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Exams
Joint Entrance Examination
Mathematics
Trigonometric Ratios and Identities
Question 239
Question 239 of 298
Medium
MCQ
Let
f
k
(
x
)
=
1
k
(
sin
k
x
+
cos
k
x
)
f_k(x) = \dfrac1k(\sin^kx+\cos^kx)
f
k
(
x
)
=
k
1
(
sin
k
x
+
cos
k
x
)
for
k
=
1
,
2
,
3
,
…
k=1,2,3,\dots
k
=
1
,
2
,
3
,
…
. Then for all
x
∈
R
x\in\mathbb{R}
x
∈
R
, the value of
f
4
(
x
)
−
f
6
(
x
)
f_4(x)-f_6(x)
f
4
(
x
)
−
f
6
(
x
)
is equal to:
A
1
12
\dfrac1{12}
12
1
B
1
4
\dfrac14
4
1
C
−
1
12
-\dfrac1{12}
−
12
1
D
5
12
\dfrac5{12}
12
5
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Solutions · 1
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Official solution
\[f_4(x) = \frac14\left[1-2\sin^2x\cos^2x\right] = \frac14 - \frac18\sin^22x\] \[f_6(x) = \frac16\left[1-3\sin^2x\cos^2x\right] = \frac16 - \frac18\sin^22x\] \[f_4(x)-f_6(x) = \frac14-\frac16 = \frac1{12}\]
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