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UPSC NDA 2 2016 Mathematics Solved Paper
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© examsnet.com
Question : 72
Total: 120
Consider the following in respect of the function
f
(
x
)
=
{
2
+
x
,
x
≥
0
2
−
x
,
x
<
0
1.
lim
x
→
1
f
(
x
)
does not exist.
2. f(x) is differentiable at
x
=
0
3. f(x) is continuous at
x
=
0
Which of the above statements is/are correct?
[NDA II 2016]
1 only
3 only
2 and 3 only
1 and 3 only
Validate
Solution:
For
x
≥
0
lim
x
→
1
f
(
x
)
=
lim
x
→
1
2
+
x
=
2
+
1
=
3
For
x
<
0
lim
x
→
1
f
(
x
)
=
lim
x
→
1
2
−
x
=
2
−
1
=
1
So,
lim
x
→
1
f
(
x
)
does not exist.
At
x
=
0
RHL
:
lim
h
→
0
+
f
(
0
+
h
)
=
lim
h
→
0
2
+
h
=
2
LHL:
lim
h
→
0
−
f
(
0
−
h
)
=
lim
h
→
0
2
−
h
=
2
f
(
0
)
=
2
+
0
=
2
So,
RHL
=
LHL
=
f
(
0
)
f(x) is continuous at
x
=
0
Differentiability at
x
=
0
L
H
D
:
lim
h
→
0
−
f
(
0
−
h
)
−
f
(
0
)
−
h
=
lim
h
→
0
−
2
+
h
−
2
−
h
=
−
h
h
=
−
1
R
H
D
:
lim
h
→
0
+
f
(
0
+
h
)
−
f
(
0
)
h
=
lim
h
→
0
+
2
+
h
−
2
h
=
1
Since
LHD
≠
RHD
So,f(x) is not differentiable at
x
=
0.
© examsnet.com
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