questionr
Early access
Exams
Joint Entrance Examination
Mathematics
Trigonometric Ratios and Identities
Question 253
Question 253 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
)
where
x
∈
R
x\in\mathbb{R}
x
∈
R
and
k
≥
1
k\ge1
k
≥
1
. Then
f
4
(
x
)
−
f
6
(
x
)
f_4(x)-f_6(x)
f
4
(
x
)
−
f
6
(
x
)
equals:
A
1
4
\dfrac14
4
1
B
1
12
\dfrac1{12}
12
1
C
1
6
\dfrac16
6
1
D
1
3
\dfrac13
3
1
Check answer
Press A-D to select · Enter to check
Solutions · 1
Post your solution
Official solution
As in Question 89:
f
4
(
x
)
−
f
6
(
x
)
=
[
1
4
−
1
8
sin
2
2
x
]
−
[
1
6
−
1
8
sin
2
2
x
]
=
1
4
−
1
6
=
1
12
f_4(x)-f_6(x) = \left[\tfrac14-\tfrac18\sin^22x\right] - \left[\tfrac16-\tfrac18\sin^22x\right] = \tfrac14-\tfrac16 = \tfrac1{12}
f
4
(
x
)
−
f
6
(
x
)
=
[
4
1
−
8
1
sin
2
2
x
]
−
[
6
1
−
8
1
sin
2
2
x
]
=
4
1
−
6
1
=
12
1
Answer the question to unlock
Previous
Next