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Byju's Answer
Standard XII
Mathematics
Geometrical Representation of Conjugate of a Complex Number
The function ...
Question
The function f (x) = |cos x| is
(a) differentiable at x = (2n + 1) π/2, n ∈ Z
(b) continuous but not differentiable at x = (2n + 1) π/2, n ∈ Z
(c) neither differentiable nor continuous at x = n ∈ Z
(d) none of these
Open in App
Solution
(b) continuous but not differentiable at x = (2n + 1) π/2, n ∈ Z
We
have
,
f
x
=
cos
x
⇒
f
x
=
cos
x
,
2
n
π
≤
x
<
4
n
+
1
π
2
0
,
x
=
4
n
+
1
π
2
-
cos
x
,
4
n
+
1
π
2
<
x
<
4
n
+
3
π
2
0
,
x
=
4
n
+
3
π
2
cos
x
,
4
n
+
3
π
2
<
x
≤
2
n
+
2
π
When
,
x
is
in
first
quadrant
,
i
.
e
.
2
n
π
≤
x
<
4
n
+
1
π
2
,
we
have
f
x
=
cos
x
which
being
a
trigonometrical
function
is
continuous
and
differentiable
in
2
n
π
,
4
n
+
1
π
2
When
,
x
is
in
second
quadrant
or
in
third
quadrant
,
i
.
e
.
,
4
n
+
1
π
2
<
x
<
4
n
+
3
π
2
,
we
have
f
x
=
-
cos
x
which
being
a
trigonometrical
function
is
continuous
and
differentiable
in
4
n
+
1
π
2
,
4
n
+
3
π
2
When
,
x
is
in
fourth
quadrant
,
i
.
e
.
,
4
n
+
3
π
2
<
x
≤
2
n
+
2
π
,
we
have
f
x
=
cos
x
which
being
a
trigonometrical
function
is
continuous
and
differentiable
in
4
n
+
3
π
2
,
2
n
+
2
π
Thus
possible
point
of
non
-
differentiability
of
f
x
are
x
=
4
n
+
1
π
2
,
4
n
+
3
π
2
Now
,
LHD
at
x
=
4
n
+
1
π
2
=
lim
x
→
4
n
+
1
π
2
-
f
x
-
f
4
n
+
1
π
2
x
-
4
n
+
1
π
2
=
lim
x
→
4
n
+
1
π
2
-
cos
x
-
0
x
-
4
n
+
1
π
2
=
lim
x
→
4
n
+
1
π
2
-
-
sin
x
1
-
0
By
L
'
Hospital
rule
=
-
1
And
RHD
at
x
=
4
n
+
1
π
2
=
lim
x
→
4
n
+
1
π
2
+
f
x
-
f
4
n
+
1
π
2
x
-
4
n
+
1
π
2
=
lim
x
→
4
n
+
1
π
2
+
-
cos
x
-
0
x
-
4
n
+
1
π
2
=
lim
x
→
4
n
+
1
π
2
+
sin
x
1
-
0
By
L
'
Hospital
rule
=
1
∴
lim
x
→
4
n
+
1
π
2
-
f
x
≠
lim
x
→
4
n
+
1
π
2
+
f
x
So
f
x
is
not
differentiable
at
x
=
4
n
+
1
π
2
Now
,
LHD
at
x
=
4
n
+
3
π
2
=
lim
x
→
4
n
+
1
π
2
-
f
x
-
f
4
n
+
3
π
2
x
-
4
n
+
3
π
2
=
lim
x
→
4
n
+
3
π
2
-
-
cos
x
-
0
x
-
4
n
+
3
π
2
=
lim
x
→
4
n
+
3
π
2
-
sin
x
1
-
0
By
L
'
Hospital
rule
=
1
And
RHD
at
x
=
4
n
+
3
π
2
=
lim
x
→
4
n
+
3
π
2
+
f
x
-
f
4
n
+
3
π
2
x
-
4
n
+
3
π
2
=
lim
x
→
4
n
+
3
π
2
+
cos
x
-
0
x
-
4
n
+
3
π
2
=
lim
x
→
4
n
+
3
π
2
+
-
sin
x
1
-
0
By
L
'
Hospital
rule
=
-
1
∴
lim
x
→
4
n
+
3
π
2
-
f
x
≠
lim
x
→
4
n
+
3
π
2
+
f
x
So
f
x
is
not
differentiable
at
x
=
4
n
+
3
π
2
Therefore
,
f
x
is
neither
differentiable
at
4
n
+
1
π
2
nor
at
4
n
+
3
π
2
i
.
e
.
f
x
is
not
differentiable
at
odd
multiples
of
π
2
i
.
e
.
f
x
is
not
differentiable
at
x
=
2
n
+
1
π
2
Therefore
,
f
(
x
)
is
everywhere
continuous
but
not
differentiable
at
2
n
+
1
π
2
.
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Similar questions
Q.
The function f (x) = |cos x| is
(a) everywhere continuous and differentiable
(b) everywhere continuous but not differentiable at (2n + 1) π/2, n ∈ Z
(c) neither continuous nor differentiable at (2n + 1) π/2, n ∈ Z
(d) none of these
Q.
Let f (x) = |cos x|. Then,
(a) f (x) is everywhere differentiable
(b) f (x) is everywhere continuous but not differentiable at x = n π, n ∈ Z
(c) f (x) is everywhere continuous but not differentiable at
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Let f (x) = |sin x|. Then,
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(b) f (x) is everywhere continuous but not differentiable at x = n π, n ∈ Z
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,
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∈
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.
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⎩
(
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a
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)
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