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Joint Entrance Examination
Mathematics
Limits, Continuity and Differentiability
Question 159
Question 159 of 401
Medium
MCQ
If
f
(
x
)
=
{
∫
0
x
(
5
+
∣
1
−
t
∣
)
d
t
,
x
>
2
5
x
+
1
,
x
≤
2
f(x)=\begin{cases}\int_0^x(5+|1-t|)\,dt, & x>2\\5x+1, & x\le2\end{cases}
f
(
x
)
=
{
∫
0
x
(
5
+
∣1
−
t
∣
)
d
t
,
5
x
+
1
,
x
>
2
x
≤
2
, then:
A
f
(
x
)
f(x)
f
(
x
)
is not continuous at
x
=
2
x=2
x
=
2
B
f
(
x
)
f(x)
f
(
x
)
is everywhere differentiable
C
f
(
x
)
f(x)
f
(
x
)
is continuous but not differentiable at
x
=
2
x=2
x
=
2
D
f
(
x
)
f(x)
f
(
x
)
is not differentiable at
x
=
1
x=1
x
=
1
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Official solution
For
x
>
2
x>2
x
>
2
:
∫
0
x
(
5
+
∣
1
−
t
∣
)
d
t
=
∫
0
1
(
6
−
t
)
d
t
+
∫
1
x
(
4
+
t
)
d
t
=
1
+
4
x
+
x
2
2
\int_0^x(5+|1-t|)\,dt=\int_0^1(6-t)\,dt+\int_1^x(4+t)\,dt=1+4x+\dfrac{x^2}{2}
∫
0
x
(
5
+
∣1
−
t
∣
)
d
t
=
∫
0
1
(
6
−
t
)
d
t
+
∫
1
x
(
4
+
t
)
d
t
=
1
+
4
x
+
2
x
2
. At
x
=
2
x=2
x
=
2
: LHL
=
5
(
2
)
+
1
=
11
=5(2)+1=11
=
5
(
2
)
+
1
=
11
; RHL
=
1
+
8
+
2
=
11
=1+8+2=11
=
1
+
8
+
2
=
11
; so
f
f
f
is continuous at
x
=
2
x=2
x
=
2
.
f
′
(
x
)
=
{
4
+
x
,
x
>
2
5
,
x
≤
2
f'(x)=\begin{cases}4+x, & x>2\\5, & x\le2\end{cases}
f
′
(
x
)
=
{
4
+
x
,
5
,
x
>
2
x
≤
2
. LHD at
x
=
2
x=2
x
=
2
is
5
5
5
; RHD is
4
+
2
=
6
4+2=6
4
+
2
=
6
--- not equal, so
f
f
f
is not differentiable at
x
=
2
x=2
x
=
2
.
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