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Byju's Answer
Standard XII
Mathematics
Differentiation Using Substitution
If f x=1-cos ...
Question
If
f
x
=
1
-
cos
x
x
sin
x
,
x
≠
0
1
2
,
x
=
0
then at x = 0, f (x) is
(a) continuous and differentiable
(b) differentiable but not continuous
(c) continuous but not differentiable
(d) neither continuous nor differentiable
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Solution
(a) continuous and differentiable
we have,
f
x
=
1
-
cos
x
x
sin
x
,
x
≠
0
1
2
,
x
=
0
f
x
=
1
-
cos
x
x
sin
x
,
x
≠
0
1
2
,
x
=
0
C
o
n
t
i
n
u
i
t
y
a
t
x
=
0
(
L
H
L
a
t
x
=
0
)
=
l
i
m
x
→
0
-
f
(
x
)
=
l
i
m
h
→
0
f
(
0
-
h
)
=
l
i
m
h
→
0
f
(
-
h
)
=
l
i
m
h
→
0
1
-
cos
(
-
h
)
(
-
h
)
sin
(
-
h
)
=
l
i
m
h
→
0
1
-
cos
h
h
sin
h
=
l
i
m
h
→
0
1
-
cos
h
l
i
m
h
→
0
1
h
sin
h
=
1
-
cos
(
0
)
.
1
0
sin
0
=
0
(
R
H
L
a
t
x
=
0
)
=
l
i
m
x
→
0
+
f
(
x
)
=
l
i
m
h
→
0
f
(
0
+
h
)
=
l
i
m
h
→
0
f
(
h
)
=
l
i
m
h
→
0
1
-
cos
(
h
)
(
h
)
sin
(
h
)
=
l
i
m
h
→
0
1
-
cos
h
h
sin
h
=
l
i
m
h
→
0
1
-
cos
h
l
i
m
h
→
0
1
h
sin
h
=
1
-
cos
0
.
1
0
sin
0
=
0
Hence, f(x)is continuous at x = 0.
For differentiability at x = 0
(
L
H
D
a
t
x
=
0
)
=
l
i
m
x
→
0
-
f
(
x
)
-
f
(
0
)
x
-
0
=
l
i
m
h
→
0
f
(
0
-
h
)
-
f
(
0
)
0
-
h
-
0
=
l
i
m
h
→
0
f
(
-
h
)
-
1
2
-
h
=
l
i
m
h
→
0
1
-
cos
(
-
h
)
-
h
sin
(
-
h
)
-
1
2
-
h
=
1
h
l
i
m
h
→
0
1
-
cos
h
h
sin
h
-
l
i
m
h
→
0
1
2
=
1
2
-
0
=
1
2
R
H
D
a
t
x
=
0
)
=
l
i
m
x
→
0
+
f
(
x
)
-
f
(
0
)
x
-
0
=
l
i
m
h
→
0
f
(
0
+
h
)
-
f
(
0
)
0
-
h
-
0
=
l
i
m
h
→
0
f
(
h
)
-
1
2
-
h
=
l
i
m
h
→
0
1
-
cos
(
h
)
-
h
sin
(
h
)
-
1
2
-
h
=
-
1
h
l
i
m
h
→
0
1
-
cos
h
h
sin
h
-
l
i
m
h
→
0
1
2
=
1
2
-
0
=
1
2
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Similar questions
Q.
If
f
(
x
)
=
⎛
⎝
1
−
cos
x
x
sin
x
,
x
≠
0
1
2
,
x
=
0
then at x = 0, f(x) is