the interval , where

and .


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Answer


f, being a polynomial function, is continuous in [1, 4] and is differentiable in (1, 4)

whose derivative is 2x 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.

Question 5:

Verify Mean Value Theorem, if in the interval [a, b], where a= 1 and

b= 3. Find all for which

Answer


f, being a polynomial function, is continuous in [1, 3] and is differentiable in (1, 3)

whose derivative is 3x2 10x 3.


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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

Question 6:

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

Answer

Mean Value Theorem states that for a function , if

(a) fis continuous on [a, b]

(b) fis 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.


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(i)







= n


It is observed that fdoes not satisfy all the conditions of the hypothesis of Mean Value

Theorem.

Hence, Mean Value Theorem is not applicable for .

(ii)



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= n


It is observed that fdoes 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 fsatisfies all the conditions of the hypothesis of Mean Value Theorem.


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Hence, Mean Value Theorem is applicable for .

It can be proved as follows.


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Miscellaneous Solutions

Question 1:

Answer


Question 2:

Answer

Question 3:

Answer


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Question 4:

Answer



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Question 5:

Answer


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Question 6:

Answer

8/13/2019 Chapter 5 Contin

Chapter 5 Continuity and Differentiability

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Transcript of Chapter 5 Continuity and Differentiability