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Byju's Answer
Standard XII
Mathematics
Trigonometric Functions
If cos x 2 ...
Question
If
cos
x
2
+
cos
x
2
2
+
cos
x
2
3
+
…
+
cos
x
2
n
=
f
(
x
)
, then
1
2
tan
x
2
+
1
2
2
tan
x
2
2
+
…
+
1
2
n
tan
x
2
n
=
A
f
′
(
x
)
f
(
x
)
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B
f
(
x
)
f
′
(
x
)
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C
−
f
′
(
x
)
f
(
x
)
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D
0
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Solution
The correct option is
C
−
f
′
(
x
)
f
(
x
)
cos
x
2
cos
x
2
2
cos
x
2
3
.
.
.
.
cos
x
2
n
=
f
(
x
)
LEt
1
2
tan
x
2
+
1
2
2
tan
x
2
2
+
.
.
.
.
1
2
n
tan
x
2
n
=
=
g
(
x
)
cos
x
2
=
f
1
;
cos
x
2
2
=
f
2
;
cos
x
2
3
=
f
3
;.......
cos
x
2
n
=
f
n
Differentiating all
f
1
till
f
n
d
d
x
(
cos
x
2
)
=
−
1
2
sin
x
2
.........................................
d
d
x
(
cos
x
2
n
)
=
−
1
2
n
sin
x
2
n
=
f
′
n
⇒
−
f
′
1
f
1
=
−
(
−
1
2
sin
x
2
)
cos
(
x
2
)
=
1
2
tan
x
2
=
g
1
--------------------------------------
−
f
′
n
f
n
=
−
(
−
1
2
n
sin
x
2
n
)
cos
(
x
2
n
)
=
1
2
n
tan
x
2
n
=
g
n
g
′
1
.
.
.
.
.
g
n
=
−
f
′
1
f
1
+
.
.
.
.
f
′
n
f
n
g
n
=
f
′
(
x
)
f
(
x
)
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0
Similar questions
Q.
If
cos
x
2
cos
x
2
2
cos
x
2
3
…
cos
x
2
n
=
f
(
x
)
, then
1
2
tan
x
2
+
1
2
2
tan
x
2
2
+
…
+
1
2
n
tan
x
2
n
=
Q.
1
2
tan
x
2
+
1
4
tan
x
4
+
.
.
.
+
1
2
n
tan
x
2
n
=
1
2
n
cot
x
2
n
-
cot
x
for all n ∈ N and
0
<
x
<
π
2
Q.
If
f
is strictly increasing and positive function, such that
x
x
∫
0
(
1
−
t
)
sin
(
f
(
t
)
)
d
t
=
2
x
∫
0
t
sin
(
f
(
t
)
)
d
t
,
where
x
>
0.
Then the value of
f
′
(
x
)
cot
f
(
x
)
+
3
1
+
x
in the domain of
f
(
x
)
is
Q.
If
y
=
cos
x
2
⋅
cos
x
2
2
⋯
∞
, then
−
y
′
y
is
Q.
If
f
(
x
)
is a twice differentiable function such that
f
(
a
)
=
0
,
f
(
b
)
=
2
,
f
(
c
)
=
−
1
,
f
(
d
)
=
2
,
f
(
e
)
=
0
when
a
<
b
<
c
<
d
<
e
then the minimum number of zeros of
g
(
x
)
=
(
f
(
x
)
)
2
+
f
′
(
x
)
f
(
x
)
in the interval
[
a
,
e
]
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