continuity and differ en ti ability

8/2/2019 Continuity and Differ en Ti Ability

1/40

CONTINUITY AND DIFFERENTIABILITY

1

CONTINUITY

1 Definition

A function f(x) is said to be continuous at x = a , if axLim

f (a) = axLim

f (x) = f(a)

i.e. LHL = RHL = value of the function at a i.e.ax

Lim

f (x) = f (a) .

If f (x) is not continuous at x = a, we say that f (x) is discontinuous at x = a .

NOTES:

All Polynomials, Trigonometrical functions, exponential and Logarithmic functionsare continuous in their domain .

We never talk about continuity/discontinuity at a points at which we cantapproach from either side of the point. These points are called isolated points .

e.g. f(x) = x + x at x = .

2 Single point of Continuity :

There are some functions which are continuous only at one point.

e.g. f (x) =

Qxifx

Qxifxand g (x) =

Qxif0

Qxifxare both continuous only at x =

0

3 Reasons of Discontinuity

(a) One or more than one of the three quantities , LHL , RHL and f (a) is not defined . Lets

consider some examples

(i) f (x) =x

1around x = 0

LHL = , RHL = + , f (0) is not defined ,

f (x) =x

1is discontinuous at x = 0 which is

obvious from the graph .

(ii) f (x) =

1xfor1x

1x2

around x = 1

LHL = RHL = 2 but f (x) is not defined . Therefore this function graph has a hole at x

= 1 , it is discontinuous at x = 1

(b) All the three quantities are defined , but any pair of form is unequal (or all three are unequal).

(i) f (x) = [x] around any integer I

LHL = I 1 , RHL = I , f (I) = 1

LHL RHL = f (I) , so this frunction is also discontinuous at all integers

(ii) f (x) = {x} around any integer I

LHL = 1 , RHL = 0 , f (I) = 0

LHL RHL = f (I) , so this frunction is also discontinuous at all integers


2/40

DIFFERENTIAL CALCULUS

2

(iii) f (x) =

Zx,0

Zx,1around any integer I

From the figure , we notic that at nay integer I ,

LHL = 1 , RHL = 1 , f (I) = 0

LHL =RHL f (I) , so this frunction is again discontinuous

(iv) f (x) =

0x,0

0x,xx

around x = 0

At x = 0 we see that

LHL = 1 , RHL = 1 , f (0) = 0

LHL RHL f (0) and the function is discontinuous

To summarise , if we intend to evaluate the continuity of a function a x = a which means thatwe want to determine whether f (x) will be continuous at x = a or not , we have to evaluate all

the three quantities , LHL , RHL and f (a) . If these three quantities are finite and equal , f (x)

is continuous at x = a . In all other cases it is discontinuous at x = a

LHL (at x = a) = RHL , (at x = a) = f (a) , for continuity at x = a .

ILLUSTRATION

Discuss the continuity of the function ,

f (x) at x =2

1, where f (x) =

1x2

1

,x2

321x,1

21x0,x

21

Solution :

We have :

2

1xatLHL =

21x

Limf (x) =

21x

Lim

x2

1

2

1x0forx

2

1)x(f

=2

1

2

1= 0 [ Using direct substitution method ]

and

2

1xatRHL =

2

1x

Limf (x) =

21x

Lim

x2

3

1x2

1forx

2

3)x(f

=2

3

2

1= 1 [ Using direct substitution method ]

21x

Lim

21x

Lim

Hence f (x) is not continuous at x =2

1. Clearly f (x) is discontinuious at x =

2

1


3/40


3

ILLUSTRATION

Discuss the continuity of the function , f (x) at x = 2 .

f (x) =

2x,x2

2x,x2

Solution :

We have : (LHL at x = 2) = 2xLim

f (x) = 2xLim

2 x . [ f (x) = 2 x for x < 2 ]

= 2 2 = 0

and (LHL at x = 2) = 2xLim

f (x) = 2xLim

2 + x . [ f (x) = 2 + x for x 2 ]

= 2 + 2 = 4

2xLim 2x

Limf (x) . Hence f (x) is not continuous at x = 2 .

Conceptual Exercise 01

Test the continuity of the following functions .

1. f (x) =

0x,10x,x

x

at x = 0

2. f (x) =

0x,0

0x,x1sinx

at x = 0

3. f (x) =

0x,0

0x,1e

1ex/1

x/1

at x = 0

4. f (x) =

2x1,x3x4

1x0,4x5

3 at x = 1

5. f (x) =

1x21,x1

21x,

21

21x0,x

at x =2

1

6. f (x) =

0x,2

0x,xcosx

xsinat x = 0

3 Continuity In An Open IntervalA function f(x) is said to be continuous in an open interval (a, b) if it is continuous at each

and every point of (a, b) i.e. y = {x} is continuous in (1, 2)

4 Continuity In A Closed Interval

A function f(x) is said to be continuous in a closed interval [a , b] if

It is continuous in (a, b)

Value of the function at b is equal to left hand limit at b i.e., f (b) = bxLim f (x)

Value of the function at a is equal to right hand limit at a i.e., f (a) = axLim

f(x)


4/40


4

illustration

Show that the function , f (x) =

1x0,2x

0x2,1x

2x3,3x2

is discontinuous at x = 0 and

continuous at every point in interval [ 3 , 1 ] .

Solution :

f (x) =

1x0,2x

0x2,1x

2x3,3x2

is plotted as shown .

Here , if we observe in the graph , we could conclude that at x = 0 ,

0xLim

f (x) = 1 and 0xLim

f (x) = 2

which shows that the function is discontinuous at x = 0 and

continuous at every other point in [ 3 , 1] .

5 Geometrical Meaning of Continuity

A function f(x) will be continuous at x = a if and only if there is no break in the graph of the

function f(x) at the point (a, f(a)). In an interval function is said to be continuous if the is no

break in graph of function in the entire interval.

Examples :f (x) = sin x is continuous in its entire domain

0

y

x23 2

f (x) = tan x is discontinuous at x = (2n + 1)2

, where n I.

2

32

O

2

2

3

x

y

f(x) will be discontinuous at x = a, in any of the following cases:

(i) axLim f (x) and axLim f (x) exist but are not equal

For example y = [x] at x I is discontinuous.

1

2

1

1

2

2

1

1

2

2 3x

y

O

x

y


5/40


5

(ii) axLim f (x) = ax

Lim f (x) but not equal to f (a)

5

x = 2For example f (x) =

2x,5

2x,2x4x2

at x = 2

Here LHL = 2 and RHL = 2 , but at x = 2 , f (x) = 5 .

(iii) Atleast one of the limits

(L.H.L. or R.H.L.) does not exist .

For example :

y = sin

x

1at x = 0

illustration

If f (x) =

0x,3x

0x,0

0x,3x2

2. Discuss the continuity .

Solution :

Here 0xLim

f (x) = 3

0x

Limf (x) = 3

f (0) = 0

Thus 0xLim

f (x) = 0xLim

f (x) = 3 f (0)

Hence f (x) is discontinuous at x = 0

illustration

If f (x) =1x

1x2

. Discuss the continuity at x 1 .

Solution :

Which shows , 1xLim

f (x) = 2 but f (1) is not defined .

1x

1x2

So f (x) is discontinuous at x = 1 .


1. Find all possible values of a and b so that f (x) is continuous for all x R if


6/40


6

f (x) =

x,3xcos

x0,b2xx2sinb

0x1,ax3

1x,3xa

2

2. Let f (x) =

0x,x2tan1n

1e

0x,1x1

xcosn

x4sin

24

Is it possible to define f (0) to make the function continuous at x = 0 . If yes what

is the value of f (0) , if not then indicate the nature of discontinuity .

3. The function f (x) =

xifxcos1

xif2b

x0if56

2

2

2

b

xtana

x5tan

x6tan

Determine the values of a and b , if f is continuous at x =2

4. Determine the value of a , b and c for which the function is continuous at x = 0 .

f (x) =

0xfor

0xforc

0xfor

2/3

2/12/12

xb

xxbx

x

xsinx)1a(sin

5. Find the locus of (a , b) for which the function f (x) =

2xforaxb

2x1forx3

1xforbxa

2

is continuous at x = 1 but discontinuous at x = 2 .

6 Algebra of Continuous Functions

Let f (x) and g (x) are continuous functions at x = a. Then,

(i) f (x) is continuous at x = a where c is any constant

(ii) f (x) g (x) is continuous at x = a

(iii) f (x). g (x) is continuous at x = a

(iv))x(g

)x(fis continuous at x = a, provided g (a) 0 .

illustration

f (x) =

0x,k

0x,x4

tanx/1

. For what value of k , f (x) is continuous at x = 0 ?

Solution :


7/40


7

Here , 0xLim

f (x) = 0xLim

x/1

x4

tan

0xLim

f (x) = 0xLim

x/1

xtan1

xtan1

( 1 form )

0xLim

f (x) = 0xLim

x/1

1xtan1

xtan11

0xLim

f (x) =x

1

xtan1

xtan2Lim

0xe

0xLim

f (x) =)xtan1(x

xtanLim2

0xe

=e2

Here , f (x) is continuous at x = 0 , when 0xLim

f (x) = f (0) k = e2

NOTES:

If f(x) is discontinuous and g(x) is also discontinuous at x = a , then the sum of thefunctions is continuous.

Example : f (x) = [x] and g (x) = {x} , then (f + g) (x) = x is a continuous function.

If f(x) is continuous and g(x) is discontinuous at x = a then the product function

)x(g).x(f)x( is not necessarily be discontinuous at x = a.

Example : f(x) = x and g (x) =

0x,0

0x,x1sin

(x) = 0 for all x R

If f(x) and g(x) both are discontinuous at x = a then the product function

)x(g).x(f)x( is not necessarily be discontinuous at x = a.

Example : f (x) =

0x,1

0x,1and g (x) =

0x,1

0x,1

Continuity of an inverse Function : If the function y = f(x) is defined, continuous andstrictly monotonic on the interval X, then there exist a single valued inverse function

x = (y) defined, continuous and also strictly monotonic in the range of the function

y = f (x).


1. Find the value of f (0) so that the function f (x) =x

x1x1 3 is continuous at x = 0.


8/40


8

2. If the function f (x) =x

)xb1(n)xa1(n is undefined at x = 0. Then find the value

which should be assigned to f at x = 0 so that it is continuous at x = 0.

3. If f (x) =)x2(sin

4x2 , x 0 is continuous function at x = 0, then find the value of f(0)

4. If f(x) = x + { x} + [x] , where [x] is the integral part and {x} is the fractionalpart of x . Discuss the continuity of f in [ 2 , 2] .

5. Discuss the continuity of the function; f (x) = [[x]] [x 1] (where [.] denotes thegreatest integral function).

6. Examine the continuity or discontinuity of the function f (x) = [x] + [x].

7 Continuity of Composite Function

If f is continuous at x = c and g is continuous at x = f(c) then the composite g(f(x)) is

continuous at x = c. e.g. f(x) =2x

xsinx2

and g(x) = x are continuous at x = 0, hence

the composite (gof)(x) =2x

xsinx2

will also be continuous at x = 0 .

illustration

Find the point(s) of discontinuity of y =2uu

12

, where u =1x

1

.

Solution:

The function u = f (x) =1x

1

is discontinuous at the point x = 1. . . . (i)

The function y = g(x) =2uu

12

=)()( 1u2u

1

is discontinuous at u = 2 and u = 1.

when u = 2 ,1x

1

= 2 x =

2

1

u = 1 1x

1

= 1 x = 2

Hence, the composite function y = g (f (x)) is discontinuous at three points x =2

1, 1, 2


1. Draw the graph of the function f (x) = x x x2 , 1 x 1 and discuss itscontinuity or discontinuity at f in the interval 1 x 1 .

2. Given the function g (x) = x26 and h (x) = 2 x2 3 x + a . Then

(a) Evaluate h (g (2))


9/40


9

(b) If f (x) =

1x,)x(h

1x,)x(g, find a so that f is continuous .

3. Let f (x) =

3x2,x3

2x0,x1. Determine the form of g(x) = f (f(x)) and hence

find the point of discontinuity of g , if any .

4. If f (x) = 1 + x 1, 1 x 3 and g(x) = 2 x + 1, 2 x 2 .then calculate f(g(x)) and g(f(x)). Discuss the continuity of f(g(x)).

5. Let f (x) =

1x0,x

0x1,1x2 and g (x) = sin x . Further let

h(x) = f(g(x)) + f (g(x)). Discuss the continuity of h(x) in [1, 1].

8 Intermediate Value Theorem

Suppose f (x) is continuous on an interval I, and a and b are two points of I. Then if y0

is a

number between f (a) and f (b), there exists a number c between a and b such thatf(c) = y

0.

illustration

Let f : [0 , 1] [1 , e] be a continuous function, then prove that f(c) = ec for some c [0 , 1]Solution :

Let h(x) = f(x) ex

h(0) = f(0) 1 0h (1) = f (1) e 0h(c) = 0 for at least one c [0 , 1]

illustration

Let f , g : [0 , 1] [0 , ) be continuous function satisfying 1x0.Max

f (x) = 1x0.Max

g (x) .

Prove that there exists [0 , 1] with 2)(f + 3 f () = 2)(g + 3 g () .Solution :

Let 1x0.Max

f (x) at x = , [0 , 1] and 1x0.Max

g (x) at x = , [0 , 1]

Now consider , h (x) = f (x) g(x) x [ , ]h () = f () g () 0h () = f () g () 0

h () = 0 for atleast one [0 , 1] f () g () = 0

2)(f + 3 f () = 2)(g + 3 g () for atleast one [0 , 1]

illustration

f : (0 , 1) [0 , 1] be a continuous function .(A) if it is onto then it must be many to one (B) f must be onto

(C) f must be oneone (D) f can be bijective

Solution :Let f to be one to one then range of f will be open and as it cant be onto . Hence if it is

onto then it must be many to one .


10/40


10

NOTES:

If f is a continuous function in [a, b] and is any real number such thatf(a) < < f(b), then there exists at least one solution of the equation f(x) = inthe open interval (a, b).

In general odd number of solution of f(x) = in the open interval (a, b). Inparticular if f(a) and f(b) possess opposite signs, then there exists at least one

solution of the equation f(x) = 0 in the open interval (a, b). In general odd number

of solutions of f(x) = 0 in the open interval (a, b). If f is continuous at every point of a closed interval I, then f assumes both an

absolute maximum value M and an absolute minimum value m somewhere in I.

That is there are numbers x1

and x2

in I with f(x1) = m, f(x

2) = M, and

M)x(fm for every other I.

In other words if m =bxa

min f(x), M =bxa

max f(x), then for any A satisfying the inequalities

m A M there exist a point x0

[a, b] for which f(x0) = A.

A continuous function whose domain is closed must have a range also in closedinterval but it is not necessary that domain is open then range is open (range can

be closed). f(x) has the minimum and maximum values on [a, b].


1. Let f : [1 , e] [0 , 1] is a continuous but need not be differentiable function thenprove that f (x) = n x has atleast one solution in [1, e] .

2. Let f : (1 , 10) [2 , 11] be a continuous function , but need not be differentiable,

then the correct statement is :

(A) it can be an odd function (B) it cant be an invertible function

(C) it can be an invertible function (D) none of these

3. Let f be a continuous function on R and periodic with fundamental period 1 i.e.

f (x + 1) = f (x), then prove that there will be a real number x0, such thatf(x

0+ ) = f(x

0) .

4. Let f : [0 , 1] R be a continuous function such that f(0) = f(1), then prove that

there is a solution of the equation f (x) f

n

1x = 0 , in

n

1n,0 for every

natural number n

5. Let f : [ 1 , 1] [ 1 , 1] be a continuous function, then prove thatf (c) = c3 for some c [ 1 , 1]

.


11/40


11

DIFFERENTIABILITY

1. Definition of The Derivative

The derivative f (x) of a function y = f (x) at a given point x is defined as

f (x) = 0xLim

x

y

= 0xLim

x

)x(f)xx(f

= finite . If this limit exists finitely then the

function f (x) is called differentiable at the point x . The number f (x)

finitex

)x(f)xx(fLim

0xis called the left hand derivative at the point x .

Similarly the number f+ (x)

finitex

)x(f)xx(fLim

0xis called the right hand

derivative at the point x . The necessary and sufficient condition for the existence of the

derivative f (x) is the existence of the finite right and left hand derivatives , and also of the

equality

f+ (x) = f

(x) = finite .

2. Geometrical Meaning of The Derivat iveLet us consider the function f(x) and the corresponding curve y = f(x). Clearly line joining

two points M0(x , y) and M

1(x + x , y + y) on the curve will be the secant to the curve

and the slope of this secant is given by tan = xy

(Where is the angle made by the

secant with the positive direction of the xaxis). In the limiting case when x 0 thepoint M

1approaches M

0and the secant joining these two points will become the tangent

at M0

whose slope will be given by tan = 0xLim

x

y

= f (x)

which means that slope of the tangent to the curve y = f (x) at

any argument is equal to the value of the derivative at that

argument.

O x x+x

y

yxy

x

y=f(x)M1

M0

Geometrically, a function is not differentiable in the following cases :

3 Differentiability on An Interval

A function y = f (x) is differentiable on an interval (finite or infinite) if it has a derivative at

each point of the interval . It is differentiable on a closed interval [a , b] if it differentiableat every point of the open interval (a, b) and if the limits

0hLim

h

)a(f)ha(f = finite (Right hand derivative at a)

0hLim h)hb(f)b(f = finite (Left hand derivative at b) exist finitely

4 Relation Between Derivability And Continuity

(a) If )a(f exists then f(x) is derivable at )x(fax is continuous at x = a .Ingeneral a function f is derivable at x then f is continuous at x. i.e. if f(x) is derivable for

every point of its domain of definition, then it is continuous in that domain. The converse

of the above


12/40


12

result is need not be true e.g. the functions f (x) = |x| and g (x) =

0x,0

0x,x1sinx

both are continuous at x = 0 but not derivable at x = 0.

(b) Let f+

(a) = and f

(a) = where and are finite then :(i) = f is derivable at x = a f is continuous at x = a.(ii) f is not derivable at x = a but f is continuous at x = a. If a

function f not differentiable but is continuous at x = a, it geometricallyimplies a sharp corner or kink at x = a .

(iii) If f is not continuous at x = a then it is not differentiable at x = a.

5. Reason of non differentiability

Case I(i) a corner, where the onesided derivatives differ

(ii) a cusp, where the slope of PQ approaches from one side and from the other

(iii) a vertical tangent, where the slope of PQ approaches from both sides or

a p p r o a c h e s from both sides (here, )

Q +

Q

P

(iv) a discontinuity

(a) (b)

Remarks :


13/40


13

There are functions which are continuous at every point but differentiable at no point.

e.g. f (x) =

0n

n

3

2

cos (9n x) .

A formula that expresses f as an infinite sum of cosines with increasing higher frequencies.

By adding wiggles to wiggles infinitely many times, so to speak, the formula produces a

graph that is too bumpy in the limit to have a tangent anywhere.

6. Algebra of a Differentiable Function

(i) If f(x) and g(x) are derivable at x = a then the functions f(x) + g(x), f(x) g(x),f(x).g(x) will be derivable at x = a and if g (a) 0 , then the function f(x)/g(x) will alsobe derivable at x = a.

(ii) If f(x) is differentiable at x = a and g(x) is not differentiable at x = a, then the product

function f(x) g(x) can still be differentiable at x = a. e.g., f(x) = x and g(x) = |x| at

x = 0.

(iii) If f(x) is differentiable at x = a and g(x) is not differentiable at x = a, then the sum

function f(x) + g(x) is not differentiable at x = a.

(iv) If f(x) and g(x) both are not differentiable at x = a, then the sum function may be adifferentiable function. e.g. f(x)= x and g(x) = x

(v) If f(x) and g(x) both are not differentiable at x = a, then the product function f(x) g(x)

can still be differentiable at x = a i.e., f (x) = x and g(x) = x at x = 0 .(vi) If f(x) is derivable at x = a then it need not be true that )x(f is continuous at x = a.

e.g. f (x) =

0xif0

0xifx1sinx2

.


1. If f (x) is differentiable at x = a , find axLim

ax

)x(fa)a(fx 22

.

2. Discuss the differentiability of f (x) =

0x,0

0x,exx1

x1

at x = 0 .

3. Show that the function f (x) =

0xwhen0

0xwhensinxx1

is continuous but not differ

entiable at x = 0 .

4. A function f is defined by f (x2) = x3 for all x > 0 . Show that f is differentiable at 4.

5. Find whether the function f (x) = x3 is differentiable or not .

6. Let f (x) =

1xforbxa

1xforx1

2. If f (x) is continuous and differentiable every

where, then find the value of a and b .


14/40


14

Functional Equation :

If x, y are independent variables, then :

(i) f(xy) = f(x) + f(y) f (x) = k ln x or f(x) = 0 .

(ii) f(xy) = f(x) . f(y) f (x) = xn , n R

(iii) f(x + y) = f(x) . f(y) f (x) = akx .

(iv) f(x + y) = f(x) + f(y) f (x) = kx, where k is a constant .

illustration

(a) Let f be a function such that f(x + f (y)) = f (x) + y x , y R , then find f (0).

(b) Now if it is given that there exists a positive real , such that f (h) = h for 0 < h < then find f(x) and hence f(x).

Solution :(a) Let x = 0, y = 0 in f (x) + f (y) = f (x) + y

f (0) + f (0) = f (0) + 0 2 f (0) = f (0) f (0) = 0

(b) Given f (h) = h

then f (x) = 0hLim

h

)x(f)hx(f for 0 < h <

f(x) = 0hLim

h

)x(f)h(fxf )( (given f (h) = h)

f(x) = 0hLim

h

)x(fh)x(f f(x) = 0hLim

h

h

f(x) = 1 , integrating both sides we get f (x) = x + c where f (0) = 0 c = 0

So, f (x) = xThus f (x) = 1 and f (x) = x

illustration

Let f : R R is a function satisfies conditionf (x + y3) = f (x) + [f (y)]3 for all x , y R. If f(0) 0 . Find f (10) .

Solution :Given f (x + y3) = f (x) + [f (y)]3 ................. (i)

and f (0) 0 ................. (ii)Replacing x , y by 0

f (0) = f (0) + f(0)

3

f (0) = 0 ................. (iii)also f (0) = 0h

Lim h

)0(f)h0(f = 0h

Lim h

)h(f

Let I = f (0) = 0hLim

33/1

33/1

)(

)(

h

)0(fh0f

= 0hLim

33/1

33/1

)(

)(h

hf= 0h

Lim

3

3/1

3/1

)(

)(

h

hf

= I3


15/40


15

I = I3

or I = 0 , 1 , 1 as f (0) 0 f(0) = 0 , 1 ................ (v)

Thus, f (x) = 0hLim

h

)x(f)hx(f = 0h

Lim 33/1

33/1

)(

)(

h

)x(fhxf

f(x) = 0hLim

33/1

33/1

)(

)(

h

)x(fhf)x(f [ using (i)]

f(x) = 0hLim

)()(

3/1

3/1

h

hf= (f (0))3

f(x) = 0 , 1 [as f (0) = 0, 1 using (v)]Integrating both sides,

f(x) = c or x + c as f(0) = 0 f(x) = 0 or xThus f(10) = 0 or 10


1. Let f

2

yx=

2

)y(f)x(f for all real x and y. If f (0) exists and equals to

1 and f (0) = 1 , find f (x) .

2. I f f (x) + f (y) = f

yx1

yxfor all x , y R (xy 1) and 0x

Lim x

)x(f= 2 .

Find f

3

1and f (1)

3. Let f (x + y) = f(x) + f(y) 2xy 1 for all x and y . If f (0) exists andf (0) = sin , then find f{ f (0)} .

4. If f

3

yx=

3

)y(f)x(f2 for all real x and y and f (2) = 2 , then determine

y = f (x) .

5. Let f

2

yx=

2

)y(f)x(ffor all x and y . If f (1) = f (1) , show that

f(x) + f(1 x) = constant for all nonzero real x.

6. If f

y

x=

)y(f

)x(f x, y R, y 0 and f(t) 0, if t 0 and f(1) = 3 then find f(x).

Maxima/Minima/Middle function :

A function f (x) = max { } where max means the expression whose values lies above of

other expressions .


16/40


16

A function f (x) = min { } where min means the expression whose values lies below of other

expressions .

A function f (x) = mid { } where mid means the expression whose values lies between those

of other two expressions .

illustration

Let f(x) = min.{tan x, cot x} x R. Find(i) Range of f(x) (ii) Period (if periodic)(iii) Points of discontinuity of f(x) (iv) Points of nondifferentiability of f(x)

Solution :

x = 3 /2x = - x = x = - x =

O X

We know,

f(x) = min{tan x, cot x} can be plotted in two steps

(i) We should plot the graph of tan x and cot x

(ii) We should find their points of intersection

and neglect the area above their point of

intersection

It can seen from the graph. That,

(a) range of f(x) = (, 1] [0, 1](b) period of f(x) =

(c) Points of discontinuity are , 2

, 0 ... which can be put in the form of

2

n, nI

(d) Also the points of non differentiability are ,4

3 ,

2

, 0 , ... which can be put in

form of4

n, n I.

Here darked lines of the curves represents min {tan x , cot x} and dotted lines is max. {tan

x , cot x}

illustration

A function f is given by f (x) = mid {1 x , x , 2 x + 1} where mid means the expressionwhose values lies between those of other two expressions . Find the function and test its

continuity in entire number scale .

Solution :

y = 2 x + 1

y = x

y = 1 x

11

12

13

The bold lines denotes the mid of the function


17/40


17

Hence f (x) = mid {1 x , x , 2 x + 1} =

x21,x

21x0,x1

0x31,1x2

31x,x


1. On how many points the functions ,

f (x) = max { x2 , (x 1)2 , 2 x (1 x)} ,0 x 1 , is not differentiable.

2. Let f (x) = sin x and g (x) =

xfor2

xcos1x0forxt0,)t(f.max

Discuss the continuity and differentiability of g (x) in (0 , ) .

3. Let f (x) = x4 8 x3 + 22 x2 24 x and g (x) =

1x,10x

1x1,1xtx,)t(fmin

Discuss the continuity and differentiability of g (x) in [ 1 , ) .

4. Let f (x) = 1 + 4x x2 , x R , g (x) = max. { f (t) ; x t (x + 1) ; 0 x < 3 }

= min. {(x + 3) ; 3 x 5} . Verify continuity of g (x) for all x [0 , 5] .


18/40


18

ANSWERS


1. f (x) is not continuous 2. f (x) is continuous 3. f (x) is not continuous

4. f (x) is continuous 5. f (x) is continuous 6. f (x) is continuous


1. a = 0 , b = 1 2. f (0+) = 2 ; f (0) = 2 , hence f (0) are not possible to define

3. a = 0 , b = 1 4. a = 2

3, b 0 , c =

2

1

5. Locus (a , b) x , y is y = x 3 excluding the points where y = 3 intersects it .


1.6

12. a + b 3.

8

14. Discontinuous at all integral values in

[2, 2]

5. Continuous on R 6. f (x) is discontinuous at x I

Conceptual Exercise 041. f is continuous at 1 x 1 2. (a) 4 3 2 + a (b) a = 3

3. g (x) =

3x2,x4

2x1,x2

1x0,2x

and g(x) is discontinuous at x = 1 and x = 2

4. Continuous everywhere 5. h (x) is not continuous at x = 0


2. B


1. 2 a f (a) a2 f (a) 2. Not differentiable at x = 0 5. differentiable every-where

6. a = 2

1, b =

2

3


1. f(x) = 1 2. f(1) = 1, f1

33

3. f { f (0)} = 1

4. f (x) = 2 x + 2 6. f (x) = x3


1. 2 2. continuous and differentiable for all x (0 , )

3. g (x) is cont. in [ 1 , ) but not diff. at x = 1 4. continuous for all x [0 , 5] except x = 3


19/40

DIFFERENTIATION

19

DIFFERENTIATION

Methods of Differentiation

1 Derivative of f(x) From The First Principle/ab Initio Method

If f (x) is a derivable function then,

0xLim

x

y

= 0xLim

x

)x(f)xx(f

= f (x) =xd

yd

or simply f (x) = 0hLim

h

)x(f)hx(f

illustration

Using the definition of the derivative, find the derivative of the function cos ax at x.

Solution :

Let y = cos a x , we have

y = cos a (x + x) cos ax = 2 sin

x2

axa sin

2

xa

x

y

=

x

2xa

sinx2axasin2

Hence y = 0xLim

x

y

= 2 0xLim

sin

x2

axa

0xLim

x

xxasin

= a sin ax .

In particular, if a = 1, then y = cos x and y = sin x

illustration

Find the derivative of tan1 x with respect to x by using first principle.

Solution :

Let tan1 x = ,

2,

2 x = tan

and tan1 (x + h) = + . . . (i)

x + h = tan ( + ) . . . (ii)

Let 0xLim

h

xtan)hx(tan 11 = L

L = 0xLim

h

= 0x

Lim h

from (i) and (ii)

= 0hLim

tan)(tan= 0

Lim

tan)(tan

= 0Lim

sin

cos)(cos= cos2 =

2sec

1= 2x1

1


20/40


20

2 Algebra of Derivatives

If u and v are derivable function of x , then :

(i)xd

d(K u) = K

xd

ud, where K is any constant

(ii)xd

d(u v) =

xd

ud

xd

vd

(iii) Product Rule :xd

d(u . v) = u

xd

vd v

xd

ud

(iv) Quotient Rule :xd

d

v

u=

2v

xdvd

uxdud

v

where v 0

(v) Chain Rule : If y = f (u) and u = g (x) thenxd

yd=

ud

yd.

xd

ud

illustration

If y = x1/2 + log5x + xcos

xsin+ 2x , find xd

yd?

Solution :Here y = x1/2 + log

5x + tanx + 2x on differentiating w.r.t. x we get,

xd

yd=

xd

d(x)1/2 +

xd

d(log

5x) +

xd

dtan x +

xd

d(2x)

= 2

1(x)1/2 1 +

5logx

1

e+ sec2 x + 2x n 2 =

2

1(x)3/2 +

5logx

1

e

+ sec2 x + 2x n 2

I llustration 43 :

Differentiate: y =12x1tan

esinn2

.

Solution :

y =12x1tan

esinn2

xd

yd=

12x1tanesinn2

n 2 .

1xtan 21esin

1cos

1xtan 21e

. 1xtan21

e .

1x112 1x2

1

2 . 2 x

=12x1tan

esinn2

.

n 2 .

1xtan 21esin

1cos

1xtan 21e . 1xtan

21

e .


21/40

DIFFERENTIATION

21

1xx

1

2

3 Some Standard Formulae of Differentiat ion

xd

d(constant) = 0

xd

dxn = nxn1

xdd ax = ax n a

xdd ex = e x

xd

d(log

a|x| ) =

x

1log

ae

xd

d n x =

x

1

xd

dsinx = cosx

xd

dcos x = sin x

xd

dtan x = sec2x

xd

dsec x = sec x tan x

xd

dcot x = cosec2x

xd

dcosec x = cosec x cot x


Differentiate the following functions w.r.t. x .

1. ex log x tan x 2. log sin x2

3. (x2 + x + 1)4 4. log (sec x + tan x)

5. If y = log

x

1x , prove that

xd

yd=

)1x(x2

1x

6. If y = x

x

e1

e1

show that

xd

yd=

x2xx

e1e1

e

4 Inverse Function And Their Derivat ives

Theorem :

If the inverse functions f and g are defined by y = f (x) and x = g (y) and if f (x) exists and

f (x) 0 then g (y) = )x(f1

. This result can also be written as , if xdyd

exists and xd

yd

0

,

thenxd

yd= xdyd

1or

xd

yd.

yd

xd= 1 or

xd

yd= xdyd

1where

0

yd

xd

xd

dsin1 x =

2x1

1

, 1 < x < 1

xd

dcos1 x =

2x1

1

, 1 < x < 1


22/40


22

xd

dtan1 x = 2x1

1

, x R

xd

dcot1 x = 2x1

1

, x R

xd

dsec1 x =

1xx

1

2 , x > 1

xd

dcosec1 x =

1xx

1

2 , x > 1

5 Differentiation of A Function Defined Parametrically

Let x and y be the functions of parameter t , i.e., x = f (t) , y = (t) , then

xd

yd=

tdxd

tdyd

=)t(f

)t(

illustration

If x =2te

and y = tan1 (2 t + 1), find

xd

yd?

Solution :

Here x =2te so ,

tdxd = 2 . t

2te and y = tanan1 (2 t + 1)

On differentiating both sides, we get td

yd=

2)1t2(1

1

(2)

xd

yd=

tdxd

tdyd

=

2t

2

e

t2

1t4t412

Hence ,

xd

yd= 1t2t2t2

e2

t2


Findxd

ydin the following cases .

1. x = a

2

ttanlog

2

1tcos 2 and y = a sin t .

2. x = a ( sin ) and y = a (1 cos )

3. x = a e (sin cos ) , y = a e (sin + cos )

4. x =2

ee tt and y =

2

ee tt

5. x = e

1and y = e

1

6. x = cos12t1

1

and y = sin1

2t1

t

, t R


23/40

DIFFERENTIATION

23

6 Logarithmic Differentiation : To find the derivative of :

(i) a function which is the product or quotient of a number of functions

OR

(ii) a function of the form [f(x)]g(x) where f and g are both derivable, it will be found

convenient to take the logarithm of the function first and then differentiate. This is

called Logarithmic Differentiation.

illustration

If xy

. yx

= 1, find xd

yd

?

Solution :

Taking n on both sides;

y n x + x n y = n 1

Differentiating both sides, we get

y .xd

d(n x) +

yxd

d. n x +

xxd

d. ny + x

ynxd

d = 0

or y .x

1+ n x .

yd

xd+ 1 . n y + x .

y

1.

xd

yd= 0

y

xxn

xd

yd=

ynx

y

xd

yd=

xnyx

ynxy

.x

y


Differentiatiate the following functions w.r.t. x .

1. xx (x > 0) 2. (sin x)log x 0 x2

3.xxx (x > 0)

4. (sin x)tan x + (cos x)sec x 0 x2

5. xy = ex y (x > 0)

6. ex + ey = ex + y . Prove thatxd

yd+ ey x = 0

7 Differentiation with substitution

Following are some substitutions useful in finding derivatives .

Expression Substitution

a2 + x2 x = a tan or a cot a2 x2 x = a tan or a cot x2 a2 x = a sec or a cosec

xa

xa

orxa

xa

x = a cos 2

22

22

xa

xa

or22

22

xa

xa

a2 = a2 cos 2


24/40


24

illustration

Differentiate y = sin1 (3 x 4x 3)

Solution :

Let y = sin1 (3x 4x3) . Putting x = sin , we get

y = sin1 (3 sin 4 sin3 ) = sin1 (sin 3 ) = 3

or y = 3 sin1 x [ x = sin = sin1 x ]

xdyd = xd

d(3 sin1 x) = 3 xd

d(sin1 x) = 3 . 2x1

1

=2x1

3

illustration

Differentiate y = sin1

2x1

x2

Solution :

Let y = sin1

2x1

x2. Putting x = tan , we get

y = sin1

2tan1

tan2 = sin1 (sin 2 ) = 2 .

or y = 2 tan1 x [ x = tan = tan1 x ]

xd

yd=

xd

d(2 tanan1 x) = 2

xd

d(tan1 x) = 2 . 2x1

1

= 2x1

2


1. Differentiate the following function w.r.t. x .

(i) sin1 (sin x) , x [0 , 2 ] (ii) cos1 (cos x) , x [0 , 2 ]

(iii) tan1

(tan x) , x [0 , ] 2

2. Differentiate cos1 (2 x2 1) with respect to x , if :

(i) 0 < x < 1 (ii) 1 < x < 0

3. Differentiate the following functions w.r.t. x .

(i) tan1

xcos1

xcos1, 0 < x < (ii) tan1

xsin1

xsin1, 2

< x < 2

4. If y = sin1

2x1xx1x and 0 < x < 1 , then find

xd

yd.

5. Differentiate sin1

2x1x2 + cos1

2

2

x1x1 w.r.t. x , if :

(i) x (0 , 1) (ii) x ( , 1)

6. Differentiate tan1

2x1

x2+ cos1

2

2

x1

x1w.r.t. x , when :

(i) x (0 , 1) (ii) x ( , 1)


25/40

DIFFERENTIATION

25

8 Differentiation of Implicit Function

If the relation between the variables x and y is given by an equation containing both, and

this equation is not immediately solvable for y, then y is called an implicit function of x.

Implicit functions are given by (x , y) = 0.

(i) In order to find xdyd

, in the case of implicit functions, we differentiate each term

w.r.t. x regarding y as a functions of x and then collect terms in xdyd

together on

one side to finally find xdyd

.

(ii) In answers of xdyd

in the case of implicit functions, both x and y are present.

illustration

If x2 + y2 + xy = 2, findxd

yd?

Solution :x2 + y2 + xy = 2, Differentiating both sides w.r.t. x we get,

xd

d(x2) +

xd

d(y2) +

xd

d(x y) =

xd

d(2)

or 2 x + 2 yxd

yd+

xd

xdy + x

yxd

d= 0 or 2 x + 2 y

xd

yd+ 1 . y + x .

xd

yd= 0

(2 y + x)xd

yd= (2 x + y)

xd

yd=

)xy2(

)yx2(

illustration

If 6x1 + 6y1 = a3 (x3 y3) , prove thatxd

yd=

2

2

y

x

6

6

x1

y1

Solution :

Here , 6x1 + 6y1 = a3 (x3 y3)

Let x3 = sin , y3 = sin then we get , 2sin1 + 2sin1 = = a3 (sin sin )

cos + cos = a3 (sin sin )

2 cos

2. cos

2= a3

2sin

2cos2

cos

2 = a3 sin

2 = cot

2 = a3

= 2 cot1 (a3)

or sin1 x3 sin1 y3 = 2 cot1 (a3), Differentiating both sides w.r.t. x we get,

6x1

1

. 3 x2

6y1

1

. 3 y2 .

xd

yd= 0 Hence ,

xd

yd=

62

62

x1y

y1x


26/40


26


Findxd

ydfor the following .

1. a x2 + 2 h x y + b y2 + 2g x + 2 f y + c = 0

2. log (x2 + y2) = 2 tan1

x

y3. x y = c2

4. (x2 + y2)2 = x y 5. ex y = log

y

x

6. If sin y = x sin (a + y) , prove thatxd

yd=

asin

)ya(sin2

9 Differentiation of A Functions With Respect To Another Function

)x(d

)x(fd

=

)x(xd

d

)x(fxd

d

=

)x(

)x(f

illustration

Differentiate : sin1

2x1

x2with respect to tan1 x

Solution :

xtanxd

d

x1x2sinxdd

1

12

= xtan

xdd

xtan2xd

d

1

1

= xtanxd

d2

xtanxd

d2

1

1

= 2

illustration

Differentiate ln tan x with respect to sin1 (ex) .

Solution :

)( x1 esindxtannd )(

=

)( x1 esin

xtann

xdd

xdd

=

x2e1

1.e

xsecxcotx

2

=xcos.xsin

e1e x2x


Differentiate

1. tan1

x

1x1 2w.r.t. tan1x , x 0 .


27/40

DIFFERENTIATION

27

2. sin1

2x1

x2w.r.t. tan1 x , 1 < x < 1 . 3. xx w.r.t. x log x .

4. x2 w.r.t. x3 . 5. log (1 + x2) w.r.t. tan1 x .

6. sin1

2

x1

x2w.r.t. cos1

2

2

x1

x1, if 0 < x < 1 .

7. tan1

xa1xa1

w.r.t. 22 xa1

8. cos1 (4 x3 3 x) w.r.t. tan1

x

x1 2

, if2

1< x < 1

9. tan1

2x1x

w.r.t. sin1

2x1x2 , if

2

1< x 0 x R .

Solution :Putting x = 0 = y in the given functional equation, we get f (0) = 1

f (0) = 0hLim h

)x(f)hx(f = 0h

Lim h

)x(f1hx2)h(f)x(f

= 2x 0hLim h

)0(f)h(f = 2 x f (0) = 2 x 2aa3

f (x) = x2 2aa3 x + C

Putting x = 0 we have C = 1. Hence f (x) = x2 2aa3 x 1

The discriminant = 3 + a a2 + 4 = a2 + a + 1


35/40

EXAMPLES

35

=

4

3

2

1a

2

< 0. Hence f (x) > 0 for all x R .

Example 5.

Draw the graph of the function f defined by f (x) =

4x1,x4

1x1,3x. Discuss the

continuity and differentiability of f at x = 1 .

Solution :The graph of the function is shown in the adjacent figure.

It is clear from the graph that it is continuous for

all x [ 1, 4] and not differentiable at x = 1,because at x = 1 ; LHD > 0 while RHD = 1 < 0.

Y

x4

1

1

3

01

Example 6.The function f is defined by y = f (x) where x = 2 t t, y = t2 + t t, t R . Draw the graphoff for the interval 1 x 1 . Also discuss its continuity and differentiability at x = 0.

Solution :When t 0; |t| = t x = 2 t t = t x 0and y = t2 + t2 = 2t2

i.e. y = 2 x2, when x 0W hen t < 0 ; t= t x = 2 t ( t) = 3t x < 0and y = t2 + t( t) = 0

i.e. y = 0, when x < 0

x

(1, 2)

x1 1O

y

yThus f (x) =

0x10x,0

1x00x,x2 2

Now the student can prove easily that the function is continuous and differentiable in [ 1,

1].

Example 7.

Let f (x) = x3 x2 + x + 1 and g (x) =

2x1,x3

1x0,xt0;)t(fmax

Discuss the continuity and differentiability of the function g (x) in the interval (0 , 2) .

Solution :f (t) = t3 t2 + t + 1 f (t) = 3 t2 2 t + 1its disc = ( 2)2 4.3.1 = 8 < 0

and coefficient of t2 = 3 > 0

Hence f (t) > 0 for all real t. f (t) is always increasingThus f(t) is maximum when t is maximum and t

max= x

max f(t) = f(x) g(x) =

2x1,x3

1x0,1xxx 23

Now it can be easily seen that f(x) is continuous in (0, 2) and differentiable in (0, 2) except


36/40


36

at x = 1. because at x = 1 LHD > 0 while RHD = 1 < 0.

Example 8.If 2 x = y1/5 + y1/5 , then express y as an explicit function of x and prove that

(x2 1) +2

2

xd

yd+ x

xd

yd= 25 y .

Solution :

25/15/1 yy = 25/15/1 yy 4 = 4 x2 4 { y1/5 y1/5 = 2 x }

2 y1/5 = 5/15/1 yy + 5/15/1 yy = 2 x + 2 1x2 y =

52 1xx

Differentiating w.r.t. x ,

xd

yd

= 5

42

1xx

1x

x2

.2

1

1 2 = 5 1x

1xx

2

52

= 1x

y5

2 { Using

(i) }

(x2 1)2

xd

yd

= 25 y2

Differentiating w.r.t. x , 2 x

2

xd

yd

+ (x2 1) . 2

xd

yd.

2

2

xd

yd= 50 y

xd

yd

or ,xd

yd+ (x2 1) .

2

2

xd

yd= 25 y

ttanconsaisythen,0xd

yd

Example 9.

If f (x) = 3x

xcosBxsinAx2sin is continuous at x = 0 . Find the values of A and B . Also

find f(0).

Solution :As f (x) is continuous at x = 0,

0xLim f (x) = f (0) and both f (0) and 0x

Lim f (x) are finite .

f (0) = 0xLim 3xxcosBxsinAx2sin

As denominator 0 , when x 0 .Numerator should also 0 when x 0 which is possible only if sin 2 (0) + A sin (0) + B cos (0) = 0 B = 0

f (0) = 0xLim 3x

xsinAx2sin

f(0) = 0xLim

x

xsin

2x

Axcos2= 0x

Lim

2x

Axcos2


37/40

EXAMPLES

37

Again we can see that denominator 0 as x 0 Numerator should also approach 0 as x 0 2 + A = 0 A = 2

f (0) = 0xLim

2x

2xcos2= 0x

Lim

2

2

x

2/xsin4= 0x

Lim

4x

2xsin2

2

= 1

So, we get A = 2 , B = 0 and f ( 0) = 1

Example 10.

Show that 2xLim

2x

xtan

+ xLim

x

2x

11

> 3 .

Solution :

2xLim

2x

xtan

= 0yLim y

)2y(tan =

and xLim

)x/1(x

2

2

x

11

=

x

1

xLim2x

2x x

11Lim

e

= e0 = 1

xLim

2x

xtan

+ xLim

x

2x

11

= + 1 > 3 .

CONCEPT BUILDING OBJECTIVE EXAMPLES

Example 1.

0xLim

xtanx2

xsinx21

1

is equal to :

(A)31 (B)

21 (C) 0 (D) None of these

Solution :

0xLim

....3xxx2

....6xxx2

3

3

=3

1

Hence (A) is the correct answer.

Example 2.

1xLim 1x

}x{sinx

, where { x } denotes the fractional part of x, is equal to

(a) 1 (B) 0 (C) 1 (D) does not exist

Solution :

01xLim

{x} = 01xLim

(x [ x ]) = 1 0 = 1

01xLim

{x} = (x [ x ]) = 1 1 = 0


38/40


38

01xLim

1x

}x{sinx

= 01xLim

1x

x

sin {x} =

01xLim

}x{

}x{sinx

1x

1x

= 1 1 1 = 1

Since, L.H. limit R. H. limitHence (D) is the correct answer.

Example 3.

Let f (x) = xLim sin2n x , then number of point(s) where f (x) is discontinuous is :

(A) 0 (B) 1 (C) 2 (D) infinitely many

Solution :

f (x) = nLim sin2 n x = n

Lim (sin2 x)n =

In,2

)1n2(x,0

In,2

)1n2(x,1

Clearly , f (x) is discontinuous at x = (2 n + 1) 2

, n I .Hence (D) is the correct answer .

Example 4.

Let f (x) =

0x,0

0x,x1sinxp

Then f (x) is continuous but not differentiable at x = 0 if :

(A) p < 0 (B) p = 0 (C) 0 < p 1 (D) p 1Solution :

f(0) = 0

For 0xLim f (x) = 0 0x

Lim x

p sinx1 = 0 .

This is possible only when p > 0 ... (i)

f (0) = 0hLim h

)0(f)h0(f = 0h

Lim h

0h1sinhp

= 0hLim h

p 1 sinh

1

f (0) will exist only when p > 1 f (x) will not be differentiable if p 1 ... (ii)From (i) and (ii), for f(x) to be not differentiable but continuous at x = 0 , possible values

of p are given by 0 < p 1 .Hence (C) is the correct answer.

Example 5.

nn22n 2tan

2

1.....

2tan

2

1

2tan

2

1tanlim

(A)1

(B)1

2 cot 2 (C) 2 cot 2 (D) None of these

Solution :tan = cot 2 cot 2


39/40

EXAMPLES

39

2

1tan

2

=

2

1cot

2

cot

n2

1tan n2

= n2

1cot n2

1n2

1 cot 1n2

Required limit = nLim

Sn = nLim

2cot22

.2

2tan2

1

nn

nn =

1

2 cot 2

Hence (B) is correct answer.

Example 6.In order that the function f(x) = (x + 1)cotx is continuous at x = 0, f(0) must be defined as

(A) 0 (B) e (C)e

1(D) None of these

Solution :

0xLim f (x) = 0xLim xtan/x

x/1)x1( = e1

So , f (0) = e

Hence (B) is the correct answer .

Example 7.

For m , n I+ , 0xLim m

n

)x(sin

xsinis equal to :

(A) 1 , if n < m (B) 0 , if n > m (C) n/m (D) 0 , if n = m

Solution :Writing the given expression in the form

n

n

x

xsin

m

n

x

x m

xsin

x

and noting that the 0

Lim

sin= 1 , we see that the required

limit equals to 1 , if n = m , and 0 if n > m.


Example 8.

If f (x) is continuous and f (0) = 2 , then 0xLim x

ud)u(fx

0

is :

(A) 0 (B) 2 (C) f (2) (D) f (1)

Solution :

0xLim x

ud)u(fx

0

= 0xLim 1

)x(f

0xatcontinuousis)x(f

Rules'Hospital'LgsinU

= f (0) = 2.

Hence (b) is the correct answer.

Example 9.


40/40


Let g (x) be the inverse of the function f (x) and f (x) =3x1

1

. Then g (x) is equal to :

(A) 3)x(g1

1

(B)

3)x(f1

1

(C) 1 + (g(x))3 (D) 1 + (f (x))3

Solution :

Since g (x) is the inverse of f (x) , therefore g (x) = f1 (x) f {g (x)} = xDifferentiate both sides w.r.t. x

f { g(x) } g (x) = 1 g (x) =))x(g(f

1

= 1 + (g(x))3

Hence (C) is the correct answer.

Example 10.

1xLim 1x

1xlogxlogxx2

23

is equal to :

(A) 2/3 (B) 3/2 (C) 0 (D) None of these

Solution :

The given l imit = 1xLim 1x

xlog1x1x

2

23 )()(

= 1x

Lim

)()( 1x1x

xlog1x1x1xx)1x( )()()( 3

= 1xLim )1x()1x(

xlog)1x(1xx)1x( ][ 2

=)11(

1log)11(1112

=2

3


continuity and differ en ti ability

Documents