5_continuity_and_differentiability.pdf

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By:-

DR. VIKRAM SINGH

TANUSHREE SINGH

YEAR OF PUBLICATION-2010

All rights reserved. No part of this publication may be reproduced,

stored in a retrieval system, transmitted in any form or by any means-

Electronic, Mechanical, Photocopying, Recording or otherwise, without

prior permission of the Authors and Publisher

SAVANT INSTITUTE

TM

CLASS XII

MATHEMATICS


2/12

Mathematics Continuity and Differentiability

SAVANT EDUCATION GROUP E-17, East of Kailash, New Delhi 110065. Ph.: +91-11-26224417 www.savantgroup.org

5

CONTINUITY AND DIFFERENTIABILITY

____________________________________________________________________________________

Slide 1

Continuity

Definition:

Suppose f is a real function on a subset of the real

numbers and let c be a point in, the domain of f. then f is

continuous at c if ( ) ( )x climf x f c

=

or ( ) ( ) ( )x c x clim f x lim f x f c

+ = =

or L.H.L. = R.H.L. = f(c)

___________________ Slide 2___________________

Solved Example-:

Check the continuity of the function f given by f(x) = 2x + 3

at x = 1.Solution-:

First note that the function is defined at the given pointx = 1 and its value is 5. Then find the limit of the function

at x = 1, clearly ( ) ( ) ( )x 1 x 1limf x lim 2x 3 2 1 3 5.

= + = + =

Thus ( ) ( )x 1limf x 5 f 1

= =

Hence, f is continuous at x = 1.

___________________ Slide 3_______________

Illustration-:

Discus the continuity of the function f given by f(x) =

x = 0.

___________________ Slide 4_______________

Illustration-:

Show that the function f given by ( )3x 3,if x

f x1, if x

+ =

not continuous at x = 0.

___________________ Slide 5_______________

A real function f is said to be continuous if i t is conti

at every point in the domain of f.


3/12

48 Continuity and Differentiability Mathem


Suppose f is a function defined on a closed interval [a, b]

then for, f to be continuous, it needs to be continuous at

every end points a and b.

Continuity of f at point a means.

( ) ( )x al imf x f a

+= and continuity of f at b means ( ) ( )

x b

l imf x f b .

=

___________________ Slide 6___________________

Solved Example-:

Is the function defined by f(x) =|x|, a continuous function?Solution-:

We may rewrite f as ( )x if x 0

f xx, if x 0

0. Then f(c) = c.

Also ( )x c x climf x limx c = = Since ( ) ( )

x climf x f c

= , f is continuous at all positive real

numbers. Hence, f is continuous at all point.

___________________ Slide 7___________________

Illustration-:

Discuss the continuity of the function f defined by

( )x 2, if x 1

f xx 2, if x 1

+ = >

___________________ Slide 8___________________

Solved Example-:

Discuss the continuity of the function defined by

( )x 2, if x 0

f x .x 2, if x 0

+

Solution-:

Observe that the function is defined at all real numbers

except at 0. Domain of definition of this function is

}1 2 1 2D D whereD x R : x 0 a ndD x R : x 0 = < = >

Case 1: If1

c D , then ( ) ( )x c x climf x lim x 2

= + = c + 2 = f

(c) and hence f is continuous in D1.

Case2: If2

c D , then ( ) ( )x c x climf x lim x 2

= + = c + 2 =

f(c) and hence f is continuous in D2. Since f is continuous

at all points in the domain of f, we deduce that f is

continuous. Graph of this function is given in the fig. note

that to graph this function we need to lift the pen from the

plane of the paper, but we need to do that only for

points where the function is not defined.

___________________ Slide 9_______________

Solved Example-:

Find all the points of discontinuity of the greatest in

function defined by f(x) = [x], where [x] denote

greatest integer less than or equal to x.

Solution-:

First observe that f is defined for all real numbers.

of the function is given in fig. from the graph it looks

is discontinuous at every integral point. Below we ex

if this is true.

Case 1: Let c be a real number which is not equal

integer. It is evident from the graph that for all real nu

close to c the value of the function is equal to [c

( ) [ ] [ ]x c x climf x lim x c

= = Also f (c) = [c] and hence func

continuous at all real numbers not equal to integers

Case 2: Let c be an integer. Then we can

sufficiently small real number r > 0 such that [c r] =

where as [c + r] = c.

This, in terms of limits mean that

( ) ( )x c x cl imf x c 1, l imf x c

+ = =

Since these limits cannot be equal to each other for the function is discontinuous at every integral point


4/12



__________________ Slide 10___________________

Algebra of continuous functions

Theorem 1: Suppose f and g be two real functions

continuous at a real number c. then

(i) f + g is continuous at x = c.

(ii) f g is continuous at x = c.(iii) f.g is continuous at x = c.

(iv)f

g

is continuous at x = c.

Provided g(c) 0.__________________ Slide 11___________________

Solved Example-:

Discuss the continuity of sine function.

Solution-:

To see this we use the following factsx 0limsinx 0

=

We have not provided it, but is intuitively clear from the

graph of sin x near 0.

Now, observe that f(x) = sin x is defined for every real

number. Let c be a real number. Put x = c + h. If x c weknow that h0. Therefore ( )

x c x climf x limsinx

=

( )

[ ]h 0

h 0

limsin c h

lim sinccosh coscsinh

= +

= +

[ ] [ ]h 0 h 0lim sinccosh lim coscsinh

= +

( )

sinc 0

sinc f c

= +

= =

Thus ( ) ( )x climf x f c

= and hence f is a continuous function.

Remark A similar proof may be given for the continuity of

cosine function.

Slide 12

Illustration-:

Prove that the function defined by f(x) = tan x

continuous function.

__________________ Slide 13_______________

Continuity of composite functions

Theorem (2):

Suppose f and g are real valued functions such that is defined at c. if g is continuous at c and if f is conti

at g (c) then (f o g) is continuous at c.

__________________ Slide 14_______________

Solved Example-:

Show that the function defined by f(x) = sin (x2

continuous function.

Solution-:

Observe that the function is defined for every real nuThe function f may be thought of as a composition g

the two functions g and h, where g (x) = sin x and h (xSince both g and h are continuous functions, by theoit can be deduced that f is a continuous function.

__________________ Slide 15_______________

Illustration-:

Show that the function f defined by ( )f x 1 x= +

Where x is any real number, is a continuous function

__________________ Slide 16_______________

Differentiability

Suppose f is a real function and c is a point in its do

The derivation of f at c is defined by( ) (

h 0

f c h f clim

h

+

Provided this limit exists. Derivative of f at c is deno

f(c) at ( )c

df x .

dx

The function defined by ( ) ( ) ( )

h 0

f x h f xf ' x lim

h

+ =

__________________ Slide 17_______________

Algebra of derivatives

If u and v are true different functions then.

(i) (u + v) = u + v

(ii) (u v)= u v(iii) (uv)= u v + uv (Leibnitz or product rule0

(iv)2

u u v uv,

v v

= where ever v ? 0 (Quotient rule


5/12



Slide 18

Derivatives of composite function

Chain rule: let f be a real valued function which is a

composite of two functions u and v, that is f = v o u.

Suppose t = u(x) and if bothdt dv

anddx dt

exist we have

df dv dt.

dx dt dx=

Chain rule can be extended as follows. Suppose f is a real

valued function which is a composition of three functions

u, v and w that is

F = (w o u) o v. If t = v(x) and s = u(t) then

( )d w o udf dt dw ds dt. . .

dx dt ds ds dt dx= =

__________________ Slide 19___________________

Solved Example-:

Find the derivative of the function given by f(x) = sin (x2).

Solution-:

Observe that the given function is a composite of two

functions. Indeed, if t = u(x) = x2and v(t) = sin t, then

f(x) = (v o u) (x) = v(u(x)) = v (x2) = sin x

2

Put t = u(x) = x2. Observe that

dvc o s t

dt= and

dt2x

dx=

exist. Hence, by chain ruledf dv dt

. cost.2xdx dt dx

= =

It is normal practice to express the final result only in

terms of x. thus 2df

cost.2x 2 xc os xdx

= =

__________________ Slide 20___________________

Illustration-:

Find the derivative of tan(2x + 3).

__________________ Slide 21___________________

Solved Example-:

Differentiate sin (cos (x2)) with respect to x.

Solution-:

The function f(x) = sin (cos (x2)) is a composition f(x) = (w o

v o u) (x) of the three functions u, v and w, where u(x) = x2,

v (t) = cos t and w (s) = sin s. put t = u(x) = x2and s = v(t) =

cos t. observe that

dw ds dt

coss, sintand 2xds dt dx= = = exist for all real x. hence by a generalization of chain rule,

we havedf dw ds dt

. .dx ds dt dx

= = (cos s). (sin t).(2x) = 2x sin

x2. cos (cos x

2)

Slide 22

Implicit and Explicit Functions

Until now we have been differentiating various fun

given in the form y = f(x). But it is not necessaryfunctions are always expressed in this form. For exaconsider one of the following relationships between x

x y = 0x + sin xy y = 0

In the first case, we can solve for y and rewritrelationship as y = x . In the second case, it doseem that there is an easy way to solve Nevertheless, there is no doubt about the depende

y on x in either of the cases. When a relationship bex and y is expressed in a way that it is easy to solveand write y = f(x), we say that y is given as an e

function of x. In the latter case it is implicit that yfunction of x and we say that the relationship second type, above, gives function implicitly. I

subsection, we learn to differentiate implicit functions

__________________ Slide 23_______________

Derivative of Implicit FunctionsSolved Example-:

Finddy

,if y siny cosx.dx

+ =

Solution-:

We differentiate the relationship directly with respec

i.e.

( ) ( )dy d d

siny cosxdx dx dx

= =

which implies using chain rule

dy dy

cosy. sinxdx dx+ =

This givesdy sinx

dx 1 cosy

=+

Where y (2n + 1) .

__________________ Slide 24_______________

Derivative of Explicit Functions.

Solved Example-:

Finddy

if x y .dx

=

Solution-:

One way is to solve for y and rewrite the above as

y = x dy

1.dx

=


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Slide 25

Derivatives of inverse trigonometric functions

Solved Example-:

Find the derivative of f given by f(x) = sin1

x assuming it

exists.

Solution-:

Let y = sin1

x. then, x = sin y.

Differentiating both sides w.r.t.x, we get 1 = cos ydy

dx

Which implies that( )1

dy 1 1

dx cosy cos sin x= =

Observe that this is defined only for cos y ? 0, i.e., sin1

x

? ,2 2

, i.e., x ? 1, 1, i.e., x (1, 1).

To make this result a bit more attractive, we carry out the

following manipulation. Recall that for x (1, 1), sin(sin

1x) = x and hence cos

2y = 1 (sin y)2 = 1 (sin

(sin1x))

2= 1 x

2.

Also, since y , ,cosy2 2

is positive and hence cos

2y 1 x=

Thus, for x (1, 1),2

dy 1 1

dx cosy 1 x= =

__________________ Slide 26___________________

Solved Example-:

Differentiation Using Logarithms.

If y =

( )( )

( )

2

2

x 3 x 4 dy.Find dx3x 4x 5

+

+ +

Solution-:

Given:( ) ( )

( )

2

2

x 3 x 4 dyy ,Find

dx3x 4x 5

+=

+ +

Taking logarithm on both sides of (i), we get

( ) ( ) ( ){ }2 21

logy log x 3 log x 4 log 3x 4x 5 .2

= + + + +

Differentiating both sides w.r.t.x, we get

( ) ( )

( )

( )2 26x 41 dy 1 1 2x

. . y dx 2 x 3 x 4 3x 4x 5

+

= + + + + +

( ) ( )

( )

( )2 26x 4dy 1 1 2x

y . dx 2 x 3 x 4 3x 4x 5

+ = + + + + +

( ) ( )

( )

( ) ( )( )

( )

2

2

2 2

x 3 x 41.

2 3x 4x 5

6x 41 2x.

x 3 x 4 3x 4x 5

+=

+ +

+ + + + + +

__________________ Slide 27_______________

Illustration-:

If y = (x)cosx

+ (cosx)sinx

, finddy

dx.

__________________ Slide 28_______________

Derivatives of functions in parametric forms:

A relation expressed between two variables x and y

from x = f(t), y = g(t) is said to be parametric form wi

a parameter.

In order to find derivative of function in such form, we

chain rule.

or

dy dy dx

.dt dx dt=

dydy dxdt [whenever 0]

dxdx dt

dt

=

Thus( )

( ) ( ) ( )

g tdy dy dxas g t and f ' t

dx f t dt dt

= = = [Provided f(t) 0]

__________________ Slide 29_______________

Solved Example-:

Finddy

dx, if x = a(+ sin), y = a(1 cos).

Solution-:

We have ( ) ( )dx dy

a 1 cos , a sind d

= + =

Therefore( )

( )

dya sindy d tan

dxdx a 1 cos 2

d

= = =+

__________________ Slide 30_______________

Illustration-:

Find2 2 2

3 3 3dy ,if x y a .dx

+ =


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Slide 31

Second order Derivative

Let y = f(x). Then ( )dy

f xdx

=

If f(x) is differentiable, we may differentiate (1) again w.r.t.x.

Then, the left hand side becomesd dy

dx dx

which is called

the second order derivative of y w.r.t.x and is denoted by2

2

d y.

dx The second order derivative of f(x) is denoted by f(x).

It is also denoted by D2y or y or y2if y = f (x).

__________________ Slide 32___________________

Solved Example-:

Find2

2

d y,

dxif y = x

3+ tan x.

Solution-:

Given that y = x3+ tan x. Then 2 2

dy3x sec x

dx

= +

Therefore ( )2

2 2

2

d y d3x sec x

dx dx= +

2

6x 2secx.secxtanx

6x 2sec xtanx

= +

= +

__________________ Slide 33___________________

Illustration-:

If y = sin1

x, show that (1 x2)

2

2

d y dy x 0

dx dx= .

__________________ Slide 34___________________

Rolles Theorem:

Let f: [a, b] R be continuous function on [a, b] anddifferentiable on (a, b). Then there exists some c in (a, b)

such that f(c) = 0.

__________________ Slide 35___________________

Solved Example-:

Verify Rolles theorem for the function y = x2+ 2, a = 2

and b = 2.

Solution-:

The function y = x2 + 2 is continuous in [2, 2] and

differentiable in (2, 2). Also f (2) = f (2) = 6 and hence thevalue of f(x) at 2 and coincide. Rolles theorem states thatthere is point c (2, 2) where f(c) = 0. Since f(x) = 2x, weget c = 0. Thus at c = 0, we have f(c) = 0 and c = 0 (2, 2).

Slide 36

Mean Value theorem

Let f ; [a, b] R be a continuous function on [a, bdifferentiable on (a, b). Then there exists some c in

such that( ) ( )f b f a

f cb a

=

__________________ Slide 37_______________

Solved Example-:

Verify Mean Value Theorem for the function f(x) = x2

interval [2, 4].

Solution-:

The function f(x) = x2 is continuous in [2, 4

differentiable in (2, 4) as its derivative f(x) = 2x is d

in (2, 4).

Now, f(2) = 4 and f(4) = 16 Hence ( ) ( )f b f a 16 4

b a 4 2

=

MVT states that there is a point c (2, 4) such that f(But f(x) = 2x which implies c = 3. Thus at c = 3

we have f(c) = 6.


8/12



CURRICULUM BASED WORKSHEET

Topics for Worksheet I

Continuity at a point

Continuity at interval

Algebra of continuous functions

Worksheet I

1. Test the continuity of the function f(x) at the origin:

( )x

; x 0f x x

1 ; x 0

=

=

2. The sine function is everywhere continuous.

3. The cosine function is everywhere continuous.

4. Discuss the continuity of the function

( )sin2x

, i f x 0f x x

x 2,if x 0


9/12



2. If( )

( )

3/ 2

4 /3

x x 4 dyy ,find .

dx4 x 3

+=

3. Differentiate ( )( )( )( )x 1 x 2 x 3 x 4 w.r.t. x.

4. If y = (2x + 3)(3x 5), find

dy.

dx

5. Differentiate1sin xx w.r.t. x.

6. Differentiate (sin x)log x

w.r.t. x.

7. If ( ) ( )cosx sinx dyy x cosx ,find .dx

= +

8. If y = (sin x)tan x

+ (cos x)sec x

, finddy

.dx

Topics for Worksheet V

Derivatives of function in parametric form

Second order derivatives

Worksheet V

1. Find the second-order derivative of:(i) x10

(ii) log x

(iii) tan1x

2. If y = (tan x + sec x), prove that( )

2

22

d y cosx.

dx 1sinx=

3. If y = e4x

sin 3x, find2

2

d y.

dx

4. Find the second derivative of log (log x) w.r.t. x.

5. If y = sin (log x), find2

2

d y.

dx

6. If y = ex(sin x + cos x), prove that

2

2

d y dy 2

dx dx

+ 2y =

0.

7. If y = sin1

x, prove that ( )2

2

2

d y dy1 x x 0.

dxdx=

8. If y = 3e2x

+ 2e3x

, prove that2

2

d y dy 5 6y 0.

dxdx+ =

Topics for Worksheet VI

Mean value theorem

Worksheet VI

1. Verify Rolles theorem for the function f(x) = x36x

2+ 11x 6 in the interval [1, 3].

2. Verify Rolles theorem for the function f(x) = x(x 1)

2in the interval [0, 1].

3. Verify Rolles theorem for each of the follfunctions:

(i) ( )f x sin2xin 0,2

=

(ii) ( ) ( )f x sinx cosx in 0,2

= +

(iii) ( )f x cos2 x in 0,4 2

=

(iv) ( ) ( ) [ ]f x sinxsin2x in 0, .= 4. Verify Rolles theorem for each of the fol

functions:

(i) f(x) = sin2x in 0 x

(ii) ( ) x xf x e cos i n x2 2

=

(iii) ( ) xsinx

f x i n0 xe

=

5. Verify Rolles theorem for the function f(x) =3) e

(x/2)in [3, 0].

6. Verify Rolles theorem for the function f(x) (x

2+ 2) log 3} on [1, 1].

7. Verify Rolles theorem for the function f(x) =x

24x 2.

8. If Rolles theorem holds for the function f(x)

bx2+ ax + 5 on [1, 3] with

1c 2 ,

3

= +

fi

values of a and b.

CURRICULUM BASED CHAPTER ASSIGNM

2 Marks Questions

Discuss the continuity of the following func

at the indicated points.1. Prove that the function f defined by f(x) = |x

R is differentiable.2. Prove that the greatest integer function defin

f(x) = [x], 0 < x >3 is not differentiable at x =

x = 2.

3. Show that the function ( )x 1, x

f x2x 3, x


10/12



3 Marks Questions

Discuss the continuity of the following functions

at the indicated points

10. ( )

4x 16, x 2

f x at x 2.x 2

16, x 2

= = =

11.( )

x 1, x 1f x at x 1.

2x, x 1

+

= =


11/12



2. If f be a function such that f (9) = 9, f (9) = 3, then

( )x 9

f x 3Lt

x 3is equal to

(a) 3 (b) 5

(c) 4 (d) 2.

3. Let f (x) = x |x| then f (0) is equal to

(a) 1 (b) 1(c) 0 (d) none of these.

4. If ( ) =+ + >

2

2

ax b b 0,x 1f xbx ax c x 1

then f(x) is

continuous and differentiable at x = 1 relation a

and b.

(a) a = 2b and c = 0 (b) a = 3b and c = 1(c) a = 4b and c = 1 (d) none of these.

5. if f(x) = exsin x in [0, ], then c in Rolles theorem

is

(a) [ ]0,4

(b) [ ]

30,

4

(c) [ ]

2

0,4

(d) none of these.

6. Let ( ) +

=+ > 2

x a, x 1f x

ax 1, x 1and f is continuous at 1,

then the value of a so that f is derivable at 1, is

(a)1

2 (b)

1

4

(c)1

3 (d) none of these.

7. If a > 0, a 1, then logan (x) =

(a)a

nlog x (b) x1

log a

n

(c) a1

log xn

(d) none of these.

8. If f (x)[ ] [ ]x x , x 0

, x 0

+ = =

then f is continuous at x

= 0 find value of .(a) 2 (b) 0

(c) 1 (d) none of these.

9. Let ( )

= =

1xs in , x 0

f x x

k , x 0

then f is continuous at

x = 0 find k.

(a) 0 (b) 1


10.

= + 1

2

d 2xsin

dx 1 x

(a)1

1 x2+ (b)

3

1 x2+

(c)2

1 x2+ (d) none of these.

11. If +

= + +

1 11 x 1 xy sin sec1 x 1 x

thendy

dx

is,

(a) 0 (b) 1

(c) 2 (d) 3.

12. If the function ( ) ( )sinxx e 1

f x1cosx

= is continu

0, then f(0) =

(a) 0 (b) 1


13. ( )+ + =2d

log x x 1dx

(a)21

x 1+ (b)

22

x 1+

(c) 2x 1+ (d) none of these.

14. ( )

= + <


12/12



19. Differentiatex

yx 1

=+

with respect to x.

(a) ( )2

x 1+ (b)( )

2

2

x 1

(c)( )

2

1

x 1+ (d) none of these.

20. Examine the continuity of function f given by f(x) =

2x + 5 at x = 1.(a) 0 (b) 1


21. If f and g be two real functions continuous at a

real number c then f + g is continuous at x = c.

(a) a (b) b

(c) c (d) none of these.

22. Find the point of discontinuity of f where f is

defined by ( ) +

=>

2x 3,if x 2f x

2 x 3 i f x 2

(a) 0 (b) 1

(c) 2 (d) none of these.23. Differentiate cos (sin x) with respect to x.

(a) ( )sin sinx co sx (b) ( )sin sinx co sx

(c) ( )sin sinx cosx (d) none of these.

24. Differentiate y = e2x

with respect to x.

(a) 2x2e (b) 2x2e

(c) 2e (d) none of these.

25. Differentiate y = sin (xy) with respect to x.

(a)( )

( )

ycos xy

1xcos xy (b)

( )

( )

ycos xy

1 xcos xy+

(c) ( )1xcos xy (d) none of these.

26. If y = 5 cos x 3 sin x, prove that + =2

2

d yy 0.

dx

(a) 0 (b) 1

(c) 2 (b) 4.

27. If the function ( ) ( ) =

=

1 /xcosx x 0

f xk x 0

is

continuous at x = 0 then the value of k is

(a) 0 (b) 1

(c) 2 (d) 4.

28. If f(x) = |cos 2x| then + f 0

4is equal to.

(a) 0 (b) 1

(c) 2 (b) none of these.

29. If ( ) ( )( )= + + =1/4 1/2 1/4dy

y 1 x 1 x 1 x ,dx

(a) 1 (b) 1(c) 2 (d) none of these.

30. If( ) }xx

2 3

e 1 x 2 logxe logx dyy ,

dxx x

+= =

(a)( ){x

3

e 1 x 2 logx

x

+(b)

( )x

3

e 1 x 2 lo

x

(c)( ){x

3

e 1 x 2 logx

x

+(d) none of these.

31. Find2

2

d y

dxif y = sin

1x

(a)

( )3

2 2

x

1 x

(b)

( )3

2 2

x

1 x

(c) ( )3

2 21 x (d) none of these.

32. Find

=

31

2

dy 3 x xif y tan .

dx 13x

(a)2

3

1 x (b)

2

4

1 x+

(c)2

3

1 x+ (d) none of these.

33. If ( )

= =

kcosxif x

2 x 2f x

3 if x2

find the value k

continuous at

=x .2

(a) 2 (b) 4

(c) 6 (d) 8.

34. Differentiate: = + + +y logx logx log .......

(a)( )

1

x 2 y 1 (b)

( )

2

x 2 y 1

(c)( )

3

x 2 y 1 (d) none of these.

35. If x 1 y y 1 x 0+ + + = for 1 < x < 1, prov

( )2

dy 1

dx 1 x=

+

(a) ( )2x 1+ (b)( )

21x 1

+

(c)( )

2

1

x 1+ (d) none of these.

5_continuity_and_differentiability.pdf

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