Page 1
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 β β β β
100
=3
for = 0x
at = 0x
Ans. ( )100
0 =3
f .... (given)
( )
1log100 + log +
100lim = lim
3
x
f xx
0 0x xβ ββ ββ ββ β
( )log 1+100 100
= lim100 3
x
x0xββββΓΓΓΓ
( )log 1+100
= 1 lim =13
x
x
0xββββ
ΓΓΓΓ β΅β΅β΅β΅
100
=3
Thus ( ) ( )lim = 0f x f0xββββ
β΄β΄β΄β΄ f is continuous at x = 0
ii) ( ) ,
log β log7= for 7
at = 7β 7
=7, for = 7
xf x x
xx
x
β β β β
Ans. ( )7 = 7f .... (given)
( )log β log 7
lim = limβ 7
xf x
x0 0x xβ ββ ββ ββ β
Put β 7 =x h , then = 7 +x h , as
7, 0x hβ ββ ββ ββ β
( )( )( )
log +7 β log 77 + = lim
+7 β 7
hf h
h0xββββ
+ 7log
7=
h
h
1 +7 + 7= lim = limlog
7 7
h h
h
17 7
0 0
h
h hβ ββ ββ ββ β
β΄β΄β΄β΄ ( ) ( )1 1
= log = 77 7
f x e f
β β β β
Since ( ) ( )lim 7 ,f x f7xββββ
β β β β f is discontinuous
at x = 7
iii) ( )1
,
β sin= for
2β
at =22
=3, for =2
x xf x x
xx x
x
xx
β β β β
Ans. = 32
f
ΟΟΟΟ.... (given)
( )1 β sin
lim = lim
β2
xf x
xΟ
2
2 2x x
Ο ΟΟ ΟΟ ΟΟ Οβ ββ ββ ββ β
.... (i)
1 β sinR.H.S = lim
β2
x
x
2
2x
ΟΟΟΟββββ ΟΟΟΟ
Put β =2
x hΟΟΟΟ
β΄β΄β΄β΄ = β2
x hΟΟΟΟ
As , 02
x hΟΟΟΟ
β ββ ββ ββ β
β΄β΄β΄β΄ R.H.S1 β sin β
2= lim
h
h
20hββββ
ΟΟΟΟ
1β cos= lim
h
h20hββββ
Page 2
Continuity
2 Mahesh Tutorials Science
1=2
β΄β΄β΄β΄ From equation (i),
( )1
lim =2 2
f x fΟ
2x
ΟΟΟΟββββ
β β β β
β΄β΄β΄β΄ f is continuous at 2
ΟΟΟΟ.
iv) ( ) ,
= for 0at = 0
= , for = 0
xf x x
x x
c x
β β β β
Ans. ( ) =x
f xx
.... (given)
Thus, =x x if 0x +ββββ
= βx if 0x βββββ
β΄β΄β΄β΄ ( ) = =1x
f xx
if 0x +ββββ
= β1 if 0x βββββ
Now ( ) ( )lim = lim β1= β1= 0f x fβ β0 0x xβ ββ ββ ββ β
.... (say)
Then, ( ) ( )lim = lim 1=1 0f x f+ +0 0x xβ ββ ββ ββ β
β β β β
Since Left hand limit β β β β Right hand limit,
f is discontinuous at x = 0
v) ( )2
2
β 9= for 0 3
β 3
= +3 for3 6
β 9= for6 9
+ 3
xf x x
x
x x
xx
x
< << << << <
β€ < β€ < β€ < β€ <
β€ <β€ <β€ <β€ <
at = 3x and = 6x
Ans. ( )3 = 3+3 = 6f
( )β 9
lim = limβ 3
xf x
x
2
3β 3x xβ ββ ββ ββ β
( ) ( )
( )
+ 3 β 3lim=
β 3
x x
x3xββββ
( )= lim + 3x3xββββ
= 6
( )= 3f
β΄β΄β΄β΄ f is left continuous at 3
( )lim lim +3=f x x3+ 3x xβ ββ ββ ββ β
= 6
( )= 3f
β΄β΄β΄β΄ f is right continuous at 3
Hence f is continuous at 3
( )6 β 9 27
6 = = 3=6+3 9
f2
( ) ( )lim lim +3=f x x6β 3x xβ ββ ββ ββ β
= 9
( )6fβ β β β
β΄β΄β΄β΄ f is not left continuous at x = 6
β΄β΄β΄β΄ f is not continuous at 6
vi) ( )( )
( )
( )
2 3
2
2
+ 3 +5 +
... + 2 β1 β= , 1
at =1β1
β1= , =1
3
n
x x x
n x nf x x
xx
n nx
β β β β
Ans. ( )( )β1
1 =3
n nf
2
.... (given)
Consider the sum ( )1+3+5...+ 2 β1n
This is sum of an A.P with =1, = 2a d and
the number of terms is n.
β΄β΄β΄β΄ [ ] ( )= + = 1+ 2 β1 =2 2
n nS a t n n
2
n n
Thus, ( )=1+3+5+. . . + 2 β1n n2.
We, shall replace 2n by R.H.S. in the
following limit
Now,
( )
( )
( )
+ 3 +5 ...+ 2 β1
βlim = lim
β1
x x x n
x nf x
x
2 3
2
1 1
n
x xβ ββ ββ ββ β
( )
( )
( )
+3 +5 +...+ 2 β1
β 1+3+5... 2 β1= lim
β1
x x x n x
n
x
2 3
1
n
xββββ
( ) ( ) ( )( ) ( )
( )
β1 + 3 β 3 + 5 β 5 +...+
2 β1 β 2 β1= lim
β1
x x x
n x n
x
2 3
1
n
xββββ
Page 3
Mahesh Tutorials Science 3
Continuity
( ) ( ) ( )( ) ( )
( )
β1 +3 β1 +5 β 5 +...+
2 β1 β1= lim
β1
x x x
n x
x
2 3
1
n
xββββ
( )( ) ( )
( ) ( )( )
1+3 +1 + + +1 +...+β1
2 β1 + + ...+1= lim
β1
x x xx
n x x
x
2
β1 β2
1
n n
xββββ
( ) ( ) ( ) ( ) ( )=1+3 2 +5 2 +7 4 +...+ 2 β1n n
( ) ( )= 2 β1r rβ=1
n
r
= 2 βr rβ β2
1 1
n n
( ) ( ) ( )2 +1 2 +1 +1= β
6 2
n n n n n
( ) ( )+1 2 +1= β1
2 3
n n n n
( ) ( )+1 4 β1=
6
n n n
β΄β΄β΄β΄ ( ) ( )lim 1f x f1xββββ
β β β β
β΄β΄β΄β΄ 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) ( )
β3 β 3= , for 0,
at 0sin
= , for 0
x x
f x xxx
k x =
β β β β =
Ans. ( )0 =f k .... (given)
( )3 β 3
lim = limsin
f xx
β
0 0
x x
x xβ ββ ββ ββ β
( ) ( )3 β1 β 3 β1= lim
sin
x x
x x
β
0
x
xββββΓΓΓΓ
3 β1 3 β1= lim β lim lim
sin
x
x x x
β
0 0 0
x x
x x xβ β ββ β ββ β ββ β βΓΓΓΓ
( )= log 3 β log 3 1
β1 ΓΓΓΓ
= log 3 + log 3
= 2 log 3
Since f is continuous at x = 0
( ) ( )lim = 0f x f0xββββ
β΄β΄β΄β΄ 2log 3=k
β΄β΄β΄β΄ 2log 3=log 9k =
ii) ( )
= β 3 , for 3at = 3
= , for = 3
f x x xx
k x
β β β β
Ans. ( )3 =f k .... (given)
( )lim = lim β 3f x x3 3x xβ ββ ββ ββ β
= 0
Since, f is continuous at x = 3
( ) ( )lim = 0f x f3xββββ
β΄β΄β΄β΄ 0 = k
i.e., = 0k
iii) ( )
2
1 β cos4= , for 0
= , for = 0 at = 0
= β 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 fβ0xββββ
β΄β΄β΄β΄1 β cos4
lim =x
kx20xββββ
β΄β΄β΄β΄1 β cos 4
lim 16 =16
xk
x20xββββΓΓΓΓ
β΄β΄β΄β΄1
16 =2
kΓΓΓΓ
i.e., = 8k
iv) ( ) ( )
2cot2= sec , for 0
at = 0= , for = 0
x
f x x xx
k x
β β β β
Ans. ( )0 =f k .... (given)
( ) ( )lim = lim secf x x2cot
2
0 0
x
x xβ ββ ββ ββ β
( )= lim 1+ tan x2cot
2
0
x
xββββ
Page 4
Continuity
4 Mahesh Tutorials Science
( )= lim 1+ =e h e
1
0
h
xβββββ΅β΅β΅β΅
Since, f is continuous at 0,
( ) ( )lim = 0f x f0xββββ
β΄β΄β΄β΄ k = e
v) ( )3 β tan
= , forβ 3 3
xf x x
x
ΟΟΟΟβ β β β
ΟΟΟΟ
= , for =3
k xΟΟΟΟ
at = .3
xΟΟΟΟ
Ans. ( )3 β tan
lim = limβ 3
xf x
x3 3
x xΟ ΟΟ ΟΟ ΟΟ Ο
β ββ ββ ββ β ΟΟΟΟ
Put = +3
x hΟΟΟΟ
As , 03
x hΟΟΟΟ
β ββ ββ ββ β
So,
( )3 β tan β
3lim = lim
β 3 β3
h
f x
h
0
3
hx
ΟΟΟΟ ββββββββ
ΟΟΟΟ
ΟΟΟΟΟΟΟΟ
tan + tan3lim 3 β
1 β tan + tan3
=β3
h
h
h
0hββββ
ΟΟΟΟ
ΟΟΟΟ
3 β tan3 β 3 β
1 β 3 tan= lim
β3
h
h
h0hββββ
3 β 3 tan β 3 β tan= lim
β3
h h
h0hββββ
4tan= lim
β3
h
h0hββββ
4 tan 1= lim lim3 1+ 3 tan
h
h h0 0h hβ ββ ββ ββ βΓΓΓΓ
4=3
β΄β΄β΄β΄ ( )4
lim =3
f x
3x
ΟΟΟΟββββ
.... (i)
Also =3
f k
ΟΟΟΟ.... (ii)
Since f (x) is continuous at =3
xΟΟΟΟ
( )lim =3
f x f
3x
ΟΟΟΟββββ
ΟΟΟΟ
β΄β΄β΄β΄4=
3k
β΄β΄β΄β΄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 β= , for 0
at = 2
= 2 , for = 0
x xf x x
xx
x
β β β β
Ans. ( )0 = 2f .... (given)
( )( )sin β
lim = limx x
f xx
2
0 0x xβ ββ ββ ββ β
( )( )
( )sin β1
= lim β1β1
x xx
x x0xββββΓΓΓΓ
( )=1 0 β1
= β1
Thus ( ) ( )lim 0f x f0xββββ
β β β β
β΄β΄β΄β΄ f is a discontinuity at x = 0
However, the discontinuity is removable.
To remove the discontinuity, we defind f as
( )( )sin β
= for 0x x
f x xx
2
β β β β
= β1 for =0x
ii) ( )( )
1 β sin= , for
β 2 2
2= ,for =7 2
xf x x
x
x
ΟΟΟΟβ β β β
ΟΟΟΟ
ΟΟΟΟ
Page 5
Mahesh Tutorials Science 5
Continuity
Ans.2
=2 7
fΟ
.... (given)
( )( )1 β sin
lim = limβ 2
xf x
x2 2
x xΟ ΟΟ ΟΟ ΟΟ Ο
β ββ ββ ββ β ΟΟΟΟ
Put β = ,2
x hΟΟΟΟ
β΄β΄β΄β΄ = β2
x hΟΟΟΟ
As , 02
x hΟΟΟΟ
β ββ ββ ββ β . Also β 2 = 2x hΟΟΟΟ
β΄β΄β΄β΄ ( )( )
20
2
hx
ΟΟΟΟ ββββββββ
ΟΟΟΟ1 β sin β
2lim = lim
2
h
f xh
1 β cos
= lim4
h
h 20hββββ
1 1 β cos
= lim4
h
h 20hββββ
1 1
=4 2
1=8
β΄β΄β΄β΄ ( )lim2
f x f
2x
ΟΟΟΟββββ
ΟΟΟΟβ β β β
β΄β΄β΄β΄ f is discontinuous at 2
ΟΟΟΟ. However, the
discontinuity can be removed by
redefining f as
( )( )
1β sin= ,for
2β 2
xf x x
x2
ΟΟΟΟβ β β β
ΟΟΟΟ
1= ,for =8 2
xΟΟΟΟ
iii) ( )
=
4 β, for 0=
6 β1
2= log , for 0
3
x x
x
ef x x
x
β β β β
Ans. ( )2
0 = log3
f
.... (given)
( )4 β
lim = lim6 β1
ef x
0 0
x x
xx xβ ββ ββ ββ β
4 β= lim
6 β1
e x
x0
x x
xxββββΓΓΓΓ
4 β1 β1= lim β lim
6 β1
e x
x x
0 0
x x
xx xβ ββ ββ ββ βΓΓΓΓ
( )1
= log 4 β loglog 6
e ΓΓΓΓ
4log
=log 6
e
4
= loge
6
( )0fβ β β β
β΄β΄β΄β΄ f has discontinuity at x = 0 . However,
this discontinuity is removable. It can
be removed by redefining f as
( )4 β
= for 06 β1
ef x x
x x
xβ β β β
4
= loge
6
iv) ( )( )
[ ] { }
8 β1, β1,1 β 0=
sin log 1+4
x
f x xx
x
ββββ
Define f(x) in [β1,+1] so that it is continuous
in [β1, +1].
Ans. ( )( )8 β1
lim lim
sin log 1+4
f xx
x
2
0 0
x
x xβ ββ ββ ββ β
( )8 β1= lim
sinlog 1+
4
x x
xxx
2
20ββββΓ ΓΓ ΓΓ ΓΓ Γ
x
x
8 β1 1= lim lim
sinlog 1+
4lim
x
xx x
x
2
0 0
0
β ββ ββ ββ β
ββββ
Γ ΓΓ ΓΓ ΓΓ Γx
x x
x
( )
1
1= log 8 14
2Γ ΓΓ ΓΓ ΓΓ Γ
( )= 4 log 82
β΄β΄β΄β΄ f would be continuous in [ ]β1,1 if
Page 6
Continuity
6 Mahesh Tutorials Science
( )( )8 β1
=
sin log 1+4
xf x
xx
2
for [ ] { }β1,1 β 0x ββββ
( )= 4 log 82
for x = 0
Q-4) If ( ) 2
2 β 1+ sin= ,
cos
xf x
x for
2x
ΟΟΟΟβ β β β is
continuous at =2
xΟΟΟΟ find
2
fΟΟΟΟ
.
Ans. ( )2 β 1+ sin
lim = limcos
xf x
x2
2 2
Ο ΟΟ ΟΟ ΟΟ Οβ ββ ββ ββ βx x
2 β 1+ sin 2 + 1+sin= lim
cos 2 + 1+sin
x x
x x2
2
ΟΟΟΟββββ
ΓΓΓΓx
( )
( )2 β 1+ sin
= limcos 2 + 1+sin
x
x x2
2
ΟΟΟΟββββx
1 β sin 1= lim lim
1 β sin 2 + 1+sin
x
x x2
2 2
Ο ΟΟ ΟΟ ΟΟ Οβ ββ ββ ββ β
ΓΓΓΓx x
( ) ( )1 β sin 1
= lim1+ sin 1 β sin 2 + 2
x
x x2
ΟΟΟΟββββ
ΓΓΓΓx
( ) ( )1 1
= lim1+sin 2 2x
2
ΟΟΟΟββββ
ΓΓΓΓx
( )1 1
=1+12 2
1=4 2
Q-5) If ( )( )β
2
1β sin= ,
2
xf x
xΟΟΟΟ for
2x
ΟΟΟΟβ β β β is continuous
at =2
xΟΟΟΟ find
2
fΟΟΟΟ
.
Ans. ( )( )
1 β sinlim = lim
β 2
xf x
x2
2 2x x
Ο ΟΟ ΟΟ ΟΟ Οβ ββ ββ ββ β ΟΟΟΟ
Put β = ,2
x hΟ
β΄β΄β΄β΄ ( )= β β 2 = 22
x h x hΟΟΟΟ
ΟΟΟΟ
As , 02
x hΟΟΟΟ
β ββ ββ ββ β
β΄β΄β΄β΄ ( )( )
20
2
ΟΟΟΟ ββββββββ
1 β sin β2
lim = lim2
h
f xh
Ο
hx
0
1 β cos 1= lim
4
h
h 2ββββΓΓΓΓ
h
1 1=4 2
ΓΓΓΓ
1=8
Since f is continuous at =2
xΟΟΟΟ
( )= lim2
f f xΟ
2
ΟΟΟΟββββx
1=8
Q-6) If ( )+14 β 2 +1
= ,1 β cos
x x
f xx
for 0x β β β β is continuous
at x = 0 find f (0).
Ans. ( )4 β 2 +1
lim = lim1β cos
f xx
+1
0 0
x x
x xβ ββ ββ ββ β
( )2 β 2.2 +1= lim
1 β cos x
2
0
x x
xββββ
( )2 β1= lim
1 β cos x
2
0
x
xββββ
2 β1= lim
1 β cos
x
x x
22
0
x
xββββΓΓΓΓ
( )= log 2 22
ΓΓΓΓ
Since, f is continuous at x = 0
( ) ( )0 = limf f x0xββββ
( )= 2 log 22
Q-7) If ( )sin 4
= +5
Ξ±Ξ±Ξ±Ξ±x
f xx
, for > 0x
= + 4 β Ξ²Ξ²Ξ²Ξ²x , for < 0x
=1 , for = 0x
is continuous at = 0x , find Ξ±Ξ±Ξ±Ξ± and Ξ²Ξ²Ξ²Ξ²
Page 7
Mahesh Tutorials Science 7
Continuity
Ans. ( )0 =1f .... (given)
Since, f is continuous at 0, it is right
continuos at 0
β΄β΄β΄β΄ ( ) ( )lim = 0f x f+0ββββx
β΄β΄β΄β΄sin4
lim + =15
x
x0ββββΞ±Ξ±Ξ±Ξ±
x
β΄β΄β΄β΄sin4 4
lim + =14 5
x
x0ββββΓ Ξ±Γ Ξ±Γ Ξ±Γ Ξ±
x
β΄β΄β΄β΄4+ =1
5Ξ±Ξ±Ξ±Ξ±
Also f is left continuous at 0.
β΄β΄β΄β΄ ( ) ( )β0xββββ
lim = 0f x f
β΄β΄β΄β΄ lim + β =1x0ββββ
Ξ²Ξ²Ξ²Ξ²x
β΄β΄β΄β΄ 4 β =1Ξ²Ξ²Ξ²Ξ²
β΄β΄β΄β΄ = 3Ξ²Ξ²Ξ²Ξ²
Thus 1
= , = 35
Ξ± Ξ²Ξ± Ξ²Ξ± Ξ²Ξ± Ξ²
Q-8) If ( )sin
= +β1
ΟΟΟΟΞ±Ξ±Ξ±Ξ±
xf x
x, for 0β€β€β€β€x
= 2ΟΟΟΟ , for = 0x
( )2
1+ cos= +
1 β
ΟΟΟΟΞ²Ξ²Ξ²Ξ²
ΟΟΟΟ
x
x, for > 0x
is continuous at =1x , find Ξ±Ξ±Ξ±Ξ± and Ξ²Ξ²Ξ²Ξ²
Ans. ( )1 = 2f ΟΟΟΟ .... (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 fβ1ββββx
β΄β΄β΄β΄sin
lim + = 2β1
x
x1ββββ
ΟΟΟΟΞ± ΟΞ± ΟΞ± ΟΞ± Ο
x
β΄β΄β΄β΄( )sin 1+
lim + = 2h
h0ββββ
ΟΟΟΟΞ± ΟΞ± ΟΞ± ΟΞ± Ο
h
(Putting β1 , 1 ,x h x h= β΄ = + As 1, 0x hβ ββ ββ ββ β )
β΄β΄β΄β΄( )sin +
lim + = 2h
h0ββββ
Ο ΟΟ ΟΟ ΟΟ ΟΞ± ΟΞ± ΟΞ± ΟΞ± Ο
h
β΄β΄β΄β΄β sin
lim + = 2h
h0ββββ
ΟΟΟΟΓ Ο Ξ± ΟΓ Ο Ξ± ΟΓ Ο Ξ± ΟΓ Ο Ξ± Ο
ΟΟΟΟh
β΄β΄β΄β΄ β + = 2Ο Ξ± ΟΟ Ξ± ΟΟ Ξ± ΟΟ Ξ± Ο
β΄β΄β΄β΄ = 3Ξ± ΟΞ± ΟΞ± ΟΞ± Ο
Since f is right continuous at x = 0
( ) ( )lim = 0f x f+1ββββx
β΄β΄β΄β΄ ( )
1+ coslim + = 2
1 β
x
x21ββββ
ΟΟΟΟΞ² ΟΞ² ΟΞ² ΟΞ² Ο
ΟΟΟΟx
β΄β΄β΄β΄( )1+ cos 1+
lim + = 2h
h 20ββββ
ΟΟΟΟΞ² ΟΞ² ΟΞ² ΟΞ² Ο
ΟΟΟΟh
(Putting β1 , 1 ,x h x h= = +β΄β΄β΄β΄ As 1,x ββββ )
β΄β΄β΄β΄( )1 cos +
lim + = 2h
h 20ββββ
Ο ΟΟ ΟΟ ΟΟ ΟΞ² ΟΞ² ΟΞ² ΟΞ² Ο
ΟΟΟΟh
+
β΄β΄β΄β΄1 cos
lim + = 2h
bh
Ο20ββββ
ΟΟΟΟΓ ΟΓ ΟΓ ΟΓ Ο
ΟΟΟΟh
β
β΄β΄β΄β΄1+ = 2
2b
Ο ΟΟ ΟΟ ΟΟ Ο
β΄β΄β΄β΄3
= 2 β =2 2
bΟ ΟΟ ΟΟ ΟΟ Ο
ΟΟΟΟ
Thus, 3
= 3 , =2
a bΟ
Ο
Q-9) Find the values of a and b so that the
function define by:
( )
5 , if 2
= + if 2 < <10
21 , if 10
β€β€β€β€
β₯β₯β₯β₯
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+2ββββx
β΄β΄β΄β΄ ( ) ( )lim = 2f x f+2ββββx
β΄β΄β΄β΄ 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 fβ10ββββx
Page 8
Continuity
8 Mahesh Tutorials Science
β΄β΄β΄β΄ lim + = 21ax b10ββββx
β΄β΄β΄β΄ 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 + β, 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 leftβ
continuous = 0x
i.e. ( ) ( )lim = 0f x fβ0ββββx
β΄β΄β΄β΄( )sin +1 +sin
lim =a x x
cx0ββββx
β΄β΄β΄β΄
( )+1 +2sin cos
2 2lim =
a x x ax
cx
0ββββx
β΄β΄β΄β΄( )
( )( )
sin + 22lim +2 limcos =
+2 2
a x axa c
a x0 0β ββ ββ ββ βΓ ΓΓ ΓΓ ΓΓ Γ
x x
β΄β΄β΄β΄ ( )2 +2 =a c
β΄β΄β΄β΄ 2 β = β 4a c .... (i)
Also f is right continuous at = 0x
β΄β΄β΄β΄ ( ) ( )lim = 0f x f+0ββββx
β΄β΄β΄β΄+ β
lim =x bx x
cb x
2
0ββββx
β΄β΄β΄β΄( )1+ β1
lim =x bx
cb x0ββββx
β΄β΄β΄β΄1+ β1
lim =bx
cb0ββββx
β΄β΄β΄β΄ = 0c (provided) 0b β β β β
β΄β΄β΄β΄ 2 = β 4a
β΄β΄β΄β΄ = β 2a
β΄β΄β΄β΄ = β2a , b any nonβzero number, = 0c
GROUP (A)- HOME WORK PROBLEMS :
Q-1) Examine the continuity of the following
functions at given points.
i) ( )
5 2β, 0
= sin3
1 , = 0
β β β β x xe e
xf x x
x
at x = 0.
Ans. ( )0 =1f .... (given)
βlim = lim
sin3
e e
x
5 2
0 0β ββ ββ ββ β
x x
x x
( ) ( )β1 β β1 3 1= lim
sin3 3
e e x
x x
5 2
0ββββΓ ΓΓ ΓΓ ΓΓ Γ
x x
x
1 β1 β1 3= lim β lim lim3 sin3
e e x
x x x
5 2
0 0 0β β ββ β ββ β ββ β βΓΓΓΓ
x x
x x x
[ ]1
= 5 β 2 13
ΓΓΓΓ
=1
ii) ( ) =
=
2
β1, for 1
β1
, for =1
nxf x x
x
n x
β β β β at x = 1.
Ans. ( )1 =f n2.... (given)
β1lim = lim
β1
x
x1 1β ββ ββ ββ β
n
x x
=n
Thus ( ) ( )lim 1f x f1ββββ
β β β β x
f is discontinuous at =1x
iii) ( ) ( )=
=
1
2
1+2 , for 0
,for = 0
xx xf x
e x
β β β β at x = 0.
Ans. ( )0f e= 2.... (given)
( ) ( )lim = lim 1+ 2f x x1
0 0
x
x xβ ββ ββ ββ β
e2=
Page 9
Mahesh Tutorials Science 9
Continuity
Thus ( ) ( )lim = 0f x f0xββββ
f is continuous at = 0x .
iv) ( )==
=
10 +7 β14 β 5for 0
12 at210
for = 07
x x x x
f x x
x
x
β β β β
Ans. ( )10
0 =7
f .... (given)
( )10 +7 β14 β 5
lim = lim1 β cos4
f xx0 0
x x x x
x xβ ββ ββ ββ β
10 β 5 β14 +7= lim
1β cos4x0
x x x x
xββββ
( ) ( )5 2 β1 β 7 2 β1= lim
1 β cos4x0
x x x x
xββββ
( ) ( ) ( )2
0ββββΓ Γ ΓΓ Γ ΓΓ Γ ΓΓ Γ Γ
2 β1 5 β 7 4 1= lim
1 β cos4 16
x
x x x
x x x
x
( )
1 2 β1 5 β 7= lim lim16
4lim
1 β cos4
x x
x
x
0 0
2
0
x x x
x x
x
β ββ ββ ββ β
ββββΓΓΓΓ
( )1 5
= log 2 log 216 7
ΓΓΓΓ
( )1 5
= log 2 log8 7
β΄β΄β΄β΄ ( ) ( )lim 0f x f0xββββ
β β β β
β΄β΄β΄β΄ f is discontinuous at = 0x
v) ( )=
= sin β cos for 0at = 7
= β1 , for 1
f x x x , xx
x
β β β β
Ans. ( )0 = β1f .... (given)
( ) ( )lim = lim sin β cosf x x x0 0x xβ ββ ββ ββ β
= sin0 β cos0
= 0 β1
= β1
Thus ( ) ( )lim = 0f x f0xββββ
β΄β΄β΄β΄ f is continuous at = 0x
vi) ( )
=
1
1
β 1= for 0
at = 0+1
= β1 , for 0
x
x
ef x , x
xe
x
β β β β
Ans. ( )0 =1f .... (given)
( )β1
lim = lim
+1
ef x
e
1
10 0
x
x xx
β ββ ββ ββ β
1 β= lim
1+
e
e
1β
10 β
x
xx
ββββ
Suppose 0x ββββ
As 1 β1
0, , βxx x
ββ β β β΄ ββ β β β΄ ββ β β β΄ ββ β β β΄ β
β΄β΄β΄β΄ lim = 0e1
β
0ββββ
x
x
β΄β΄β΄β΄ ( )1 β 0
lim = =11+0
f x0ββββx
( )= 0f
Let 0x βββββ . then 1
βx
β ββββ
β΄β΄β΄β΄ ( )0 β1
lim f =0 +1
xβ0ββββx
= β1
( )0fβ β β β
β΄β΄β΄β΄ f is not left continuous at = 0x
and hence not continuous at = 0x
vii) ( )=
1= , for 0
12at
1 2=1 β for 1
2
f x x x
x
x , x
β€ <β€ <β€ <β€ <
β€ <β€ <β€ <β€ <
Ans.1 1 1=1 β =
2 2 2f
( )lim = limf x x
β
1 1
2 2x xβ ββ ββ ββ β
1
=2
1
2f
=
Page 10
Continuity
10 Mahesh Tutorials Science
β΄β΄β΄β΄ f is left continuous at 1
2
( ) ( )lim = lim 1βf x x
+
1 1
2 2x xβ ββ ββ ββ β
1
=2
1
2f
=
β΄β΄β΄β΄ f is right continuous at 1
2
Thus, f is continuous at 1
2
viii) ( )sin2
= , for 021 β cos2
cos= for 1
β 2 2
at =2
xf x x
x
x, x
x
x
ΟΟΟΟ< β€< β€< β€< β€
ΟΟΟΟ< β€< β€< β€< β€
ΟΟΟΟ
ΟΟΟΟ
Ans.sin
= = 02 1 β cos
f
Ο ΟΟ ΟΟ ΟΟ Ο
ΟΟΟΟ.... (i)
( )cos
lim = limβ 2
xf x
x+
22xx
ΟΟΟΟΟΟΟΟ ββββββββ ΟΟΟΟ
In R.H.S., put β =2
x hΟΟΟΟ
β 2 = 2x hΟΟΟΟ
As , 02
x hΟΟΟΟ
β ββ ββ ββ β
( )+ 0
2
ΟΟΟΟ ββββββββ
ΟΟΟΟcos β
2lim = lim
2
h
f xh
hx
sin= lim
2
h
h0hββββ
0ββββ
1 sin= lim2
h
h
h
( )1
= 12
1=2
.... (ii)
β΄β΄β΄β΄ from (i) and (ii)
( )+
2
ΟΟΟΟββββ
ΟΟΟΟβ β β β lim
2f x f
x
β΄β΄β΄β΄ f is not continuous at
2x
ΟΟΟΟ=
ix) ( )( ) ( )log 2 + β log 2 β
= , for 0tan
=1 for = 0
at = 0
x xf x x
x
, x
x
β β β β
Ans. ( )0 =1f .... (given)
( )( ) ( )log 2+ β log 2 β
lim = lim2
x xf x
0 0x xβ ββ ββ ββ β
2+log
2 β= lim
tan
x
x
x
0xββββ
1+2log
1 β2
= limtan
x
x
x
0xββββ
log 1+ β log 1 β2 2
= limtan
x x
x
x x
0xββββΓΓΓΓ
log 1+ log 1 β2 2
= lim β lim
x x
x x
0 0x xβ ββ ββ ββ β
limtan
x
x0xββββΓΓΓΓ
1 β1= β 12 2
ΓΓΓΓ
=1
Thus ( ) ( )lim = 0f x f0xββββ
β΄β΄β΄β΄ f is continuous at = 0x
Page 11
Mahesh Tutorials Science 11
Continuity
Q-2) Find the value of k, so that the function f (x)
is continuous at the indicated point
i) ( )( )
2
β 1 sin= for 0 at = 0
= 4 for = 0
kxe kxf x x
xx
x
β β β β
Ans. ( )0 = 4f .... (given)
( )( )β1 sin
lim = lime kx
f xx20 0
kx
x xβ ββ ββ ββ β
β1 sin= lim
e kxk
kx kx
2
0
kx
xββββΓΓΓΓ
=1 1 k2Γ ΓΓ ΓΓ ΓΓ Γ
=k2
Since, f is continuous at = 0x
( ) ( )lim = 0f x f0xββββ
β΄β΄β΄β΄ = 4k2
β΄β΄β΄β΄ = 2k Β±
ii) ( ) 2
2
= +1 for 0
= +1 + for < 0
at = 0
f x x x
x k x
x
β₯β₯β₯β₯
Ans. ( )0 = 0 +1=1f 2
( ) ( )lim = lim 2 +1+f x x kβ
2
00 xx ββββββββ
= 2 0+1+k
= 2+k
Since, f is continuous at = 0x , it is left
continuous at 0.
β΄β΄β΄β΄ ( ) ( )lim = 0f x fβ0xββββ
β΄β΄β΄β΄ + 2 =1k
β΄β΄β΄β΄ = β1k
iii) ( )( )+
log 1= , for 0
at = 0sin
= 5 , for = 0
kxf x x
xx
x
β β β β
Ans ( )0 =5f .... (given)
( )( )log 1+
lim = limsin
kxf x
x0 0x xβ ββ ββ ββ β
( )0ββββ
Γ ΓΓ ΓΓ ΓΓ Γlog 1+
= limsin
kx xk
kx xx
( )log 1+= lim lim
sin
kx x
kx x
0 0x xβ ββ ββ ββ β
( ) ( )= 1 1k
=k
Since, f is continuous at = 0x
β΄β΄β΄β΄ ( ) ( )lim = 0f x fβ0xββββ
β΄β΄β΄β΄ = 5k
iv) ( )
8 β 2= ,for 0
at = 0β1
= 2 , for = 0
x x
xf x x
xk
x
β β β β
Ans. ( )0 = 2f .... (given)
( )8 β 2
lim = limβ1
f xk0 0
x x
xx xβ ββ ββ ββ β
8 β 2= lim
β1
x
x k0
x x
xxββββΓΓΓΓ
( ) ( )8 β1 β 2 β1= lim lim
β1
x
x k0 0
x x
xx xβ ββ ββ ββ βΓΓΓΓ
8 β1 2 β1 1= lim β lim
logx x k
0 0
x x
x xβ ββ ββ ββ βΓΓΓΓ
( )log 8 β log 2=
logk
8log
2=
logk
= log 4k (change of base rule)
Since, f is continuous at = 0x
( ) ( )lim = 0f x fβ0xββββ
β΄β΄β΄β΄ log 4 = 2k
β΄β΄β΄β΄ 4 =k2
β΄β΄β΄β΄ = 2k
(β΅β΅β΅β΅base of logarithm cannot be negative
= β2k is not possible)
Page 12
Continuity
12 Mahesh Tutorials Science
v) ( ) ( )
2= β 2 , for 0at = 0
= 4 +1 , for > 0
f x k x xx
x x
β€β€β€β€
Ans. ( ) ( )0 = 0 β 2 = β2f k k
( ) ( )lim = lim 4 +1f x x+0 0x xβ ββ ββ ββ β
=1
Since, f is continuous at = 0x
( ) ( )lim = 0f x f+0xββββ
β΄β΄β΄β΄ 1= β2k
β΄β΄β΄β΄1
= β2
k
Q-3) Discuss the continuity of the following
functions, which of these functions have
a removable discontinuity ? Redefine the
function so as to remove the discontinuity
i) ( )
1 β cos3= , for 0
tan at = 0
= 9 ,for = 0
xf x x
x x x
x
β β β β
Ans. ( )0 = 9f .... (given)
( )1 β cos3
lim = limtan
xf x
x x0 0x xβ ββ ββ ββ β
1β cos3 9= lim
tan9
x x
x xx
2
20xββββΓΓΓΓ
1 β cos3= 9 lim lim
tan9
x x
xx
20 0x xβ ββ ββ ββ β
( )1
= 9 12
9=
2
β΄β΄β΄β΄ ( ) ( )lim 0f x f0xββββ
β β β β
β΄β΄β΄β΄ f has a discontinuity at = 0x
However, the discontinuity is
removable by redefining f as
( )1 β cos3
=9
xf x
x2 for 0x β β β β
9
=2
for = 0x
ii) ( )( )2
2
β1 tan= ,for 0
sin
= , for = 0
at = 0
xe xf x x
x x
e x
x
β β β β
Ans. ( )0 =f e2.... (given)
( )( )β1 tan
lim = limsin
e xf x
x x
2
0 0
x
x xβ ββ ββ ββ β
β1 tan= lim 2
2 sin
e x x
x x x
2
0
x
xββββΓ ΓΓ ΓΓ ΓΓ ΓΓΓΓΓ
β1 tan= 2 lim lim
2
e x
x x
2
0 0
x
x xβ ββ ββ ββ β
limsin
x
x
0xββββ
= 2 1 1 1Γ Γ ΓΓ Γ ΓΓ Γ ΓΓ Γ Γ = 2
β΄β΄β΄β΄ ( ) ( )lim 0f x f0xββββ
β β β β
β΄β΄β΄β΄ f has a discontinuity at = 0x
However, the discontinuity can be
removed by redefining f as
( )( )β1 tan
= for 0sin
e xf x x
x x
2x
β β β β
=2 for = 0x
iii) ( )( )3 0
2
β 1 sin= , for 0
= , for = 060
at = 0
xe xf x x
x
x
x
β β β β
ΟΟΟΟ
Ans. ( )0 =60
fΟΟΟΟ
.... (given)
( )( )β1 sin
lim = lime x
f xx
3 0
20 0
x
x xβ ββ ββ ββ β
3
0ββββ
ΟΟΟΟΟΟΟΟ
Γ Γ ΓΓ Γ ΓΓ Γ ΓΓ Γ ΓΟΟΟΟ
sinβ1 180lim 3
3 180
180
xe
xx
x
x
sin3 β1 180= lim lim180 3
180
xe
xx
3
0 0
x
x xβ ββ ββ ββ β
ΟΟΟΟΟΟΟΟ
ΓΓΓΓΟΟΟΟ
Page 13
Mahesh Tutorials Science 13
Continuity
= 1 160
ΟΟΟΟΓ ΓΓ ΓΓ ΓΓ Γ
=60
ΟΟΟΟ ( )= 0f
β΄β΄β΄β΄ f is continuous at = 0x
iv) ( ) = β1 , for1 2
= 2 +3 , for2 0
at = 0
f x x x
x x
x
β€ <β€ <β€ <β€ <
β€ β€β€ β€β€ β€β€ β€
Ans. ( ) ( )2 = 2 2 + 3 = 7f
( ) ( )lim = lim β1f x x2 2x xβ ββ ββ ββ β
= 2 β1=1
β΄β΄β΄β΄ ( ) ( )lim 2f x fβ2xββββ
β β β β
β΄β΄β΄β΄ f is not left continuous at 2 and hence
is not continuous at 2.
This discontinuity is nonβremovable,
since if we change value of f at 2, it would
be discontinuous from right at = 2x
Q-4) If ( )2
2
β cos= , for 0
xe xf x x
xβ β β β is continuous
at = 0x , find ( )0f .
Ans. ( )β cos
lim = lime x
f xx
2
20 0
x
x xβ ββ ββ ββ β
( ) ( )β1 β cos β1= lim
e x
x
2
20
x
xββββ
β1 1 β cos= lim + lim
e x
x x
2
2 20 0
x
x xβ ββ ββ ββ β
1 3=1+ =
2 2
Since f is continuous at = 0x
β΄β΄β΄β΄ ( ) ( )lim = 0f x f0xββββ
β΄β΄β΄β΄ ( )3
0 =2
f
Q-5) If ( )( ){ }
( )2
1β cos 7 β= for
5 β
xf x x
x
ΟΟΟΟβ Οβ Οβ Οβ Ο
ΟΟΟΟ is continuous
at =x ΟΟΟΟ , find ( )f ΟΟΟΟ .
Ans. ( )( ){ }
( )
1 β cos 7 βlim = lim
5 β
xf x
x2x xβΟ βΟβΟ βΟβΟ βΟβΟ βΟ
ΟΟΟΟ
ΟΟΟΟ
Put β = .x hΟΟΟΟ As , 0x hβ Ο ββ Ο ββ Ο ββ Ο β
( )1 β cos7
lim = lim5
hf x
h 2x hβΟ β0βΟ β0βΟ β0βΟ β0
1 1 β cos7= lim 49
5 49
h
h 2hβ0β0β0β0ΓΓΓΓ
1 1= 49
5 2Γ ΓΓ ΓΓ ΓΓ Γ
49=10
Since f is continuous at =x ΟΟΟΟ
( ) ( )lim =f x fxβΟβΟβΟβΟ
ΟΟΟΟ
β΄β΄β΄β΄ ( )49
=10
f ΟΟΟΟ
Q-6) If ( )( )
( )
2sin4 β1
= for 0log 1+2
x
f x xx x
β β β β is continuous
at = 0x , find ( )0f .
Ans. ( )( )
( )
4 β1lim = lim
log 1+ 2f x
x x
2sin
0 0
x
x xβ ββ ββ ββ β
( )( )
4 β1 sin= lim
log 1 2sin
x
x xx
2sin 2
20
x
xββββΓΓΓΓ
+
4 β1 sin= lim lim
sin
x
x x
2sin 2
2 20 0
x
x xβ ββ ββ ββ βΓΓΓΓ
( )lim
log 1+ 2
x
x
0xββββΓΓΓΓ
( )2
Γ ΓΓ ΓΓ ΓΓ Γ1
= log 4 12
( )1
= log 42
2
Since f is continuous at = 0x
( ) ( )0 = limf f x0xββββ
( )1
= log 42
2
Page 14
Continuity
14 Mahesh Tutorials Science
Q-7) If ( )1 β tan
= , for41 β 2 sin
xf x x
x
ΟΟΟΟβ β β β is continuous
at =4
xΟΟΟΟ, find
4
fΟΟΟΟ
.
Ans. f is continuous at 4
ΟΟΟΟ
( )1 β tan
= lim = lim4 1 β 2sin
xf f x
x
4 4x x
Ο ΟΟ ΟΟ ΟΟ Οβ ββ ββ ββ β
ΟΟΟΟ
sin1 β
cos= lim1β 2sin
x
x
x4
xΟΟΟΟ
ββββ
4
ΟΟΟΟββββ
ΓΓΓΓcos β sin 1+ 2 sin
= limcos 1+ 2 sin
x x x
x xx
( ) ( )( ) ( )2
4
ΟΟΟΟββββ
cos β sin 1+ 2 sin= lim
cos 1 β 2sin
x x x
x xx
( ) ( )( ) ( )
cos β sin 1+ 2sin= lim
cos cos + sin β 2sin
x x x
x x x x2 2 2
4x
ΟΟΟΟββββ
( ) ( )( ) ( ) ( )
cos β sin 1+ 2 sin= lim
cos cos + sin cos β sin
x x x
x x x x x4
xΟΟΟΟ
ββββ
,2
... cos β sin 0
cosβ sin 0
x
x x
x
β
β΅
β΅
ΟΟΟΟ
ββββ
β β β β
( )( ) ( )
1+ 2sin= lim
cos cos + sin
x
x x x4
xΟΟΟΟ
ββββ
( )
( )
lim 1+ 2 sin
=limcos cos + sin
x
x x x
4
4
x
x
ΟΟΟΟββββ
ΟΟΟΟββββ
11+ 2
2=
1 1 1
2 2 2
ΓΓΓΓ
1+1= = 2
1
Hence, = 24
f
ΟΟΟΟ
Q-8) If ( )1β 3tan
= , forβ 6 6
xf x x
x
ΟΟΟΟβ β β β
ΟΟΟΟ is continuous
at =6
xΟΟΟΟ, find
6
fΟΟΟΟ
.
Ans. f is continuous at 6
ΟΟΟΟ
β΄β΄β΄β΄ ( )1 β 3 tan
= lim = lim6 β 6
xf f x
x
6 6x x
Ο ΟΟ ΟΟ ΟΟ Οβ ββ ββ ββ β
ΟΟΟΟ
ΟΟΟΟ
( )cos β 3 sin
= limcos β 6
x x
x x6
xΟΟΟΟ
ββββ ΟΟΟΟ
Multiplying and dividing by 2, we get
( )
1 32 cos . β sin
2 2= lim
6 cos β 6
x x
fx x
6x
ΟΟΟΟββββ
ΟΟΟΟ
ΟΟΟΟ
( )6
ΟΟΟΟββββ
Ο ΟΟ ΟΟ ΟΟ Ο
ΟΟΟΟ
2 sin .cos β sin sin6 6
= limcos β 6
x x
x x
x
( )
sin β6
= 2 limcos β 6
x
x x
6x
ΟΟΟΟββββ
ΟΟΟΟ
ΟΟΟΟ
( )
( )
1sin β 6
2 6= limlimcos β 6
x
x x
6
6
x
x
ΟΟΟΟββββ
ΟΟΟΟββββ
ΟΟΟΟ
ΟΟΟΟ
2 1=
63
2
ΓΓΓΓ
2=
3 3
Hence 2
=6 3 3
f
ΟΟΟΟ
Page 15
Mahesh Tutorials Science 15
Continuity
Q-9) ( ) 2
2
= + , for 0
= 2 +1 + ,for 0
f x x x
x x
Ξ± β₯Ξ± β₯Ξ± β₯Ξ± β₯
Ξ² <Ξ² <Ξ² <Ξ² <
And
1= 2
2f , is continuous at = 0x , find
Ξ±Ξ±Ξ±Ξ± and Ξ²Ξ²Ξ²Ξ².
Ans.1 1
= + = 22 2
f
2
Ξ±Ξ±Ξ±Ξ± .... (given)
β΄β΄β΄β΄1
+ = 24
Ξ±Ξ±Ξ±Ξ±
β΄β΄β΄β΄7
=4
Ξ±Ξ±Ξ±Ξ±
( )7
0 = 0 + = =4
f Ξ± Ξ±Ξ± Ξ±Ξ± Ξ±Ξ± Ξ±
Since f is continuous at = 0x , it is left
continuous at 0.
β΄β΄β΄β΄ ( ) ( )lim = 0f x fβ0xββββ
β΄β΄β΄β΄7
lim2 +1+ =4
x2
0xββββΞ²Ξ²Ξ²Ξ²
β΄β΄β΄β΄7
2 1+ =4
Ξ²Ξ²Ξ²Ξ²
β΄β΄β΄β΄7 β1
= β 2 =4 4
Ξ²Ξ²Ξ²Ξ²
Thus, 7
=4
Ξ±Ξ±Ξ±Ξ± and β1
=4
Ξ²Ξ²Ξ²Ξ²
Q-10) ( )2 β 9
= +β 3
xf x
xΞ±Ξ±Ξ±Ξ± , for 3x >>>>
= 5 , for = 3x
2= 2 +3 +x x Ξ²Ξ²Ξ²Ξ²
is continuous at = 3x find Ξ±Ξ±Ξ±Ξ± and Ξ²Ξ²Ξ²Ξ².
Ans. ( )3 = 5f .... (given)
Since, f is continuous at = 3x , it is left
continuous at = 3x
i.e. ( ) ( )lim = 3f x fβ3xββββ
β΄β΄β΄β΄ lim2 + 3 + = 5x x2
3xββββΞ²Ξ²Ξ²Ξ²
β΄β΄β΄β΄ 18+ 9+ = 5Ξ²Ξ²Ξ²Ξ²
β΄β΄β΄β΄ = 5 β 27 = β22Ξ²Ξ²Ξ²Ξ²
Also, f is rightβcontinuous at = 3x
β΄β΄β΄β΄ ( ) ( )+3xββββ
lim = 3f x f
β΄β΄β΄β΄β 9
lim = 5β 3
x
x
2
3xββββΞ±Ξ±Ξ±Ξ±+
β΄β΄β΄β΄( ) ( )
3xββββΞ±Ξ±Ξ±Ξ±
+ 3 β 3lim + = 5
β 3
x x
x
β΄β΄β΄β΄ ( )lim + 3 + = 5x Ξ±3xββββ
( )β β β΄ β β β β΄ β β β β΄ β β β β΄ β β΅β΅β΅β΅ 3, 3, β 3 0x x x
β΄β΄β΄β΄ 6+ = 5Ξ±Ξ±Ξ±Ξ±
β΄β΄β΄β΄ = β1Ξ±Ξ±Ξ±Ξ±
Thus = β1, = β22Ξ± Ξ²Ξ± Ξ²Ξ± Ξ²Ξ± Ξ²
Q-11)If ( )f x is defined by
( ) = sin2 if6
= + if6
f x x x
ax b x
ΟΟΟΟβ€β€β€β€
ΟΟΟΟ>>>>
Find a and b if ( )f x and ( )f xβ² are
continuous at 2
x =ΟΟΟΟ
Ans.3
= sin2 =6 6 2
f
Ο ΟΟ ΟΟ ΟΟ Ο
Since, f is continuous at =6
xΟΟΟΟ, it is left
continuous at 6
ΟΟΟΟ.
i.e., ( )lim =6
f x f
6x
ΟΟΟΟββββ
ΟΟΟΟ
β΄β΄β΄β΄3
lim + =2
ax b
6x
ΟΟΟΟββββ
β΄β΄β΄β΄3
+ =6 2a b
ΟΟΟΟ.... (i)
Now ( ) = 2cos2 ,f x xβ² for6
xΟΟΟΟ
β€β€β€β€
=a for6
xΟΟΟΟ
>>>>
β΄β΄β΄β΄Ο ΟΟ ΟΟ ΟΟ Ο
= 2cos26 6
f
β²
Page 16
Continuity
16 Mahesh Tutorials Science
= 2cos2 =13
ΟΟΟΟ
Since f β²β²β²β² is continuous at =6
xΟΟΟΟ, it is right
continuous at 6
ΟΟΟΟ
β΄β΄β΄β΄ ( )+6
xΟΟΟΟ
ββββ
ΟΟΟΟlim =
6f x f
β² β²
β΄β΄β΄β΄lim =1a
6x
ΟΟΟΟββββ
β΄β΄β΄β΄ =1a
Putting this value of a in (i)
( )3
1 + =6 2
bΟΟΟΟ
β΄β΄β΄β΄ 3
= β2 6
bΟΟΟΟ
Thus, 3
=1, = β2 6
a bΟΟΟΟ
Q-12) Find k so that the following function f is
continuous at =1x
( ) 2=f x kx , for 1x β₯β₯β₯β₯
= 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 fβ1xββββ
lim 4 =kβ1xββββ
= 4k
GROUP (B)- CLASS WORK PROBLEMS
Q-1) If ( )f x is defined by
( ) = + 2 sinf x x a x ,04
xΟΟΟΟ
β€ <β€ <β€ <β€ <
= 2 cot + bx x ,4 2
xΟ ΟΟ ΟΟ ΟΟ Ο
β€ <β€ <β€ <β€ <
= cos2 β sina x b x , ,2
xΟΟΟΟ
< β€ Ο< β€ Ο< β€ Ο< β€ Ο
Find a and b
Ans. Since f is continuous in [0, ΟΟΟΟ]. It is continuous
at =4
xΟΟΟΟ
and =2
xΟΟΟΟ
= 2 cot +4 4 4
f b
Ο Ο ΟΟ Ο ΟΟ Ο ΟΟ Ο Ο
( )= 1 +2
bΟΟΟΟ
= +2
bΟΟΟΟ
Since f is continuous at =4
xΟΟΟΟ
, it is left
continuous at =4
xΟΟΟΟ
i.e., ( )= lim4
f f x
β4x
ΟΟΟΟββββ
ΟΟΟΟ
+ = lim + 2sin2
b x a x
4x
ΟΟΟΟββββ
ΟΟΟΟ
Ο ΟΟ ΟΟ ΟΟ Ο
= + 2 sin4 4
a
ΟΟΟΟ
= +4
a
β =4
a bΟΟΟΟ
.... (i)
= 2 cot +2 2 2
f b
Ο Ο ΟΟ Ο ΟΟ Ο ΟΟ Ο Ο
= b cot = 02
β΅
ΟΟΟΟ
Since f is continuous at =2
xΟΟΟΟ, it is right
continuous at =2
xΟΟΟΟ.
β΄β΄β΄β΄ ( )= lim2
f f x
+4x
ΟΟΟΟββββ
ΟΟΟΟ
β΄β΄β΄β΄
+2
ΟΟΟΟββββ
= lim cos2 β sinb a x b xx
= cos β sin2
a bΟΟΟΟ
ΟΟΟΟ
= β βa b
β΄β΄β΄β΄ + = βa b b
Page 17
Mahesh Tutorials Science 17
Continuity
β΄β΄β΄β΄ = β 2a b .... (ii)
β΄β΄β΄β΄ From equation (i) β3 =4
bΟΟΟΟ
β΄β΄β΄β΄β
=12
bΟΟΟΟ
β΄β΄β΄β΄ =6
aΟΟΟΟ
Thus, β3 = , = β6 12
a bΟ ΟΟ ΟΟ ΟΟ Ο
Q-2) Find Ξ±Ξ±Ξ±Ξ± and Ξ²Ξ²Ξ²Ξ² so that the functions, defined
below is continuous is [β ΟΟΟΟ, ΟΟΟΟ]
( ) = β2sinf x x, for β2
xΟΟΟΟ
Ο β€ β€Ο β€ β€Ο β€ β€Ο β€ β€
= sin + ,xΞ± Ξ²Ξ± Ξ²Ξ± Ξ²Ξ± Ξ² for β2 2
xΟ ΟΟ ΟΟ ΟΟ Ο
< << << << <
=cos x, for2
xΟΟΟΟ
β€ β€ Οβ€ β€ Οβ€ β€ Οβ€ β€ Ο
Ans. Since f is continuous in [βΟΟΟΟ, ΟΟΟΟ]. I t is
continuous at β
2
ΟΟΟΟ and
2
ΟΟΟΟ
β β= β2sin = 2
2 2f
Ο ΟΟ ΟΟ ΟΟ Ο
Since f is continuous at β
2
ΟΟΟΟ, it is right
continuous at β
=2
xΟΟΟΟ
β΄β΄β΄β΄ ( )β
β= lim
2f f x
+2
ΟΟΟΟββββ
ΟΟΟΟ
x
β2 = lim sin +x
2
ΟΟΟΟββββ
Ξ± Ξ²Ξ± Ξ²Ξ± Ξ²Ξ± Ξ²x
β
= sin +2
ΟΟΟΟΞ± Ξ²Ξ± Ξ²Ξ± Ξ²Ξ± Ξ²
= β +Ξ± Ξ²Ξ± Ξ²Ξ± Ξ²Ξ± Ξ²
Thus, β + = 2Ξ± Ξ²Ξ± Ξ²Ξ± Ξ²Ξ± Ξ² .... (i)
= cos = 02 2
f
Ο ΟΟ ΟΟ ΟΟ Ο
Since f is continuous at =2
xΟΟΟΟ, it is left
continuous at 2
ΟΟΟΟ
β΄β΄β΄β΄ ( )lim =2
f x f
β2x
ΟΟΟΟββββ
ΟΟΟΟ
β΄β΄β΄β΄lim sin + = 0x
2x
ΟΟΟΟββββ
Ξ± Ξ²Ξ± Ξ²Ξ± Ξ²Ξ± Ξ²
β΄β΄β΄β΄ ( )1 + = 0Ξ± Ξ²Ξ± Ξ²Ξ± Ξ²Ξ± Ξ²
β΄β΄β΄β΄ + = 0Ξ± Ξ²Ξ± Ξ²Ξ± Ξ²Ξ± Ξ² .... (ii)
From equation (i) and (ii)
= β1, =1Ξ± Ξ²Ξ± Ξ²Ξ± Ξ²Ξ± Ξ²
Q-3) If ( )3
3
+3 +5=
β 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.
β΄β΄β΄β΄ ( )f x is continuous for all real values of x,
except when its denominator becomes zero.
i.e., β 3 +2 = 0x x3
β΄β΄β΄β΄ ( ) ( )β1 + β 2 = 0x x x2
β΄β΄β΄β΄ ( ) ( ) ( )β1 +2 β1 = 0x x x
β΄β΄β΄β΄ ( ) ( )β1 + 2 = 0x x2
β΄β΄β΄β΄ =1or = β2x x
But [ ]β2 0,5ββββ
β΄β΄β΄β΄ ( )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 + βL f a h f a 0ββββh
( )= lim log + β loga h a 0ββββc c
h
0ββββ
+= lim log
a h
a
ch
Page 18
Continuity
18 Mahesh Tutorials Science
0ββββΓΓΓΓ
+= lim log
a hh
a
ch
1
0ββββΓΓΓΓ= lim log 1+
hh
a
h
ch
= lim log 1+h
ha
1
0ββββΓΓΓΓ
a ah
ch
( )= log lim 1+ limh
ha
1
0 0β ββ ββ ββ βΓΓΓΓ
a ah
ch h
( ) ( )1
= log 0 = log 0e ea
1
ac c = 0
Thus, ( ) ( )lim β = 0f x f a 0ββββh
β΄β΄β΄β΄ f is continuous at =x a .
But, a is any positive ral number.
β΄β΄β΄β΄ f is continuous at all positive real number
Thus, log xc where 0, 1, 0c c x> β >> β >> β >> β > is
continuous
Q-5) Test the continuity of the function
( )( ) ( )
+1=
β 2 β 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.
( ) ( )β 2 β 5 = 0 = 2, = 5x x x xββββ
None of these points lie in [0, 1] and hence
( )f x is continuous in [0, 1]
The value [ ]= 5 4,6x ββββ 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 4xβ€ β€β€ β€β€ β€β€ β€
+1
=β 3
x
x, for 4 6x< β€< β€< β€< β€
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 4xβ€ β€β€ β€β€ β€β€ β€ .
In 4 6x< β€< β€< β€< β€ , ( )+1
=β 3
xf x
x, which is a rational
function and hence it is continuous for
4 6x< β€< β€< β€< β€
When = 4x , ( )1
=+1
f xx
β΄β΄β΄β΄ ( )1 1
4 = =4 +1 5
f
β΄β΄β΄β΄ ( )+1
lim = limβ 3
xf x
x
+ +4 4x xβ ββ ββ ββ β
4 +1
=4 β 3
5=1 ( )= 5 4fβ β β β
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 β€β€β€β€
= +ax b for 2 10x< << << << <
= 21 for 10x β₯β₯β₯β₯
is continuous in its domain
Ans The function is continuous on [2, 10].
At ( ) ( ) ( )= 2 lim = = 5 = limx f x f x f xβ +2 2x xβ ββ ββ ββ β
( )lim = 5f xβ2xββββ
( )2 = 5f
( ) ( )lim = lim + = 2 +f x ax b a b+ +2 2x xβ ββ ββ ββ β
β΄β΄β΄β΄ 2 + = 5a b .... (i)
At =10x
( )lim = lim + =10 +f x ax b a b+ +10 10x xβ ββ ββ ββ β
( )10 = 21f
β΄β΄β΄β΄ 10 + = 21a b .... (ii)
Find the value of
Page 19
Mahesh Tutorials Science 19
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 ββββ .
Ans Consider the function ( ) =g x x
β΄β΄β΄β΄ ( ) = ,g x x for 0x β₯β₯β₯β₯ , and
= β ,x for 0x <<<<
Clearing ( )0 = 0g
( ) ( )lim = lim β = 0g x xβ0 0x xβ ββ ββ ββ β
Also, ( ) ( ) ( )lim = lim = 0g x x+0 0x xβ ββ ββ ββ β
β΄β΄β΄β΄ g is continuous at = 0x
( ) ( ) ( )lim = lim = 0g x g x g+0 0x xβ ββ ββ ββ β
In ( )β ,0ββββ and in [ )0,ββββ
g is polynomial and hence continuous.
Thus, =g x is continuous at every point of R.
Similarly, 1+ x is continuous over R.
β΄β΄β΄β΄ By algebra of continuous functions, + 1+x x
is continuous over R i.e. ( )f x is continuous
for all x Rββββ .
Q-9) Prove that the exponential function ax is
continuous at every point ( )0a >>>> .
Ans Let ( ) =f x ax and t Rββββ
β΄β΄β΄β΄ ( ) =f t at
( )lim = lim =f x a a+
+
0
t h t
x t hβ ββ ββ ββ β
( )lim = lim =f x a aβ +
β
0
t h t
x t hβ ββ ββ ββ β
Thus, ( ) ( ) ( )lim = lim =f x f x f t+ βtx t xβ ββ ββ ββ β
β΄β΄β΄β΄ ( ) =f x ax is continuous at =x t . But t is
arbitrary real number.
β΄β΄β΄β΄ 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 hβ β ββ β ββ β ββ β β
= sin t
( )lim = lim βf t hβ +0x t hβ ββ ββ ββ β
( )= limsin βt h0hββββ
= sin t
Thus, ( ) ( ) ( )lim = lim =f x f x f t+ βx t x tβ ββ ββ ββ β
β΄β΄β΄β΄ sin x is continuous at t.
But t is arbitary
β΄β΄β΄β΄ 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 2xβ€ <β€ <β€ <β€ <
= 3 +2,x for2 4xβ€ β€β€ β€β€ β€β€ β€
= 2 +5 ,ax b for 4 8x< β€< β€< β€< β€
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
Since f is continuous at = 2x , it is left
continuous at = 2x .
β΄β΄β΄β΄ ( ) ( )lim = 2f x fβ2xββββ
β΄β΄β΄β΄ ( )lim + + 6 = 8x ax2
2xββββ
β΄β΄β΄β΄ 4 + 2 + 6 = 8a
β΄β΄β΄β΄ 2 = β2a
β΄β΄β΄β΄ = β1a
= 4x
Since f is continuous at = 4x , it is right
continuous at = 4x
β΄β΄β΄β΄ ( ) ( )lim = 4f x f+2xββββ
β΄β΄β΄β΄ ( )2 +5 =14ax b
Page 20
Continuity
20 Mahesh Tutorials Science
β΄β΄β΄β΄ 8 +5 =14a b
β΄β΄β΄β΄ β 8 + 5 =14b
β΄β΄β΄β΄ 5 = 22b
β΄β΄β΄β΄22
=5
b
Thus, 22
= β1, =5
a b
Q-2) If the function ( )f x defined below is
continuous in [0, 3], find the value of k.
( ) = 3 β 4,f x x for 0 2xβ€ <β€ <β€ <β€ <
= 2 + ,x k for2 3xβ€ β€β€ β€β€ β€β€ β€
Ans Since f is continuous in [0, 3]. It is continuous
at = 2x
( ) ( )2 = 3 2 β 4f
= 2
Since f is continuous at = 2x , it is left
continuous at = 2x .
β΄β΄β΄β΄ ( ) ( )2 = limf f x+2xββββ
β΄β΄β΄β΄ ( )2 = lim 2 +x k2xββββ
β΄β΄β΄β΄ 2 = 4 +k
β΄β΄β΄β΄ = β 2k
β΄β΄β΄β΄ = β1a
Q-3) Discuss the continuity of the following
function in its domain
( ) = ,f x x 0x β₯β₯β₯β₯
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 fβ +0 0β ββ ββ ββ βx x
Now, ( )lim = lim = 0f x xβ +
2
0 0β ββ ββ ββ βx x
and ( ) ( ) ( )lim = lim = 0f x f x f+ +0 0β ββ ββ ββ βx x
β΄β΄β΄β΄ ( ) ( ) ( )lim = lim = 0f x f x fβ +0 0β ββ ββ ββ βx x
Hence, f is continuous at = 0x
Thus, f is continuous in its domain R.
Q-4) Discuss the continuity of the following
function in its domain
2= β 4f x for 0 2xβ€ β€β€ β€β€ β€β€ β€
= 2 + 3x for2 4x< β€< β€< β€< β€
2= β 5x for 4 6x< β€< β€< β€< β€
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 β 4 = 0f 2
( ) ( )lim = lim 2 + 3 = 7f x x+2 2β ββ ββ ββ βx x
β΄β΄β΄β΄ ( ) ( )lim 2f x f+2ββββ
β β β β x
β΄β΄β΄β΄ f is not right continuous = 2x
i.e. f is discontinuous at = 2x
( )4 = 2 4+3 =11f ΓΓΓΓ
( ) ( )lim = lim β 5f x x+
2
4 4β ββ ββ ββ βx x=16 β 5 =11
β΄β΄β΄β΄ f is right continuous at 4
( ) ( )lim = lim 2 +3 =11f x xβ4 4β ββ ββ ββ βx x
β΄β΄β΄β΄ f is left continuous at = 4x
β΄β΄β΄β΄ 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
β€ β€β€ β€β€ β€β€ β€
β€ β€β€ β€β€ β€β€ β€
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.
Page 21
Mahesh Tutorials Science 21
Continuity
( )1 = 3f .... (given)
( )lim = lim4 = 4f x+1 1x xβ ββ ββ ββ β
β΄β΄β΄β΄ ( ) ( )lim 1f x f+1xββββ
β β β β
β΄β΄β΄β΄ f is discontinuous at =1x
( )3 = 5f .... (given)
( )lim = lim4 = 4f x+1 3x xβ ββ ββ ββ β
β΄β΄β΄β΄ ( ) ( )lim 3f x fβ1xββββ
β β β β
β΄β΄β΄β΄ 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
Yβ²
Xβ²
Q-6) Discuss the continuity of the following
function in its domain
( )
β2 , if β1
= 2 , if β 1 < 1
2 , if 1
x
f x x x
x
β€β€β€β€
β€β€β€β€
>>>>
Ans f is constant (function) in (β ββββ, β1) and (1, ββββ)
and hence continuous in these sub-intervals.
f is a polynomial in (β 1, + 1] and hence
continuous over it.
Therefore, the only possible points of
discontinuity of f are β 1 and + 1.
( )β1 = β2f .... (given)
( )lim = lim 2 = β2f x x+β1 β1x xβ ββ ββ ββ β
β΄β΄β΄β΄ f is right continuous at β 1 and hence
continuous at β 1.
( ) ( )1 = 2 1 = 2f .... (given)
( )lim = lim2 = 2f x+1 1x xβ ββ ββ ββ β
β΄β΄β΄β΄ f is right continuous at 1 and hence
continuous at 1
β΄β΄β΄β΄ f is continuous in R.
Q-7) Examine the continuity of f (x) on its
domain, where
( )
+3 ,if β3
= β 2 , if β 3 < <1
6 +2 , if 1
x x
f x x x
x x
β€β€β€β€
β₯β₯β₯β₯
Ans Domain of f is (β β, + ββ, + ββ, + ββ, + β). f is a polynomial in
each of subintervals (β β, β, β, β, β 3], (β 3, 3] and [3, ββββ)
and hence continuous in each subinterval. The
possible points of discontinuity are β 3 and + 3
It remains to check whether f is right
continuous at β 3 and left continuous at = 3x
( )β3 = β3 3 = 0f +
( ) ( )lim = lim β2 = 6f x x+β3 β3x xβ ββ ββ ββ β
β΄β΄β΄β΄ ( ) ( )lim β3f x f+β3xββββ
β β β β
β΄β΄β΄β΄ f is not continuous at = 3x
( ) ( )3 = 6 3 +2 = 20f
( ) ( )lim = lim β2 = β 6f x xβ β3 3x xβ ββ ββ ββ β
β΄β΄β΄β΄ ( ) ( )lim 3f x fβ3xββββ
β β β β
β΄β΄β΄β΄ f is not continuous at = 3x . Hence f is
continuous in (β ββββ, ββββ) except at = β 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
Page 22
Continuity
22 Mahesh Tutorials Science
BASIC ASSIGNMENTS (BA) :
BA β 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 β1
=log 1+
x
f xx x
, for 0x β β β β at = 0x .
Ans. i) f (0) = 2 log 3 = log 32 .... (given)(i)
( )( )
( )
3 β1lim = lim
log 1+f x
x. x
2sin
0 0
x
x xβ ββ ββ ββ β
( )
3 β1 sin.
sin= lim
.log 1+
x
x x
x x
x
2sin
0
2
x
xββββ
( )
3 β1 sinlim .lim
sin= lim
limlog 1+
x
x x
x
2sin
0 0
10
0
x
x x
xx
x
β ββ ββ ββ β
ββββ
ββββ
β΄β΄β΄β΄ ( )( )
( )log3 1
= log3log
f xe
2
2ΓΓΓΓΓΓΓΓ .... (ii)
From (i) and (ii), we get
( ) ( )lim 0f x f0xββββ
β β β β
β΄β΄β΄β΄ f is discontinuous at = 0x
Discontinuity can be removed by redefining
the function as follows.
( )( )
( )
3 β1=
.log 1+f x
x x
2sinx for 0
at = 0for = 0
xx
x
β β β β
= (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
ββββfor 0x β β β β
=k for = 0x
at = 0x
Ans f (0) = k
As f (x ) is continuous at x = 0, f (0) = ( )0ββββ
lim f xx
β΄ k = ( ) ( )β
lim log1 2 1+2
0 x xxββββ
= ( )( )
log 1+2lim
log 1β 2
x
x0xββββ
=
( )
( )
log 1+2lim
2log 1β 2
lim2
x
xx
x
0
0
x
x
ββββ
ββββ
= 1
β1 = β1
β΄ k = β1
ii) ( ) ,1β tan
=1β 2 sin
f kΞΈΞΈΞΈΞΈ
ΞΈΞΈΞΈΞΈfor
4x
ΟΟΟΟβ β β β
=2
kfor =
4x
ΟΟΟΟ
at =4
xΟΟΟΟ
Ansk
=4 2
f
ΟΟΟΟ
As ( )f x is continuous at =4
ΟΟΟΟΞΈΞΈΞΈΞΈ ,
β΄β΄β΄β΄ ( ) ( )= limf f
4
ΟΟΟΟΞΈβΞΈβΞΈβΞΈβ
ΞΈ ΞΈΞΈ ΞΈΞΈ ΞΈΞΈ ΞΈ
β΄β΄β΄β΄ f (ΞΈΞΈΞΈΞΈ ) = ( )lim f
4
ΟΟΟΟΞΈβΞΈβΞΈβΞΈβ
ΞΈΞΈΞΈΞΈ =
4
ΟΟΟΟΞΈβΞΈβΞΈβΞΈβ
ΞΈΞΈΞΈΞΈ
ΞΈΞΈΞΈΞΈ
1 β tanlim
1 β 2 sin
= 4
ΟΟΟΟΞΈβΞΈβΞΈβΞΈβ
ΞΈΞΈΞΈΞΈ
ΞΈΞΈΞΈΞΈ
ΞΈΞΈΞΈΞΈ
sin1 β
coslim1 β 2 sin
= ( )cos β sin
limcos 1 β 2sin
4
ΟΟΟΟΞΈβΞΈβΞΈβΞΈβ
ΞΈ ΞΈΞΈ ΞΈΞΈ ΞΈΞΈ ΞΈ
ΞΈ ΞΈΞΈ ΞΈΞΈ ΞΈΞΈ ΞΈ
( )cos β sin 1+ 2sin
= lim1β 2sincos 1 β 2sin
4
ΟΟΟΟΞΈβΞΈβΞΈβΞΈβ
ΞΈ ΞΈ ΞΈΞΈ ΞΈ ΞΈΞΈ ΞΈ ΞΈΞΈ ΞΈ ΞΈΓΓΓΓ
ΞΈΞΈΞΈΞΈΞΈ ΞΈΞΈ ΞΈΞΈ ΞΈΞΈ ΞΈ
( ) ( )( )2
4
ΟΟΟΟΞΈβΞΈβΞΈβΞΈβ
ΞΈ ΞΈ ΞΈΞΈ ΞΈ ΞΈΞΈ ΞΈ ΞΈΞΈ ΞΈ ΞΈ
ΞΈ ΞΈΞΈ ΞΈΞΈ ΞΈΞΈ ΞΈ
cos β sin 1+ 2sin= lim
cos 1 β 2sin
( ) ( )( )2 2 2
4
ΟΟΟΟΞΈβΞΈβΞΈβΞΈβ
ΞΈ ΞΈ ΞΈΞΈ ΞΈ ΞΈΞΈ ΞΈ ΞΈΞΈ ΞΈ ΞΈ
ΞΈ ΞΈ ΞΈ ΞΈΞΈ ΞΈ ΞΈ ΞΈΞΈ ΞΈ ΞΈ ΞΈΞΈ ΞΈ ΞΈ ΞΈ
cos β sin 1+ 2 sin= lim
cos cos + sin β 2sin
( ) ( )
( )
cos β sin 1+ 2sin= lim
cos cos β sin2 2
4
ΟΟΟΟΞΈβΞΈβΞΈβΞΈβ
ΞΈ ΞΈ ΞΈΞΈ ΞΈ ΞΈΞΈ ΞΈ ΞΈΞΈ ΞΈ ΞΈ
ΞΈ ΞΈΞΈ ΞΈΞΈ ΞΈΞΈ ΞΈ
Page 23
Mahesh Tutorials Science 23
Continuity
( ) ( )( ) ( )
cos β sin 1+ 2sin= lim
cos cos β sin cos +sin4
ΟΟΟΟΞΈβΞΈβΞΈβΞΈβ
ΞΈ ΞΈ ΞΈΞΈ ΞΈ ΞΈΞΈ ΞΈ ΞΈΞΈ ΞΈ ΞΈ
ΞΈ ΞΈ ΞΈ ΞΈ ΞΈΞΈ ΞΈ ΞΈ ΞΈ ΞΈΞΈ ΞΈ ΞΈ ΞΈ ΞΈΞΈ ΞΈ ΞΈ ΞΈ ΞΈ
( )( )
1+ 2sin= lim
cos cos +sin4
ΟΟΟΟΞΈβΞΈβΞΈβΞΈβ
ΞΈΞΈΞΈΞΈ
ΞΈ ΞΈ ΞΈΞΈ ΞΈ ΞΈΞΈ ΞΈ ΞΈΞΈ ΞΈ ΞΈ
1,cos and
4 2
1sin ,cos β sin 0
2
ΟΟΟΟΞΈ β ΞΈ βΞΈ β ΞΈ βΞΈ β ΞΈ βΞΈ β ΞΈ β
ΞΈ β ΞΈ ΞΈ β ΞΈ β ΞΈ ΞΈ β ΞΈ β ΞΈ ΞΈ β ΞΈ β ΞΈ ΞΈ β
β΅β΅β΅β΅
( )
( )
lim 1 2sin+
=limcos cos sin+ΞΈ ΞΈ
4
4
x
x
ΞΈβΞΈβΞΈβΞΈβ
ΞΈβΞΈβΞΈβΞΈβ
ΞΈΞΈΞΈΞΈ
ΞΈΞΈΞΈΞΈ
11 2+
2=
1 1 1+
2 2 2
ΓΓΓΓ
1 1+=
1 2Γ
2 2
= 2
1 = 2
β΄β΄β΄β΄2
k = 2 i.e., = 4k
Q-3) Find k, if the functions are continuous at
the indicated points
( ) ,cos
=β 2
= 3
k xf x
xΟΟΟΟ
2
=2
β
for
for
x
x
ΟΟΟΟ
ΟΟΟΟ at x =
2
ΟΟΟΟ
Ans. i) = 32
f
ΟΟΟΟ
As f is continuous at x = ,2
ΟΟΟΟ
( )= lim2
f f x
2x
ΟΟΟΟββββ
ΟΟΟΟ
β΄ 3 = cos
limβ 2
k x
x2
xΟΟΟΟ
ββββ ΟΟΟΟ .... (i)
Put β2
xΟΟΟΟ
= ΞΈΞΈΞΈΞΈ, then x = β2
ΟΟΟΟΞΈΞΈΞΈΞΈ
As , 02
xΟΟΟΟ
β ΞΈ ββ ΞΈ ββ ΞΈ ββ ΞΈ β
β΄ 3 = 0ΞΈβΞΈβΞΈβΞΈβ
ΟΟΟΟΞΈΞΈΞΈΞΈ
ΞΈΞΈΞΈΞΈ
cos β2
lim2
k
=
0ΞΈβΞΈβΞΈβΞΈβ
ΞΈΞΈΞΈΞΈ
ΞΈΞΈΞΈΞΈ
sinlim
2
k =
2
k
β΄ k = 6
Q-4) Is the function defined by f (x) = x2 β sin x + 5
continuous at x = ΟΟΟΟ ?
Ans. f (x ) = x2 β sin x + 5
f (ΟΟΟΟ ) = ΟΟΟΟ2 β sin ΟΟΟΟ + 5 = ΟΟΟΟ2 β 0 + 5 = ΟΟΟΟ2 + 5
Also, ( )lim f xxβΟβΟβΟβΟ
= ( )lim β sin +5x x2
xβΟβΟβΟβΟ=
2Ο ΟΟ ΟΟ ΟΟ Οβsin +5 = β 0+52ΟΟΟΟ = +52ΟΟΟΟ
Since, ( )lim f xxβΟβΟβΟβΟ
= ( ),f ΟΟΟΟ the function f is
continuous at x = ΟΟΟΟ.
Q-5) A function defined by
f (x) = x + a for x < 0
= x for 0 β€β€β€β€ x < 1
= b β x for x β₯β₯β₯β₯ 1
is continuous in [β2, 2]. Show that (a + b) is
even
Ans f (x ) is continuous in [β2, 2].
β΄β΄β΄β΄ f (x ) is continuous at = 0x .
β΄β΄β΄β΄ ( ) ( ) ( )lim = 0 = limf x f f xβ +0 0x xβ ββ ββ ββ β
( )lim + = limx a x0 0x xβ ββ ββ ββ β
β΄β΄β΄β΄ 0 + a = 0 i.e. a = 0
Also, f (x ) is continuous at =1x .
β΄β΄β΄β΄ ( ) ( ) ( )lim = 1 = limf x f f xβ +1 1x xβ ββ ββ ββ β
β΄β΄β΄β΄ ( ) ( )lim = 1 = lim βx f b x1 1x xβ ββ ββ ββ β
β΄β΄β΄β΄ 1 = b β 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 β₯β₯β₯β₯
is a continuous function at x = 0.
Ans If ( ) ( ) ( )lim = 0 = limf x f f xβ +0 0x xβ ββ ββ ββ β
, then f (x ) is a
continuous function.
β΄β΄β΄β΄sin
limx
x0xββββ = 0 +1 = ( )lim +1x
+0xββββ
β΄β΄β΄β΄ 1 = 1 = 1
Hence, f (x ) is continuous at x = 0.