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Mahesh Tutorials Science 1

Continuity

GROUP (A)- CLASS WORK PROBLEMS

Q-1) Examine the continuity of the following

functions at given points

i) ( )( )log100 + log 0.01+

=3

xf x

x for 0x ≠≠≠≠

100

=3

for = 0x

at = 0x

Ans. ( )100

0 =3

f .... (given)

( )

1log100 + log +

100lim = lim

3

x

f xx

0 0x x→ →→ →→ →→ →

( )log 1+100 100

= lim100 3

x

x0x→→→→××××

( )log 1+100

= 1 lim =13

x

x

0x→→→→

×××× ∵∵∵∵

100

=3

Thus ( ) ( )lim = 0f x f0x→→→→

∴∴∴∴ f is continuous at x = 0

ii) ( ) ,

log – log7= for 7

at = 7– 7

=7, for = 7

xf x x

xx

x

≠≠≠≠

Ans. ( )7 = 7f .... (given)

( )log – log 7

lim = lim– 7

xf x

x0 0x x→ →→ →→ →→ →

Put – 7 =x h , then = 7 +x h , as

7, 0x h→ →→ →→ →→ →

( )( )( )

log +7 – log 77 + = lim

+7 – 7

hf h

h0x→→→→

+ 7log

7=

h

h

1 +7 + 7= lim = limlog

7 7

h h

h

17 7

0 0

h

h h→ →→ →→ →→ →

∴∴∴∴ ( ) ( )1 1

= log = 77 7

f x e f

≠≠≠≠

Since ( ) ( )lim 7 ,f x f7x→→→→

≠≠≠≠ f is discontinuous

at x = 7

iii) ( )1

,

– sin= for

2–

at =22

=3, for =2

x xf x x

xx x

x

xx

≠≠≠≠

Ans. = 32

f

ππππ.... (given)

( )1 – sin

lim = lim

–2

xf x

xπ

2

2 2x x

π ππ ππ ππ π→ →→ →→ →→ →

.... (i)

1 – sinR.H.S = lim

–2

x

x

2

2x

ππππ→→→→ ππππ

Put – =2

x hππππ

∴∴∴∴ = –2

x hππππ

As , 02

x hππππ

→ →→ →→ →→ →

∴∴∴∴ R.H.S1 – sin –

2= lim

h

h

20h→→→→

ππππ

1– cos= lim

h

h20h→→→→

Continuity

2 Mahesh Tutorials Science

1=2

∴∴∴∴ From equation (i),

( )1

lim =2 2

f x fπ

2x

ππππ→→→→

≠≠≠≠

∴∴∴∴ f is continuous at 2

ππππ.

iv) ( ) ,

= for 0at = 0

= , for = 0

xf x x

x x

c x

≠≠≠≠

Ans. ( ) =x

f xx

.... (given)

Thus, =x x if 0x +→→→→

= –x if 0x –→→→→

∴∴∴∴ ( ) = =1x

f xx

if 0x +→→→→

= –1 if 0x –→→→→

Now ( ) ( )lim = lim –1= –1= 0f x f– –0 0x x→ →→ →→ →→ →

.... (say)

Then, ( ) ( )lim = lim 1=1 0f x f+ +0 0x x→ →→ →→ →→ →

≠≠≠≠

Since Left hand limit ≠≠≠≠ Right hand limit,

f is discontinuous at x = 0

v) ( )2

2

– 9= for 0 3

– 3

= +3 for3 6

– 9= for6 9

+ 3

xf x x

x

x x

xx

x

< << << << <

≤ < ≤ < ≤ < ≤ <

≤ <≤ <≤ <≤ <

at = 3x and = 6x

Ans. ( )3 = 3+3 = 6f

( )– 9

lim = lim– 3

xf x

x

2

3– 3x x→ →→ →→ →→ →

( ) ( )

( )

+ 3 – 3lim=

– 3

x x

x3x→→→→

( )= lim + 3x3x→→→→

= 6

( )= 3f

∴∴∴∴ f is left continuous at 3

( )lim lim +3=f x x3+ 3x x→ →→ →→ →→ →

= 6

( )= 3f

∴∴∴∴ f is right continuous at 3

Hence f is continuous at 3

( )6 – 9 27

6 = = 3=6+3 9

f2

( ) ( )lim lim +3=f x x6– 3x x→ →→ →→ →→ →

= 9

( )6f≠≠≠≠

∴∴∴∴ f is not left continuous at x = 6

∴∴∴∴ f is not continuous at 6

vi) ( )( )

( )

( )

2 3

2

2

+ 3 +5 +

... + 2 –1 –= , 1

at =1–1

–1= , =1

3

n

x x x

n x nf x x

xx

n nx

≠≠≠≠

Ans. ( )( )–1

1 =3

n nf

2

.... (given)

Consider the sum ( )1+3+5...+ 2 –1n

This is sum of an A.P with =1, = 2a d and

the number of terms is n.

∴∴∴∴ [ ] ( )= + = 1+ 2 –1 =2 2

n nS a t n n

2

n n

Thus, ( )=1+3+5+. . . + 2 –1n n2.

We, shall replace 2n by R.H.S. in the

following limit

Now,

( )

( )

( )

+ 3 +5 ...+ 2 –1

–lim = lim

–1

x x x n

x nf x

x

2 3

2

1 1

n

x x→ →→ →→ →→ →

( )

( )

( )

+3 +5 +...+ 2 –1

– 1+3+5... 2 –1= lim

–1

x x x n x

n

x

2 3

1

n

x→→→→

( ) ( ) ( )( ) ( )

( )

–1 + 3 – 3 + 5 – 5 +...+

2 –1 – 2 –1= lim

–1

x x x

n x n

x

2 3

1

n

x→→→→


Continuity

( ) ( ) ( )( ) ( )

( )

–1 +3 –1 +5 – 5 +...+

2 –1 –1= lim

–1

x x x

n x

x

2 3

1

n

x→→→→

( )( ) ( )

( ) ( )( )

1+3 +1 + + +1 +...+–1

2 –1 + + ...+1= lim

–1

x x xx

n x x

x

2

–1 –2

1

n n

x→→→→

( ) ( ) ( ) ( ) ( )=1+3 2 +5 2 +7 4 +...+ 2 –1n n

( ) ( )= 2 –1r r∑=1

n

r

= 2 –r r∑ ∑2

1 1

n n

( ) ( ) ( )2 +1 2 +1 +1= –

6 2

n n n n n

( ) ( )+1 2 +1= –1

2 3

n n n n

( ) ( )+1 4 –1=

6

n n n

∴∴∴∴ ( ) ( )lim 1f x f1x→→→→

≠≠≠≠

∴∴∴∴ f is discontinuous at x = 1

Q-2) Find the value of k so that the function f (x)

is continuous at indicated point.

i) ( )

–3 – 3= , for 0,

at 0sin

= , for 0

x x

f x xxx

k x =

≠≠≠≠=

Ans. ( )0 =f k .... (given)

( )3 – 3

lim = limsin

f xx

–

0 0

x x

x x→ →→ →→ →→ →

( ) ( )3 –1 – 3 –1= lim

sin

x x

x x

–

0

x

x→→→→××××

3 –1 3 –1= lim – lim lim

sin

x

x x x

–

0 0 0

x x

x x x→ → →→ → →→ → →→ → →××××

( )= log 3 – log 3 1

–1 ××××

= log 3 + log 3

= 2 log 3

Since f is continuous at x = 0

( ) ( )lim = 0f x f0x→→→→

∴∴∴∴ 2log 3=k

∴∴∴∴ 2log 3=log 9k =

ii) ( )

= – 3 , for 3at = 3

= , for = 3

f x x xx

k x

≠≠≠≠

Ans. ( )3 =f k .... (given)

( )lim = lim – 3f x x3 3x x→ →→ →→ →→ →

= 0

Since, f is continuous at x = 3

( ) ( )lim = 0f x f3x→→→→

∴∴∴∴ 0 = k

i.e., = 0k

iii) ( )

2

1 – cos4= , for 0

= , for = 0 at = 0

= – 4, for 016 +

xf x x

x

k x x

xx

x

<<<<

>>>>

Ans. ( )0 =f k .... (given)

Since, f is continuous at x = 0, it is left

continuous at x = 0.

i.e., ( ) ( )lim = 0f x f–0x→→→→

∴∴∴∴1 – cos4

lim =x

kx20x→→→→

∴∴∴∴1 – cos 4

lim 16 =16

xk

x20x→→→→××××

∴∴∴∴1

16 =2

k××××

i.e., = 8k

iv) ( ) ( )

2cot2= sec , for 0

at = 0= , for = 0

x

f x x xx

k x

≠≠≠≠

Ans. ( )0 =f k .... (given)

( ) ( )lim = lim secf x x2cot

2

0 0

x

x x→ →→ →→ →→ →

( )= lim 1+ tan x2cot

2

0

x

x→→→→

Continuity


( )= lim 1+ =e h e

1

0

h

x→→→→∵∵∵∵

Since, f is continuous at 0,

( ) ( )lim = 0f x f0x→→→→

∴∴∴∴ k = e

v) ( )3 – tan

= , for– 3 3

xf x x

x

ππππ≠≠≠≠

ππππ

= , for =3

k xππππ

at = .3

xππππ

Ans. ( )3 – tan

lim = lim– 3

xf x

x3 3

x xπ ππ ππ ππ π

→ →→ →→ →→ → ππππ

Put = +3

x hππππ

As , 03

x hππππ

→ →→ →→ →→ →

So,

( )3 – tan –

3lim = lim

– 3 –3

h

f x

h

0

3

hx

ππππ →→→→→→→→

ππππ

ππππππππ

tan + tan3lim 3 –

1 – tan + tan3

=–3

h

h

h

0h→→→→

ππππ

ππππ

3 – tan3 – 3 –

1 – 3 tan= lim

–3

h

h

h0h→→→→

3 – 3 tan – 3 – tan= lim

–3

h h

h0h→→→→

4tan= lim

–3

h

h0h→→→→

4 tan 1= lim lim3 1+ 3 tan

h

h h0 0h h→ →→ →→ →→ →××××

4=3

∴∴∴∴ ( )4

lim =3

f x

3x


.... (i)

Also =3

f k

ππππ.... (ii)

Since f (x) is continuous at =3

xππππ

( )lim =3

f x f

3x


ππππ

∴∴∴∴4=

3k

∴∴∴∴4

=3

k

Q-3) Discuss the continuity of the following

functions. Which of these functions have

removable discontinuity ? Redefine the

function so as to remove the discontinuity.

i) ( )( )

2sin –= , for 0

at = 2

= 2 , for = 0

x xf x x

xx

x

≠≠≠≠

Ans. ( )0 = 2f .... (given)

( )( )sin –

lim = limx x

f xx

2

0 0x x→ →→ →→ →→ →

( )( )

( )sin –1

= lim –1–1

x xx

x x0x→→→→××××

( )=1 0 –1

= –1

Thus ( ) ( )lim 0f x f0x→→→→

≠≠≠≠

∴∴∴∴ f is a discontinuity at x = 0

However, the discontinuity is removable.

To remove the discontinuity, we defind f as

( )( )sin –

= for 0x x

f x xx

2

≠≠≠≠

= –1 for =0x

ii) ( )( )

1 – sin= , for

– 2 2

2= ,for =7 2

xf x x

x

x


ππππ

ππππ


Continuity

Ans.2

=2 7

fπ

.... (given)

( )( )1 – sin

lim = lim– 2

xf x

x2 2

x xπ ππ ππ ππ π

→ →→ →→ →→ → ππππ

Put – = ,2

x hππππ

∴∴∴∴ = –2

x hππππ

As , 02

x hππππ

→ →→ →→ →→ → . Also – 2 = 2x hππππ

∴∴∴∴ ( )( )

20

2

hx

ππππ →→→→→→→→

ππππ1 – sin –

2lim = lim

2

h

f xh

1 – cos

= lim4

h

h 20h→→→→

1 1 – cos

= lim4

h

h 20h→→→→

1 1

=4 2

1=8

∴∴∴∴ ( )lim2

f x f

2x



∴∴∴∴ f is discontinuous at 2

ππππ. However, the

discontinuity can be removed by

redefining f as

( )( )

1– sin= ,for

2– 2

xf x x

x2


ππππ

1= ,for =8 2

xππππ

iii) ( )

=

4 –, for 0=

6 –1

2= log , for 0

3

x x

x

ef x x

x

≠≠≠≠

Ans. ( )2

0 = log3

f

.... (given)

( )4 –

lim = lim6 –1

ef x

0 0

x x

xx x→ →→ →→ →→ →

4 –= lim

6 –1

e x

x0

x x

xx→→→→××××

4 –1 –1= lim – lim

6 –1

e x

x x

0 0

x x

xx x→ →→ →→ →→ →××××

( )1

= log 4 – loglog 6

e ××××

4log

=log 6

e

4

= loge

6

( )0f≠≠≠≠

∴∴∴∴ f has discontinuity at x = 0 . However,

this discontinuity is removable. It can

be removed by redefining f as

( )4 –

= for 06 –1

ef x x

x x

x≠≠≠≠

4

= loge

6

iv) ( )( )

[ ] { }

8 –1, –1,1 – 0=

sin log 1+4

x

f x xx

x

∈∈∈∈

Define f(x) in [–1,+1] so that it is continuous

in [–1, +1].

Ans. ( )( )8 –1

lim lim

sin log 1+4

f xx

x

2

0 0

x

x x→ →→ →→ →→ →

( )8 –1= lim

sinlog 1+

4

x x

xxx

2

20→→→→× ×× ×× ×× ×

x

x

8 –1 1= lim lim

sinlog 1+

4lim

x

xx x

x

2

0 0

0

→ →→ →→ →→ →

→→→→

× ×× ×× ×× ×x

x x

x

( )

1

1= log 8 14

2× ×× ×× ×× ×

( )= 4 log 82

∴∴∴∴ f would be continuous in [ ]–1,1 if

Continuity


( )( )8 –1

=

sin log 1+4

xf x

xx

2

for [ ] { }–1,1 – 0x ∈∈∈∈

( )= 4 log 82

for x = 0

Q-4) If ( ) 2

2 – 1+ sin= ,

cos

xf x

x for

2x

ππππ≠≠≠≠ is

continuous at =2

xππππ find

2

fππππ

.

Ans. ( )2 – 1+ sin

lim = limcos

xf x

x2

2 2

π ππ ππ ππ π→ →→ →→ →→ →x x

2 – 1+ sin 2 + 1+sin= lim

cos 2 + 1+sin

x x

x x2

2


××××x

( )

( )2 – 1+ sin

= limcos 2 + 1+sin

x

x x2

2

ππππ→→→→x

1 – sin 1= lim lim

1 – sin 2 + 1+sin

x

x x2

2 2

π ππ ππ ππ π→ →→ →→ →→ →

××××x x

( ) ( )1 – sin 1

= lim1+ sin 1 – sin 2 + 2

x

x x2


××××x

( ) ( )1 1

= lim1+sin 2 2x

2


××××x

( )1 1

=1+12 2

1=4 2

Q-5) If ( )( )–

2

1– sin= ,

2

xf x

xππππ for

2x

ππππ≠≠≠≠ is continuous

at =2

xππππ find

2

fππππ

.

Ans. ( )( )

1 – sinlim = lim

– 2

xf x

x2

2 2x x

π ππ ππ ππ π→ →→ →→ →→ → ππππ

Put – = ,2

x hπ

∴∴∴∴ ( )= – – 2 = 22

x h x hππππ

ππππ

As , 02

x hππππ

→ →→ →→ →→ →

∴∴∴∴ ( )( )

20

2

ππππ →→→→→→→→

1 – sin –2

lim = lim2

h

f xh

π

hx

0

1 – cos 1= lim

4

h

h 2→→→→××××

h

1 1=4 2

××××

1=8

Since f is continuous at =2

xππππ

( )= lim2

f f xπ

2

ππππ→→→→x

1=8

Q-6) If ( )+14 – 2 +1

= ,1 – cos

x x

f xx

for 0x ≠≠≠≠ is continuous

at x = 0 find f (0).

Ans. ( )4 – 2 +1

lim = lim1– cos

f xx

+1

0 0

x x

x x→ →→ →→ →→ →

( )2 – 2.2 +1= lim

1 – cos x

2

0

x x

x→→→→

( )2 –1= lim

1 – cos x

2

0

x

x→→→→

2 –1= lim

1 – cos

x

x x

22

0

x

x→→→→××××

( )= log 2 22

××××

Since, f is continuous at x = 0

( ) ( )0 = limf f x0x→→→→

( )= 2 log 22

Q-7) If ( )sin 4

= +5

ααααx

f xx

, for > 0x

= + 4 – ββββx , for < 0x

=1 , for = 0x

is continuous at = 0x , find αααα and ββββ


Continuity

Ans. ( )0 =1f .... (given)

Since, f is continuous at 0, it is right

continuos at 0

∴∴∴∴ ( ) ( )lim = 0f x f+0→→→→x

∴∴∴∴sin4

lim + =15

x

x0→→→→αααα

x

∴∴∴∴sin4 4

lim + =14 5

x

x0→→→→× α× α× α× α

x

∴∴∴∴4+ =1

5αααα

Also f is left continuous at 0.

∴∴∴∴ ( ) ( )–0x→→→→

lim = 0f x f

∴∴∴∴ lim + – =1x0→→→→

ββββx

∴∴∴∴ 4 – =1ββββ

∴∴∴∴ = 3ββββ

Thus 1

= , = 35

α βα βα βα β

Q-8) If ( )sin

= +–1

ππππαααα

xf x

x, for 0≤≤≤≤x

= 2ππππ , for = 0x

( )2

1+ cos= +

1 –

ππππββββ

ππππ

x

x, for > 0x

is continuous at =1x , find αααα and ββββ

Ans. ( )1 = 2f ππππ .... (given)

Since, f is continuous at 1, it is left as well

as right continuous at 1. Since f is left

continuous at 1.

( ) ( )lim = 1f x f–1→→→→x

∴∴∴∴sin

lim + = 2–1

x

x1→→→→

ππππα πα πα πα π

x

∴∴∴∴( )sin 1+

lim + = 2h

h0→→→→

ππππα πα πα πα π

h

(Putting –1 , 1 ,x h x h= ∴ = + As 1, 0x h→ →→ →→ →→ → )

∴∴∴∴( )sin +

lim + = 2h

h0→→→→

π ππ ππ ππ πα πα πα πα π

h

∴∴∴∴– sin

lim + = 2h

h0→→→→

ππππ× π α π× π α π× π α π× π α π

ππππh

∴∴∴∴ – + = 2π α ππ α ππ α ππ α π

∴∴∴∴ = 3α πα πα πα π

Since f is right continuous at x = 0

( ) ( )lim = 0f x f+1→→→→x

∴∴∴∴ ( )

1+ coslim + = 2

1 –

x

x21→→→→

ππππβ πβ πβ πβ π

ππππx

∴∴∴∴( )1+ cos 1+

lim + = 2h

h 20→→→→

ππππβ πβ πβ πβ π

ππππh

(Putting –1 , 1 ,x h x h= = +∴∴∴∴ As 1,x →→→→ )

∴∴∴∴( )1 cos +

lim + = 2h

h 20→→→→

π ππ ππ ππ πβ πβ πβ πβ π

ππππh

+

∴∴∴∴1 cos

lim + = 2h

bh

π20→→→→

ππππ× π× π× π× π

ππππh

–

∴∴∴∴1+ = 2

2b

π ππ ππ ππ π

∴∴∴∴3

= 2 – =2 2

bπ ππ ππ ππ π

ππππ

Thus, 3

= 3 , =2

a bπ

π

Q-9) Find the values of a and b so that the

function define by:

( )

5 , if 2

= + if 2 < <10

21 , if 10

≤≤≤≤

≥≥≥≥

x

f x ax b , x

x

is continuous at = 2x and =10x

Ans. First we use continuoity of f at = 2x

( )2 =5f .... (given)

Since, f is continuous at = 2x , it is right

continuous at 2.

i.e. ( ) ( )lim = 2f x f+2→→→→x

∴∴∴∴ ( ) ( )lim = 2f x f+2→→→→x

∴∴∴∴ 2 + = 5a b .... (i)

Now, we use continuity of f at =10x .

( )10 = 21f .... (given)

Since, f is continuous at =10x , it is left

continuous at =10x

i.e. ( ) ( )lim = 10f x f–10→→→→x

Continuity


∴∴∴∴ lim + = 21ax b10→→→→x

∴∴∴∴ 10 + = 21a b .... (ii)

Solving equations (i) and (ii) simultaneously,

we get

= 2, =1a b

Q-10) Determine the values of a,b,c so that the

following function is continuous at = 0x .

( )

( )

( )

112 22

1

2

sin +1 + sin, for < 0

= for = 0

sin + –, for > 0

a x xx

x

f x c , x

x bx xx

bx

Ans. ( )0 =f c .... (given)

Since, f is continuous at = 0x , it is left–

continuous = 0x

i.e. ( ) ( )lim = 0f x f–0→→→→x

∴∴∴∴( )sin +1 +sin

lim =a x x

cx0→→→→x

∴∴∴∴

( )+1 +2sin cos

2 2lim =

a x x ax

cx

0→→→→x

∴∴∴∴( )

( )( )

sin + 22lim +2 limcos =

+2 2

a x axa c

a x0 0→ →→ →→ →→ →× ×× ×× ×× ×

x x

∴∴∴∴ ( )2 +2 =a c

∴∴∴∴ 2 – = – 4a c .... (i)

Also f is right continuous at = 0x

∴∴∴∴ ( ) ( )lim = 0f x f+0→→→→x

∴∴∴∴+ –

lim =x bx x

cb x

2

0→→→→x

∴∴∴∴( )1+ –1

lim =x bx

cb x0→→→→x

∴∴∴∴1+ –1

lim =bx

cb0→→→→x

∴∴∴∴ = 0c (provided) 0b ≠≠≠≠

∴∴∴∴ 2 = – 4a

∴∴∴∴ = – 2a

∴∴∴∴ = –2a , b any non–zero number, = 0c

GROUP (A)- HOME WORK PROBLEMS :

Q-1) Examine the continuity of the following

functions at given points.

i) ( )

5 2–, 0

= sin3

1 , = 0

≠≠≠≠x xe e

xf x x

x

at x = 0.

Ans. ( )0 =1f .... (given)

–lim = lim

sin3

e e

x

5 2

0 0→ →→ →→ →→ →

x x

x x

( ) ( )–1 – –1 3 1= lim

sin3 3

e e x

x x

5 2

0→→→→× ×× ×× ×× ×

x x

x

1 –1 –1 3= lim – lim lim3 sin3

e e x

x x x

5 2

0 0 0→ → →→ → →→ → →→ → →××××

x x

x x x

[ ]1

= 5 – 2 13

××××

=1

ii) ( ) =

=

2

–1, for 1

–1

, for =1

nxf x x

x

n x

≠≠≠≠ at x = 1.

Ans. ( )1 =f n2.... (given)

–1lim = lim

–1

x

x1 1→ →→ →→ →→ →

n

x x

=n

Thus ( ) ( )lim 1f x f1→→→→

≠≠≠≠x

f is discontinuous at =1x

iii) ( ) ( )=

=

1

2

1+2 , for 0

,for = 0

xx xf x

e x

≠≠≠≠ at x = 0.

Ans. ( )0f e= 2.... (given)

( ) ( )lim = lim 1+ 2f x x1

0 0

x

x x→ →→ →→ →→ →

e2=


Continuity

Thus ( ) ( )lim = 0f x f0x→→→→

f is continuous at = 0x .

iv) ( )==

=

10 +7 –14 – 5for 0

12 at210

for = 07

x x x x

f x x

x

x

≠≠≠≠

Ans. ( )10

0 =7

f .... (given)

( )10 +7 –14 – 5

lim = lim1 – cos4

f xx0 0

x x x x

x x→ →→ →→ →→ →

10 – 5 –14 +7= lim

1– cos4x0

x x x x

x→→→→

( ) ( )5 2 –1 – 7 2 –1= lim

1 – cos4x0

x x x x

x→→→→

( ) ( ) ( )2

0→→→→× × ×× × ×× × ×× × ×

2 –1 5 – 7 4 1= lim

1 – cos4 16

x

x x x

x x x

x

( )

1 2 –1 5 – 7= lim lim16

4lim

1 – cos4

x x

x

x

0 0

2

0

x x x

x x

x

→ →→ →→ →→ →

→→→→××××

( )1 5

= log 2 log 216 7

××××

( )1 5

= log 2 log8 7

∴∴∴∴ ( ) ( )lim 0f x f0x→→→→

≠≠≠≠

∴∴∴∴ f is discontinuous at = 0x

v) ( )=

= sin – cos for 0at = 7

= –1 , for 1

f x x x , xx

x

≠≠≠≠

Ans. ( )0 = –1f .... (given)

( ) ( )lim = lim sin – cosf x x x0 0x x→ →→ →→ →→ →

= sin0 – cos0

= 0 –1

= –1

Thus ( ) ( )lim = 0f x f0x→→→→

∴∴∴∴ f is continuous at = 0x

vi) ( )

=

1

1

– 1= for 0

at = 0+1

= –1 , for 0

x

x

ef x , x

xe

x

≠≠≠≠

Ans. ( )0 =1f .... (given)

( )–1

lim = lim

+1

ef x

e

1

10 0

x

x xx

→ →→ →→ →→ →

1 –= lim

1+

e

e

1–

10 –

x

xx

→→→→

Suppose 0x →→→→

As 1 –1

0, , –xx x

→→ → ∞ ∴ ∞→ → ∞ ∴ ∞→ → ∞ ∴ ∞→ → ∞ ∴ ∞

∴∴∴∴ lim = 0e1

–

0→→→→

x

x

∴∴∴∴ ( )1 – 0

lim = =11+0

f x0→→→→x

( )= 0f

Let 0x –→→→→ . then 1

–x

→ ∞∞∞∞

∴∴∴∴ ( )0 –1

lim f =0 +1

x–0→→→→x

= –1

( )0f≠≠≠≠

∴∴∴∴ f is not left continuous at = 0x

and hence not continuous at = 0x

vii) ( )=

1= , for 0

12at

1 2=1 – for 1

2

f x x x

x

x , x

≤ <≤ <≤ <≤ <

≤ <≤ <≤ <≤ <

Ans.1 1 1=1 – =

2 2 2f

( )lim = limf x x

–

1 1

2 2x x→ →→ →→ →→ →

1

=2

1

2f

=

Continuity


∴∴∴∴ f is left continuous at 1

2

( ) ( )lim = lim 1–f x x

+

1 1

2 2x x→ →→ →→ →→ →

1

=2

1

2f

=


2

Thus, f is continuous at 1

2

viii) ( )sin2

= , for 021 – cos2

cos= for 1

– 2 2

at =2

xf x x

x

x, x

x

x

ππππ< ≤< ≤< ≤< ≤

ππππ< ≤< ≤< ≤< ≤

ππππ

ππππ

Ans.sin

= = 02 1 – cos

f


ππππ.... (i)

( )cos

lim = lim– 2

xf x

x+

22xx

ππππππππ →→→→→→→→ ππππ

In R.H.S., put – =2

x hππππ

– 2 = 2x hππππ

As , 02

x hππππ

→ →→ →→ →→ →

( )+ 0

2

ππππ →→→→→→→→

ππππcos –

2lim = lim

2

h

f xh

hx

sin= lim

2

h

h0h→→→→

0→→→→

1 sin= lim2

h

h

h

( )1

= 12

1=2

.... (ii)

∴∴∴∴ from (i) and (ii)

( )+

2


ππππ≠≠≠≠lim

2f x f

x

∴∴∴∴ f is not continuous at

2x

ππππ=

ix) ( )( ) ( )log 2 + – log 2 –

= , for 0tan

=1 for = 0

at = 0

x xf x x

x

, x

x

≠≠≠≠

Ans. ( )0 =1f .... (given)

( )( ) ( )log 2+ – log 2 –

lim = lim2

x xf x

0 0x x→ →→ →→ →→ →

2+log

2 –= lim

tan

x

x

x

0x→→→→

1+2log

1 –2

= limtan

x

x

x

0x→→→→

log 1+ – log 1 –2 2

= limtan

x x

x

x x

0x→→→→××××

log 1+ log 1 –2 2

= lim – lim

x x

x x

0 0x x→ →→ →→ →→ →

limtan

x

x0x→→→→××××

1 –1= – 12 2

××××

=1

Thus ( ) ( )lim = 0f x f0x→→→→



Continuity

Q-2) Find the value of k, so that the function f (x)

is continuous at the indicated point

i) ( )( )

2

– 1 sin= for 0 at = 0

= 4 for = 0

kxe kxf x x

xx

x

≠≠≠≠

Ans. ( )0 = 4f .... (given)

( )( )–1 sin

lim = lime kx

f xx20 0

kx

x x→ →→ →→ →→ →

–1 sin= lim

e kxk

kx kx

2

0

kx

x→→→→××××

=1 1 k2× ×× ×× ×× ×

=k2

Since, f is continuous at = 0x

( ) ( )lim = 0f x f0x→→→→

∴∴∴∴ = 4k2

∴∴∴∴ = 2k ±

ii) ( ) 2

2

= +1 for 0

= +1 + for < 0

at = 0

f x x x

x k x

x

≥≥≥≥

Ans. ( )0 = 0 +1=1f 2

( ) ( )lim = lim 2 +1+f x x k–

2

00 xx →→→→→→→→

= 2 0+1+k

= 2+k

Since, f is continuous at = 0x , it is left

continuous at 0.

∴∴∴∴ ( ) ( )lim = 0f x f–0x→→→→

∴∴∴∴ + 2 =1k

∴∴∴∴ = –1k

iii) ( )( )+

log 1= , for 0

at = 0sin

= 5 , for = 0

kxf x x

xx

x

≠≠≠≠

Ans ( )0 =5f .... (given)

( )( )log 1+

lim = limsin

kxf x

x0 0x x→ →→ →→ →→ →

( )0→→→→

× ×× ×× ×× ×log 1+

= limsin

kx xk

kx xx

( )log 1+= lim lim

sin

kx x

kx x

0 0x x→ →→ →→ →→ →

( ) ( )= 1 1k

=k


∴∴∴∴ ( ) ( )lim = 0f x f–0x→→→→

∴∴∴∴ = 5k

iv) ( )

8 – 2= ,for 0

at = 0–1

= 2 , for = 0

x x

xf x x

xk

x

≠≠≠≠

Ans. ( )0 = 2f .... (given)

( )8 – 2

lim = lim–1

f xk0 0

x x

xx x→ →→ →→ →→ →

8 – 2= lim

–1

x

x k0

x x

xx→→→→××××

( ) ( )8 –1 – 2 –1= lim lim

–1

x

x k0 0

x x

xx x→ →→ →→ →→ →××××

8 –1 2 –1 1= lim – lim

logx x k

0 0

x x

x x→ →→ →→ →→ →××××

( )log 8 – log 2=

logk

8log

2=

logk

= log 4k (change of base rule)


( ) ( )lim = 0f x f–0x→→→→

∴∴∴∴ log 4 = 2k

∴∴∴∴ 4 =k2

∴∴∴∴ = 2k

(∵∵∵∵base of logarithm cannot be negative

= –2k is not possible)

Continuity


v) ( ) ( )

2= – 2 , for 0at = 0

= 4 +1 , for > 0

f x k x xx

x x

≤≤≤≤

Ans. ( ) ( )0 = 0 – 2 = –2f k k

( ) ( )lim = lim 4 +1f x x+0 0x x→ →→ →→ →→ →

=1


( ) ( )lim = 0f x f+0x→→→→

∴∴∴∴ 1= –2k

∴∴∴∴1

= –2

k


functions, which of these functions have

a removable discontinuity ? Redefine the

function so as to remove the discontinuity

i) ( )

1 – cos3= , for 0

tan at = 0

= 9 ,for = 0

xf x x

x x x

x

≠≠≠≠

Ans. ( )0 = 9f .... (given)

( )1 – cos3

lim = limtan

xf x

x x0 0x x→ →→ →→ →→ →

1– cos3 9= lim

tan9

x x

x xx

2

20x→→→→××××

1 – cos3= 9 lim lim

tan9

x x

xx

20 0x x→ →→ →→ →→ →

( )1

= 9 12

9=

2

∴∴∴∴ ( ) ( )lim 0f x f0x→→→→

≠≠≠≠

∴∴∴∴ f has a discontinuity at = 0x

However, the discontinuity is

removable by redefining f as

( )1 – cos3

=9

xf x

x2 for 0x ≠≠≠≠

9

=2

for = 0x

ii) ( )( )2

2

–1 tan= ,for 0

sin

= , for = 0

at = 0

xe xf x x

x x

e x

x

≠≠≠≠

Ans. ( )0 =f e2.... (given)

( )( )–1 tan

lim = limsin

e xf x

x x

2

0 0

x

x x→ →→ →→ →→ →

–1 tan= lim 2

2 sin

e x x

x x x

2

0

x

x→→→→× ×× ×× ×× ×××××

–1 tan= 2 lim lim

2

e x

x x

2

0 0

x

x x→ →→ →→ →→ →

limsin

x

x

0x→→→→

= 2 1 1 1× × ×× × ×× × ×× × × = 2

∴∴∴∴ ( ) ( )lim 0f x f0x→→→→

≠≠≠≠

∴∴∴∴ f has a discontinuity at = 0x

However, the discontinuity can be

removed by redefining f as

( )( )–1 tan

= for 0sin

e xf x x

x x

2x

≠≠≠≠

=2 for = 0x

iii) ( )( )3 0

2

– 1 sin= , for 0

= , for = 060

at = 0

xe xf x x

x

x

x

≠≠≠≠

ππππ

Ans. ( )0 =60

fππππ

.... (given)

( )( )–1 sin

lim = lime x

f xx

3 0

20 0

x

x x→ →→ →→ →→ →

3

0→→→→

ππππππππ

× × ×× × ×× × ×× × ×ππππ

sin–1 180lim 3

3 180

180

xe

xx

x

x

sin3 –1 180= lim lim180 3

180

xe

xx

3

0 0

x

x x→ →→ →→ →→ →

ππππππππ

××××ππππ


Continuity

= 1 160

ππππ× ×× ×× ×× ×

=60

ππππ ( )= 0f


iv) ( ) = –1 , for1 2

= 2 +3 , for2 0

at = 0

f x x x

x x

x

≤ <≤ <≤ <≤ <

≤ ≤≤ ≤≤ ≤≤ ≤

Ans. ( ) ( )2 = 2 2 + 3 = 7f

( ) ( )lim = lim –1f x x2 2x x→ →→ →→ →→ →

= 2 –1=1

∴∴∴∴ ( ) ( )lim 2f x f–2x→→→→

≠≠≠≠

∴∴∴∴ f is not left continuous at 2 and hence

is not continuous at 2.

This discontinuity is non–removable,

since if we change value of f at 2, it would

be discontinuous from right at = 2x

Q-4) If ( )2

2

– cos= , for 0

xe xf x x

x≠≠≠≠ is continuous

at = 0x , find ( )0f .

Ans. ( )– cos

lim = lime x

f xx

2

20 0

x

x x→ →→ →→ →→ →

( ) ( )–1 – cos –1= lim

e x

x

2

20

x

x→→→→

–1 1 – cos= lim + lim

e x

x x

2

2 20 0

x

x x→ →→ →→ →→ →

1 3=1+ =

2 2

Since f is continuous at = 0x

∴∴∴∴ ( ) ( )lim = 0f x f0x→→→→

∴∴∴∴ ( )3

0 =2

f

Q-5) If ( )( ){ }

( )2

1– cos 7 –= for

5 –

xf x x

x

ππππ≠ π≠ π≠ π≠ π

ππππ is continuous

at =x ππππ , find ( )f ππππ .

Ans. ( )( ){ }

( )

1 – cos 7 –lim = lim

5 –

xf x

x2x x→π →π→π →π→π →π→π →π

ππππ

ππππ

Put – = .x hππππ As , 0x h→ π →→ π →→ π →→ π →

( )1 – cos7

lim = lim5

hf x

h 2x h→π →0→π →0→π →0→π →0

1 1 – cos7= lim 49

5 49

h

h 2h→0→0→0→0××××

1 1= 49

5 2× ×× ×× ×× ×

49=10

Since f is continuous at =x ππππ

( ) ( )lim =f x fx→π→π→π→π

ππππ

∴∴∴∴ ( )49

=10

f ππππ

Q-6) If ( )( )

( )

2sin4 –1

= for 0log 1+2

x

f x xx x

≠≠≠≠ is continuous

at = 0x , find ( )0f .

Ans. ( )( )

( )

4 –1lim = lim

log 1+ 2f x

x x

2sin

0 0

x

x x→ →→ →→ →→ →

( )( )

4 –1 sin= lim

log 1 2sin

x

x xx

2sin 2

20

x

x→→→→××××

+

4 –1 sin= lim lim

sin

x

x x

2sin 2

2 20 0

x

x x→ →→ →→ →→ →××××

( )lim

log 1+ 2

x

x

0x→→→→××××

( )2

× ×× ×× ×× ×1

= log 4 12

( )1

= log 42

2

Since f is continuous at = 0x

( ) ( )0 = limf f x0x→→→→

( )1

= log 42

2

Continuity


Q-7) If ( )1 – tan

= , for41 – 2 sin

xf x x

x

ππππ≠≠≠≠ is continuous

at =4

xππππ, find

4

fππππ

.

Ans. f is continuous at 4

ππππ

( )1 – tan

= lim = lim4 1 – 2sin

xf f x

x

4 4x x

π ππ ππ ππ π→ →→ →→ →→ →

ππππ

sin1 –

cos= lim1– 2sin

x

x

x4

xππππ

→→→→

4


××××cos – sin 1+ 2 sin

= limcos 1+ 2 sin

x x x

x xx

( ) ( )( ) ( )2

4


cos – sin 1+ 2 sin= lim

cos 1 – 2sin

x x x

x xx

( ) ( )( ) ( )

cos – sin 1+ 2sin= lim

cos cos + sin – 2sin

x x x

x x x x2 2 2

4x


( ) ( )( ) ( ) ( )


cos cos + sin cos – sin

x x x

x x x x x4

xππππ

→→→→

,2

... cos – sin 0

cos– sin 0

x

x x

x

→

∵

∵

ππππ

→→→→

≠≠≠≠

( )( ) ( )

1+ 2sin= lim

cos cos + sin

x

x x x4

xππππ

→→→→

( )

( )

lim 1+ 2 sin

=limcos cos + sin

x

x x x

4

4

x

x



11+ 2

2=

1 1 1

2 2 2

××××

1+1= = 2

1

Hence, = 24

f

ππππ

Q-8) If ( )1– 3tan

= , for– 6 6

xf x x

x


ππππ is continuous

at =6

xππππ, find

6

fππππ

.

Ans. f is continuous at 6

ππππ

∴∴∴∴ ( )1 – 3 tan

= lim = lim6 – 6

xf f x

x

6 6x x

π ππ ππ ππ π→ →→ →→ →→ →

ππππ

ππππ

( )cos – 3 sin

= limcos – 6

x x

x x6

xππππ

→→→→ ππππ

Multiplying and dividing by 2, we get

( )

1 32 cos . – sin

2 2= lim

6 cos – 6

x x

fx x

6x


ππππ

ππππ

( )6



ππππ

2 sin .cos – sin sin6 6

= limcos – 6

x x

x x

x

( )

sin –6

= 2 limcos – 6

x

x x

6x


ππππ

ππππ

( )

( )

1sin – 6

2 6= limlimcos – 6

x

x x

6

6

x

x



ππππ

ππππ

2 1=

63

2

××××

2=

3 3

Hence 2

=6 3 3

f

ππππ


Continuity

Q-9) ( ) 2

2

= + , for 0

= 2 +1 + ,for 0

f x x x

x x

α ≥α ≥α ≥α ≥

β <β <β <β <

And

1= 2

2f , is continuous at = 0x , find

αααα and ββββ.

Ans.1 1

= + = 22 2

f

2

αααα .... (given)

∴∴∴∴1

+ = 24

αααα

∴∴∴∴7

=4

αααα

( )7

0 = 0 + = =4

f α αα αα αα α

Since f is continuous at = 0x , it is left

continuous at 0.

∴∴∴∴ ( ) ( )lim = 0f x f–0x→→→→

∴∴∴∴7

lim2 +1+ =4

x2

0x→→→→ββββ

∴∴∴∴7

2 1+ =4

ββββ

∴∴∴∴7 –1

= – 2 =4 4

ββββ

Thus, 7

=4

αααα and –1

=4

ββββ

Q-10) ( )2 – 9

= +– 3

xf x

xαααα , for 3x >>>>

= 5 , for = 3x

2= 2 +3 +x x ββββ

is continuous at = 3x find αααα and ββββ.

Ans. ( )3 = 5f .... (given)

Since, f is continuous at = 3x , it is left

continuous at = 3x

i.e. ( ) ( )lim = 3f x f–3x→→→→

∴∴∴∴ lim2 + 3 + = 5x x2

3x→→→→ββββ

∴∴∴∴ 18+ 9+ = 5ββββ

∴∴∴∴ = 5 – 27 = –22ββββ

Also, f is right–continuous at = 3x

∴∴∴∴ ( ) ( )+3x→→→→

lim = 3f x f

∴∴∴∴– 9

lim = 5– 3

x

x

2

3x→→→→αααα+

∴∴∴∴( ) ( )

3x→→→→αααα

+ 3 – 3lim + = 5

– 3

x x

x

∴∴∴∴ ( )lim + 3 + = 5x α3x→→→→

( )→ ≠ ∴ ≠→ ≠ ∴ ≠→ ≠ ∴ ≠→ ≠ ∴ ≠∵∵∵∵ 3, 3, – 3 0x x x

∴∴∴∴ 6+ = 5αααα

∴∴∴∴ = –1αααα

Thus = –1, = –22α βα βα βα β

Q-11)If ( )f x is defined by

( ) = sin2 if6

= + if6

f x x x

ax b x

ππππ≤≤≤≤

ππππ>>>>

Find a and b if ( )f x and ( )f x′ are

continuous at 2

x =ππππ

Ans.3

= sin2 =6 6 2

f


Since, f is continuous at =6

xππππ, it is left

continuous at 6

ππππ.

i.e., ( )lim =6

f x f

6x


ππππ

∴∴∴∴3

lim + =2

ax b

6x


∴∴∴∴3

+ =6 2a b

ππππ.... (i)

Now ( ) = 2cos2 ,f x x′ for6

xππππ

≤≤≤≤

=a for6

xππππ

>>>>

∴∴∴∴π ππ ππ ππ π

= 2cos26 6

f

′

Continuity


= 2cos2 =13

ππππ

Since f ′′′′ is continuous at =6

xππππ, it is right

continuous at 6

ππππ

∴∴∴∴ ( )+6

xππππ

→→→→

ππππlim =

6f x f

′ ′

∴∴∴∴lim =1a

6x


∴∴∴∴ =1a

Putting this value of a in (i)

( )3

1 + =6 2

bππππ

∴∴∴∴ 3

= –2 6

bππππ

Thus, 3

=1, = –2 6

a bππππ

Q-12) Find k so that the following function f is

continuous at =1x

( ) 2=f x kx , for 1x ≥≥≥≥

= 4 1, for x <<<<

Ans. ( ) ( )1 = 1 =f k k2

Since f is continuous at =1x , it is left

continuous at =1x

i.e., ( ) ( )lim = 1f x f–1x→→→→

lim 4 =k–1x→→→→

= 4k

GROUP (B)- CLASS WORK PROBLEMS

Q-1) If ( )f x is defined by

( ) = + 2 sinf x x a x ,04

xππππ

≤ <≤ <≤ <≤ <

= 2 cot + bx x ,4 2

xπ ππ ππ ππ π

≤ <≤ <≤ <≤ <

= cos2 – sina x b x , ,2

xππππ

< ≤ π< ≤ π< ≤ π< ≤ π

Find a and b

Ans. Since f is continuous in [0, ππππ]. It is continuous

at =4

xππππ

and =2

xππππ

= 2 cot +4 4 4

f b

π π ππ π ππ π ππ π π

( )= 1 +2

bππππ

= +2

bππππ


xππππ

, it is left

continuous at =4

xππππ

i.e., ( )= lim4

f f x

–4x


ππππ

+ = lim + 2sin2

b x a x

4x


ππππ


= + 2 sin4 4

a

ππππ

= +4

a

– =4

a bππππ

.... (i)

= 2 cot +2 2 2

f b

π π ππ π ππ π ππ π π

= b cot = 02

∵

ππππ


xππππ, it is right

continuous at =2

xππππ.

∴∴∴∴ ( )= lim2

f f x

+4x


ππππ

∴∴∴∴

+2


= lim cos2 – sinb a x b xx

= cos – sin2

a bππππ

ππππ

= – –a b

∴∴∴∴ + = –a b b


Continuity

∴∴∴∴ = – 2a b .... (ii)

∴∴∴∴ From equation (i) –3 =4

bππππ

∴∴∴∴–

=12

bππππ

∴∴∴∴ =6

aππππ

Thus, –3 = , = –6 12

a bπ ππ ππ ππ π

Q-2) Find αααα and ββββ so that the functions, defined

below is continuous is [– ππππ, ππππ]

( ) = –2sinf x x, for –2

xππππ

π ≤ ≤π ≤ ≤π ≤ ≤π ≤ ≤

= sin + ,xα βα βα βα β for –2 2

xπ ππ ππ ππ π

< << << << <

=cos x, for2

xππππ

≤ ≤ π≤ ≤ π≤ ≤ π≤ ≤ π

Ans. Since f is continuous in [–ππππ, ππππ]. I t is

continuous at –

2

ππππ and

2

ππππ

– –= –2sin = 2

2 2f


Since f is continuous at –

2

ππππ, it is right

continuous at –

=2

xππππ

∴∴∴∴ ( )–

–= lim

2f f x

+2


ππππ

x

–2 = lim sin +x

2


α βα βα βα βx

–

= sin +2

ππππα βα βα βα β

= – +α βα βα βα β

Thus, – + = 2α βα βα βα β .... (i)

= cos = 02 2

f



xππππ, it is left

continuous at 2

ππππ

∴∴∴∴ ( )lim =2

f x f

–2x


ππππ

∴∴∴∴lim sin + = 0x

2x


α βα βα βα β

∴∴∴∴ ( )1 + = 0α βα βα βα β

∴∴∴∴ + = 0α βα βα βα β .... (ii)

From equation (i) and (ii)

= –1, =1α βα βα βα β

Q-3) If ( )3

3

+3 +5=

– 3 +2

x xf x

x x. Discuss the

continuity of ( )f x on [0, 5]

Ans. ( )f x is a rational polynomial and we know

that every rational polynomial is continuous

for all real values x except when its

denominator becomes zero.

∴∴∴∴ ( )f x is continuous for all real values of x,

except when its denominator becomes zero.

i.e., – 3 +2 = 0x x3

∴∴∴∴ ( ) ( )–1 + – 2 = 0x x x2

∴∴∴∴ ( ) ( ) ( )–1 +2 –1 = 0x x x

∴∴∴∴ ( ) ( )–1 + 2 = 0x x2

∴∴∴∴ =1or = –2x x

But [ ]–2 0,5∉∉∉∉

∴∴∴∴ ( )f x is continuous for all real values of

x, except at =1x .

Thus, ( )f x is discontinuous at =1x

Q-4) Discuss the continuity of the function

clog x where 0, 0> >> >> >> >c x

Ans Let ( ) = logf x xc

Let a be any positive real number, then

( ) = logf a ac

Let ( ) ( )= lim + –L f a h f a 0→→→→h

( )= lim log + – loga h a 0→→→→c c

h

0→→→→

+= lim log

a h

a

ch

Continuity


0→→→→××××

+= lim log

a hh

a

ch

1

0→→→→××××= lim log 1+

hh

a

h

ch

= lim log 1+h

ha

1

0→→→→××××

a ah

ch

( )= log lim 1+ limh

ha

1

0 0→ →→ →→ →→ →××××

a ah

ch h

( ) ( )1

= log 0 = log 0e ea

1

ac c = 0

Thus, ( ) ( )lim – = 0f x f a 0→→→→h

∴∴∴∴ f is continuous at =x a .

But, a is any positive ral number.

∴∴∴∴ f is continuous at all positive real number

Thus, log xc where 0, 1, 0c c x> ≠ >> ≠ >> ≠ >> ≠ > is

continuous

Q-5) Test the continuity of the function

( )( ) ( )

+1=

– 2 – 5

xf x

x x in [0, 1] and [4, 6]

Ans Since f is a rational function (division of

polynomials) it is continuous at every point

of the domain, except at the point where

denominator is zero.

( ) ( )– 2 – 5 = 0 = 2, = 5x x x x⇒⇒⇒⇒

None of these points lie in [0, 1] and hence

( )f x is continuous in [0, 1]

The value [ ]= 5 4,6x ∈∈∈∈ and hence f is

discontinuous only at = 5x .

Thus f is continuous everywhere is [ ]4,6

except at = 5x .

Q-6) Examine the continuity of ( )f x on its

domain

Where, ( )1

=+1

f xx

, for 2 4x≤ ≤≤ ≤≤ ≤≤ ≤

+1

=– 3

x

x, for 4 6x< ≤< ≤< ≤< ≤

Ans The domain of the function is [2, 6].

( )1

=+1

f xx

, which is rational function and

hence is continuous as the denominator

( )+1x is not zero in 2 4x≤ ≤≤ ≤≤ ≤≤ ≤ .

In 4 6x< ≤< ≤< ≤< ≤ , ( )+1

=– 3

xf x

x, which is a rational

function and hence it is continuous for

4 6x< ≤< ≤< ≤< ≤

When = 4x , ( )1

=+1

f xx

∴∴∴∴ ( )1 1

4 = =4 +1 5

f

∴∴∴∴ ( )+1

lim = lim– 3

xf x

x

+ +4 4x x→ →→ →→ →→ →

4 +1

=4 – 3

5=1 ( )= 5 4f≠≠≠≠

Hence f is continuous on [2, 6] except at = 4x

Q-7) a and b such that the

function defined by

( ) = 5f x for 2x ≤≤≤≤

= +ax b for 2 10x< << << << <

= 21 for 10x ≥≥≥≥

is continuous in its domain

Ans The function is continuous on [2, 10].

At ( ) ( ) ( )= 2 lim = = 5 = limx f x f x f x– +2 2x x→ →→ →→ →→ →

( )lim = 5f x–2x→→→→

( )2 = 5f

( ) ( )lim = lim + = 2 +f x ax b a b+ +2 2x x→ →→ →→ →→ →

∴∴∴∴ 2 + = 5a b .... (i)

At =10x

( )lim = lim + =10 +f x ax b a b+ +10 10x x→ →→ →→ →→ →

( )10 = 21f

∴∴∴∴ 10 + = 21a b .... (ii)

Find the value of


Continuity

Solving (i) and (ii) simultaneously, we get

= 2a and =1b

Q-8) Show that ( ) = 1+ +f x x x is continuous

for all Rx ∈∈∈∈ .

Ans Consider the function ( ) =g x x

∴∴∴∴ ( ) = ,g x x for 0x ≥≥≥≥ , and

= – ,x for 0x <<<<

Clearing ( )0 = 0g

( ) ( )lim = lim – = 0g x x–0 0x x→ →→ →→ →→ →

Also, ( ) ( ) ( )lim = lim = 0g x x+0 0x x→ →→ →→ →→ →

∴∴∴∴ g is continuous at = 0x

( ) ( ) ( )lim = lim = 0g x g x g+0 0x x→ →→ →→ →→ →

In ( )– ,0∞∞∞∞ and in [ )0,∞∞∞∞

g is polynomial and hence continuous.

Thus, =g x is continuous at every point of R.

Similarly, 1+ x is continuous over R.

∴∴∴∴ By algebra of continuous functions, + 1+x x

is continuous over R i.e. ( )f x is continuous

for all x R∈∈∈∈ .

Q-9) Prove that the exponential function ax is

continuous at every point ( )0a >>>> .

Ans Let ( ) =f x ax and t R∈∈∈∈

∴∴∴∴ ( ) =f t at

( )lim = lim =f x a a+

+

0

t h t

x t h→ →→ →→ →→ →

( )lim = lim =f x a a– +

–

0

t h t

x t h→ →→ →→ →→ →

Thus, ( ) ( ) ( )lim = lim =f x f x f t+ –tx t x→ →→ →→ →→ →

∴∴∴∴ ( ) =f x ax is continuous at =x t . But t is

arbitrary real number.

∴∴∴∴ ax is continuous at every point of R.

Q-10) Prove that the exponential function ax is

continuous at every point ( )0a >>>> .

Ans Let ( ) = sinf x x

Let t be arbitary real number.

( ) = sinf t t

( ) ( ) ( )lim = lim + = lim sin +f x f t h t h+ + +0 0x t h h→ → →→ → →→ → →→ → →

= sin t

( )lim = lim –f t h– +0x t h→ →→ →→ →→ →

( )= limsin –t h0h→→→→

= sin t

Thus, ( ) ( ) ( )lim = lim =f x f x f t+ –x t x t→ →→ →→ →→ →

∴∴∴∴ sin x is continuous at t.

But t is arbitary

∴∴∴∴ sin x is continuous at every real number.

GROUP (B)- HOME WORK PROBLEMS

Q-1) If ( )f x is continouous on [0, 8] define as

( ) 2= + +6,f x x ax for 0 2x≤ <≤ <≤ <≤ <

= 3 +2,x for2 4x≤ ≤≤ ≤≤ ≤≤ ≤

= 2 +5 ,ax b for 4 8x< ≤< ≤< ≤< ≤

Find a and b

Ans Since f is continuous in [0, 8]. It is continuous

at = 2x and = 8x

= 2x ( ) ( )2 = 3 2 +2 = 8f


continuous at = 2x .

∴∴∴∴ ( ) ( )lim = 2f x f–2x→→→→

∴∴∴∴ ( )lim + + 6 = 8x ax2

2x→→→→

∴∴∴∴ 4 + 2 + 6 = 8a

∴∴∴∴ 2 = –2a

∴∴∴∴ = –1a

= 4x

Since f is continuous at = 4x , it is right

continuous at = 4x

∴∴∴∴ ( ) ( )lim = 4f x f+2x→→→→

∴∴∴∴ ( )2 +5 =14ax b

Continuity


∴∴∴∴ 8 +5 =14a b

∴∴∴∴ – 8 + 5 =14b

∴∴∴∴ 5 = 22b

∴∴∴∴22

=5

b

Thus, 22

= –1, =5

a b

Q-2) If the function ( )f x defined below is

continuous in [0, 3], find the value of k.

( ) = 3 – 4,f x x for 0 2x≤ <≤ <≤ <≤ <

= 2 + ,x k for2 3x≤ ≤≤ ≤≤ ≤≤ ≤

Ans Since f is continuous in [0, 3]. It is continuous

at = 2x

( ) ( )2 = 3 2 – 4f

= 2


continuous at = 2x .

∴∴∴∴ ( ) ( )2 = limf f x+2x→→→→

∴∴∴∴ ( )2 = lim 2 +x k2x→→→→

∴∴∴∴ 2 = 4 +k

∴∴∴∴ = – 2k

∴∴∴∴ = –1a


function in its domain

( ) = ,f x x 0x ≥≥≥≥

2= ,x 0x <<<<

Ans Consider ( ) =f x x for 0x >>>>

Since f is a linear function, f is continuous

for all 0x >>>> .

Next, consider ( ) =f x x2 for 0x <<<<

Since f is a quadratic function, f is

continuous for all 0x <<<< .

Now, let us consider the behaviour of function

f at = 0x .

f is continuous at = 0x if

( ) ( ) ( )lim = lim = 0f x f x f– +0 0→ →→ →→ →→ →x x

Now, ( )lim = lim = 0f x x– +

2

0 0→ →→ →→ →→ →x x

and ( ) ( ) ( )lim = lim = 0f x f x f+ +0 0→ →→ →→ →→ →x x

∴∴∴∴ ( ) ( ) ( )lim = lim = 0f x f x f– +0 0→ →→ →→ →→ →x x

Hence, f is continuous at = 0x

Thus, f is continuous in its domain R.



2= – 4f x for 0 2x≤ ≤≤ ≤≤ ≤≤ ≤

= 2 + 3x for2 4x< ≤< ≤< ≤< ≤

2= – 5x for 4 6x< ≤< ≤< ≤< ≤

Ans Clearly, the domain of f is [0, 6]. Since f is

polynomial in x in each part of the domain, f

is continuous in [0, 2], (2, 4], (4, 6]. The only

possible points of discontinuity are 2 and 4.

( )2 = 2 – 4 = 0f 2

( ) ( )lim = lim 2 + 3 = 7f x x+2 2→ →→ →→ →→ →x x

∴∴∴∴ ( ) ( )lim 2f x f+2→→→→

≠≠≠≠x

∴∴∴∴ f is not right continuous = 2x

i.e. f is discontinuous at = 2x

( )4 = 2 4+3 =11f ××××

( ) ( )lim = lim – 5f x x+

2

4 4→ →→ →→ →→ →x x=16 – 5 =11


( ) ( )lim = lim 2 +3 =11f x x–4 4→ →→ →→ →→ →x x

∴∴∴∴ f is left continuous at = 4x

∴∴∴∴ x is continuous at = 4x

Thus, x is continuous in [0, 6] except at = 2x

Q-5) Discuss the continuity of f in its domain

where f is defined as

( )

3 if 0 1

= 4 if 1 < < 3

5 if 3 10

x

f x x

x

≤ ≤≤ ≤≤ ≤≤ ≤

≤ ≤≤ ≤≤ ≤≤ ≤

Justify your answer with the help of graph.

Ans The domain of f is [0, 10]. f is constant in

[0, 1]. (1, 3) and [3, 10] and hence continuous

in each of these sub-intervals

The only possible points of discontinuity are

1 and 3.


Continuity

( )1 = 3f .... (given)

( )lim = lim4 = 4f x+1 1x x→ →→ →→ →→ →

∴∴∴∴ ( ) ( )lim 1f x f+1x→→→→

≠≠≠≠

∴∴∴∴ f is discontinuous at =1x

( )3 = 5f .... (given)

( )lim = lim4 = 4f x+1 3x x→ →→ →→ →→ →

∴∴∴∴ ( ) ( )lim 3f x f–1x→→→→

≠≠≠≠

∴∴∴∴ f is discontinuous at = 3x from left and

hence discontinuous at = 3x .

Thus f is continuous at every point of

[0, 10], except at =1x and = 3x

Graph of the function :

1

1 2 3 4 5 6 7 8 9 10

2

3

4

5

0X

Y

Y′

X′



( )

–2 , if –1

= 2 , if – 1 < 1

2 , if 1

x

f x x x

x

≤≤≤≤

≤≤≤≤

>>>>

Ans f is constant (function) in (– ∞∞∞∞, –1) and (1, ∞∞∞∞)

and hence continuous in these sub-intervals.

f is a polynomial in (– 1, + 1] and hence

continuous over it.

Therefore, the only possible points of

discontinuity of f are – 1 and + 1.

( )–1 = –2f .... (given)

( )lim = lim 2 = –2f x x+–1 –1x x→ →→ →→ →→ →

∴∴∴∴ f is right continuous at – 1 and hence

continuous at – 1.

( ) ( )1 = 2 1 = 2f .... (given)

( )lim = lim2 = 2f x+1 1x x→ →→ →→ →→ →

∴∴∴∴ f is right continuous at 1 and hence

continuous at 1

∴∴∴∴ f is continuous in R.

Q-7) Examine the continuity of f (x) on its

domain, where

( )

+3 ,if –3

= – 2 , if – 3 < <1

6 +2 , if 1

x x

f x x x

x x

≤≤≤≤

≥≥≥≥

Ans Domain of f is (– ∞, + ∞∞, + ∞∞, + ∞∞, + ∞). f is a polynomial in

each of subintervals (– ∞, ∞, ∞, ∞, – 3], (– 3, 3] and [3, ∞∞∞∞)

and hence continuous in each subinterval. The

possible points of discontinuity are – 3 and + 3

It remains to check whether f is right

continuous at – 3 and left continuous at = 3x

( )–3 = –3 3 = 0f +

( ) ( )lim = lim –2 = 6f x x+–3 –3x x→ →→ →→ →→ →

∴∴∴∴ ( ) ( )lim –3f x f+–3x→→→→

≠≠≠≠

∴∴∴∴ f is not continuous at = 3x

( ) ( )3 = 6 3 +2 = 20f

( ) ( )lim = lim –2 = – 6f x x– –3 3x x→ →→ →→ →→ →

∴∴∴∴ ( ) ( )lim 3f x f–3x→→→→

≠≠≠≠

∴∴∴∴ f is not continuous at = 3x . Hence f is

continuous in (– ∞∞∞∞, ∞∞∞∞) except at = – 3x and

=3x .

Q-8) Show that the function defined by

( ) ( )2=sinf x x is a continuous function.

Ans Let f and g be real valued functions such that

fog is defined at =x c .

If g is continuous at =x c and f is continuous

at ( )=x g c ,then fog is continuous at =x c .

Now, the function ( ) = sinf x x2 is defined for

every real number,

We can treat the function f as a composite

function goh of two functions h and g, where

( ) = sing x x an ( ) =h x x2.

Since both h and g are continuous functions,

by the above theorem, it follows that f is

continuous function

Continuity


BASIC ASSIGNMENTS (BA) :

BA – 1

Q-1) Discuss the continuity of the following.

Which of these functions have removable

discontinuity ? Redefine such function at

the given point so as to remove

discontinuity.

i) ( )( )

( )

2sin3 –1

=log 1+

x

f xx x

, for 0x ≠≠≠≠ at = 0x .

Ans. i) f (0) = 2 log 3 = log 32 .... (given)(i)

( )( )

( )

3 –1lim = lim

log 1+f x

x. x

2sin

0 0

x

x x→ →→ →→ →→ →

( )

3 –1 sin.

sin= lim

.log 1+

x

x x

x x

x

2sin

0

2

x

x→→→→

( )

3 –1 sinlim .lim

sin= lim

limlog 1+

x

x x

x

2sin

0 0

10

0

x

x x

xx

x

→ →→ →→ →→ →

→→→→

→→→→

∴∴∴∴ ( )( )

( )log3 1

= log3log

f xe

2

2×××××××× .... (ii)

From (i) and (ii), we get

( ) ( )lim 0f x f0x→→→→

≠≠≠≠

∴∴∴∴ f is discontinuous at = 0x

Discontinuity can be removed by redefining

the function as follows.

( )( )

( )

3 –1=

.log 1+f x

x x

2sinx for 0

at = 0for = 0

xx

x

≠≠≠≠

= (log 3)2

Q-2) Find k, if the following are continuous at

the indicated points

i) ( ) ( ) ( )1 2= log 1+2

xf x x

––––for 0x ≠≠≠≠

=k for = 0x

at = 0x

Ans f (0) = k

As f (x ) is continuous at x = 0, f (0) = ( )0→→→→

lim f xx

∴ k = ( ) ( )–

lim log1 2 1+2

0 x xx→→→→

= ( )( )

log 1+2lim

log 1– 2

x

x0x→→→→

=

( )

( )

log 1+2lim

2log 1– 2

lim2

x

xx

x

0

0

x

x

→→→→

→→→→

= 1

–1 = –1

∴ k = –1

ii) ( ) ,1– tan

=1– 2 sin

f kθθθθ

θθθθfor

4x


=2

kfor =

4x

ππππ

at =4

xππππ

Ansk

=4 2

f

ππππ

As ( )f x is continuous at =4

ππππθθθθ ,

∴∴∴∴ ( ) ( )= limf f

4

ππππθ→θ→θ→θ→

θ θθ θθ θθ θ

∴∴∴∴ f (θθθθ ) = ( )lim f

4


θθθθ =

4


θθθθ

θθθθ

1 – tanlim

1 – 2 sin

= 4


θθθθ

θθθθ

θθθθ

sin1 –

coslim1 – 2 sin

= ( )cos – sin

limcos 1 – 2sin

4




( )cos – sin 1+ 2sin

= lim1– 2sincos 1 – 2sin

4


θ θ θθ θ θθ θ θθ θ θ××××

θθθθθ θθ θθ θθ θ

( ) ( )( )2

4


θ θ θθ θ θθ θ θθ θ θ



cos 1 – 2sin

( ) ( )( )2 2 2

4



θ θ θ θθ θ θ θθ θ θ θθ θ θ θ


cos cos + sin – 2sin

( ) ( )

( )


cos cos – sin2 2

4





Continuity

( ) ( )( ) ( )


cos cos – sin cos +sin4



θ θ θ θ θθ θ θ θ θθ θ θ θ θθ θ θ θ θ

( )( )

1+ 2sin= lim

cos cos +sin4


θθθθ


1,cos and

4 2

1sin ,cos – sin 0

2

ππππθ → θ →θ → θ →θ → θ →θ → θ →

θ → θ θ ≠θ → θ θ ≠θ → θ θ ≠θ → θ θ ≠

∵∵∵∵

( )

( )

lim 1 2sin+

=limcos cos sin+θ θ

4

4

x

x

θ→θ→θ→θ→

θ→θ→θ→θ→

θθθθ

θθθθ

11 2+

2=

1 1 1+

2 2 2

××××

1 1+=

1 2×

2 2

= 2

1 = 2

∴∴∴∴2

k = 2 i.e., = 4k

Q-3) Find k, if the functions are continuous at

the indicated points

( ) ,cos

=– 2

= 3

k xf x

xππππ

2

=2

≠

for

for

x

x

ππππ

ππππ at x =

2

ππππ

Ans. i) = 32

f

ππππ

As f is continuous at x = ,2

ππππ

( )= lim2

f f x

2x


ππππ

∴ 3 = cos

lim– 2

k x

x2

xππππ

→→→→ ππππ .... (i)

Put –2

xππππ

= θθθθ, then x = –2

ππππθθθθ

As , 02

xππππ

→ θ →→ θ →→ θ →→ θ →

∴ 3 = 0θ→θ→θ→θ→

ππππθθθθ

θθθθ

cos –2

lim2

k

=

0θ→θ→θ→θ→

θθθθ

θθθθ

sinlim

2

k =

2

k

∴ k = 6

Q-4) Is the function defined by f (x) = x2 – sin x + 5

continuous at x = ππππ ?

Ans. f (x ) = x2 – sin x + 5

f (ππππ ) = ππππ2 – sin ππππ + 5 = ππππ2 – 0 + 5 = ππππ2 + 5

Also, ( )lim f xx→π→π→π→π

= ( )lim – sin +5x x2

x→π→π→π→π=

2π ππ ππ ππ π–sin +5 = – 0+52ππππ = +52ππππ

Since, ( )lim f xx→π→π→π→π

= ( ),f ππππ the function f is

continuous at x = ππππ.

Q-5) A function defined by

f (x) = x + a for x < 0

= x for 0 ≤≤≤≤ x < 1

= b – x for x ≥≥≥≥ 1

is continuous in [–2, 2]. Show that (a + b) is

even

Ans f (x ) is continuous in [–2, 2].

∴∴∴∴ f (x ) is continuous at = 0x .

∴∴∴∴ ( ) ( ) ( )lim = 0 = limf x f f x– +0 0x x→ →→ →→ →→ →

( )lim + = limx a x0 0x x→ →→ →→ →→ →

∴∴∴∴ 0 + a = 0 i.e. a = 0

Also, f (x ) is continuous at =1x .

∴∴∴∴ ( ) ( ) ( )lim = 1 = limf x f f x– +1 1x x→ →→ →→ →→ →

∴∴∴∴ ( ) ( )lim = 1 = lim –x f b x1 1x x→ →→ →→ →→ →

∴∴∴∴ 1 = b – 1 i.e. b = 2

As a + b = 0 +2 = 2, (a + b ) is even.

Q-6) Show that ( )

sin, < 0

=

+1 , 0

xx

f x x

x x ≥≥≥≥

is a continuous function at x = 0.

Ans If ( ) ( ) ( )lim = 0 = limf x f f x– +0 0x x→ →→ →→ →→ →

, then f (x ) is a

continuous function.

∴∴∴∴sin

limx

x0x→→→→ = 0 +1 = ( )lim +1x

+0x→→→→

∴∴∴∴ 1 = 1 = 1

Hence, f (x ) is continuous at x = 0.

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