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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→→→→
Mahesh Tutorials Science 3
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
4 Mahesh Tutorials Science
( )= 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
ππππ→→→→
.... (i)
Also =3
f k
ππππ.... (ii)
Since f (x) is continuous at =3
xππππ
( )lim =3
f x f
3x
ππππ→→→→
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
ππππ
Mahesh Tutorials Science 5
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
ππππ→→→→
ππππ≠≠≠≠
∴∴∴∴ 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≠≠≠≠
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
6 Mahesh Tutorials Science
( )( )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
ππππ→→→→
××××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
ππππ→→→→
××××x
( ) ( )1 1
= lim1+sin 2 2x
2
ππππ→→→→
××××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
Mahesh Tutorials Science 7
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
8 Mahesh Tutorials Science
∴∴∴∴ 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=
Mahesh Tutorials Science 9
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
10 Mahesh Tutorials Science
∴∴∴∴ f is left continuous at 1
2
( ) ( )lim = lim 1–f x x
+
1 1
2 2x x→ →→ →→ →→ →
1
=2
1
2f
=
∴∴∴∴ f is right continuous at 1
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
π ππ ππ ππ π
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→→→→
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→→→→
∴∴∴∴ f is continuous at = 0x
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
– 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
Since, f is continuous at = 0x
∴∴∴∴ ( ) ( )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)
Since, f is continuous at = 0x
( ) ( )lim = 0f x f–0x→→→→
∴∴∴∴ log 4 = 2k
∴∴∴∴ 4 =k2
∴∴∴∴ = 2k
(∵∵∵∵base of logarithm cannot be negative
= –2k is not possible)
Continuity
12 Mahesh Tutorials Science
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
Since, f is continuous at = 0x
( ) ( )lim = 0f x f+0x→→→→
∴∴∴∴ 1= –2k
∴∴∴∴1
= –2
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 – 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
Mahesh Tutorials Science 13
Continuity
= 1 160
ππππ× ×× ×× ×× ×
=60
ππππ ( )= 0f
∴∴∴∴ f is continuous at = 0x
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
14 Mahesh Tutorials Science
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
ππππ→→→→
××××cos – sin 1+ 2 sin
= limcos 1+ 2 sin
x x x
x xx
( ) ( )( ) ( )2
4
ππππ→→→→
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
ππππ→→→→
( ) ( )( ) ( ) ( )
cos – sin 1+ 2 sin= lim
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
ππππ→→→→
ππππ→→→→
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≠≠≠≠
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
ππππ
( )6
ππππ→→→→
π ππ ππ ππ π
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
ππππ
( )
( )
1sin – 6
2 6= limlimcos – 6
x
x x
6
6
x
x
ππππ→→→→
ππππ→→→→
ππππ
ππππ
2 1=
63
2
××××
2=
3 3
Hence 2
=6 3 3
f
ππππ
Mahesh Tutorials Science 15
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
π ππ ππ ππ π
Since, f is continuous at =6
xππππ, it is left
continuous at 6
ππππ.
i.e., ( )lim =6
f x f
6x
ππππ→→→→
ππππ
∴∴∴∴3
lim + =2
ax b
6x
ππππ→→→→
∴∴∴∴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
16 Mahesh Tutorials Science
= 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
ππππ→→→→
∴∴∴∴ =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
Since f is continuous at =4
xππππ
, it is left
continuous at =4
xππππ
i.e., ( )= lim4
f f x
–4x
ππππ→→→→
ππππ
+ = lim + 2sin2
b x a x
4x
ππππ→→→→
ππππ
π ππ ππ ππ π
= + 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
Since f is continuous at =2
xππππ, it is right
continuous at =2
xππππ.
∴∴∴∴ ( )= lim2
f f x
+4x
ππππ→→→→
ππππ
∴∴∴∴
+2
ππππ→→→→
= lim cos2 – sinb a x b xx
= cos – sin2
a bππππ
ππππ
= – –a b
∴∴∴∴ + = –a b b
Mahesh Tutorials Science 17
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
π ππ ππ ππ π
Since f is continuous at –
2
ππππ, it is right
continuous at –
=2
xππππ
∴∴∴∴ ( )–
–= lim
2f f x
+2
ππππ→→→→
ππππ
x
–2 = lim sin +x
2
ππππ→→→→
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
π ππ ππ ππ π
Since f is continuous at =2
xππππ, it is left
continuous at 2
ππππ
∴∴∴∴ ( )lim =2
f x f
–2x
ππππ→→→→
ππππ
∴∴∴∴lim sin + = 0x
2x
ππππ→→→→
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
18 Mahesh Tutorials Science
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
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 ∈∈∈∈ .
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
Since f is continuous at = 2x , it is left
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
20 Mahesh Tutorials Science
∴∴∴∴ 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
Since f is continuous at = 2x , it is left
continuous at = 2x .
∴∴∴∴ ( ) ( )2 = limf f x+2x→→→→
∴∴∴∴ ( )2 = lim 2 +x k2x→→→→
∴∴∴∴ 2 = 4 +k
∴∴∴∴ = – 2k
∴∴∴∴ = –1a
Q-3) Discuss the continuity of the following
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.
Q-4) Discuss the continuity of the following
function in its domain
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
∴∴∴∴ f is right continuous at 4
( ) ( )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.
Mahesh Tutorials Science 21
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′
Q-6) Discuss the continuity of the following
function in its domain
( )
–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
22 Mahesh Tutorials Science
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
ππππ≠≠≠≠
=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
ππππθ→θ→θ→θ→
θθθθ =
4
ππππθ→θ→θ→θ→
θθθθ
θθθθ
1 – tanlim
1 – 2 sin
= 4
ππππθ→θ→θ→θ→
θθθθ
θθθθ
θθθθ
sin1 –
coslim1 – 2 sin
= ( )cos – sin
limcos 1 – 2sin
4
ππππθ→θ→θ→θ→
θ θθ θθ θθ θ
θ θθ θθ θθ θ
( )cos – sin 1+ 2sin
= lim1– 2sincos 1 – 2sin
4
ππππθ→θ→θ→θ→
θ θ θθ θ θθ θ θθ θ θ××××
θθθθθ θθ θθ θθ θ
( ) ( )( )2
4
ππππθ→θ→θ→θ→
θ θ θθ θ θθ θ θθ θ θ
θ θθ θθ θθ θ
cos – sin 1+ 2sin= lim
cos 1 – 2sin
( ) ( )( )2 2 2
4
ππππθ→θ→θ→θ→
θ θ θθ θ θθ θ θθ θ θ
θ θ θ θθ θ θ θθ θ θ θθ θ θ θ
cos – sin 1+ 2 sin= lim
cos cos + sin – 2sin
( ) ( )
( )
cos – sin 1+ 2sin= lim
cos cos – sin2 2
4
ππππθ→θ→θ→θ→
θ θ θθ θ θθ θ θθ θ θ
θ θθ θθ θθ θ
Mahesh Tutorials Science 23
Continuity
( ) ( )( ) ( )
cos – sin 1+ 2sin= lim
cos cos – sin cos +sin4
ππππθ→θ→θ→θ→
θ θ θθ θ θθ θ θθ θ θ
θ θ θ θ θθ θ θ θ θθ θ θ θ θθ θ θ θ θ
( )( )
1+ 2sin= lim
cos cos +sin4
ππππθ→θ→θ→θ→
θθθθ
θ θ θθ θ θθ θ θθ θ θ
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→→→→
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.