Let f _ k (x) = (1)/(k)( ^ k x + ^ k x) where x ∈ R and k ≥ 1 . Then f _ 4 (x) - f _ 6 (x) equals

Correct answer: B. (1)/(1 2) f _ 4 (x) = (1)/(4)( ^ 4 x + ^ 4 x) = (1)/(4)[ ( ^ 2 x + ^ 2 x) ^ 2 - 2 ^ 2 x ^ 2 x] = (1)/(4)[1 - 2 ^ 2 x ^ 2 x] f _ 6 (x) = (1)/(6)[ ^ 6 x + ^ 6 x] = (1)/(6)[ ( ^ 2 x + ^ 2 x) ^ 3 - 3 ^ 2 x ^ 2 x( ^ 2 x + ^ 2 x)] = (1)/(6)[1 - 3 ^ 2 x ^ 2 x(1)] f _ 4 (x) - f _ 6 (x) = (1)/(4) - (1)/(6) = (1)/(12)

JEE Main 20146 April

Exam: JEE Main
Year: 2014
Date: 6 April
Subject: Mathematics
Chapter: Trigonometric Ratios & Identities

JEE Main 2014 · 6 April · Mathematics · Trigonometric Ratios & Identities

Let f _ k (x) = (1)/(k)( ^ k x + ^ k x) where x ∈ R and k ≥ 1 ....

Let

{f}_{k}\left(x\right) = \dfrac{1}{k}\left(\sin^{k}x + \cos^{k}x\right)

where

x \in R

and

k \ge 1

. Then

{f}_{4}\left(x\right) - {f}_{6}\left(x\right)

equals

Options

A $\dfrac{1}{4}$
B $\dfrac{1}{1 2}$
C $\dfrac{1}{6}$
D $\dfrac{1}{3}$

Correct answer

B. $\dfrac{1}{1 2}$

Step-by-step solution

{f}_{4}\left(x\right) = \dfrac{1}{4}\left(\sin^{4}x + \cos^{4}x\right)

= \dfrac{1}{4}\left[{\left(\sin^{2}x + \cos^{2}x\right)}^{2} - 2\sin^{2}x\cos^{2}x\right]

= \dfrac{1}{4}\left[1 - 2\sin^{2}x\cos^{2}x\right]

{f}_{6}\left(x\right) = \dfrac{1}{6}\left[\sin^{6}x + \cos^{6}x\right]

= \dfrac{1}{6}\left[{\left(\sin^{2}x + \cos^{2}x\right)}^{3} - 3\sin^{2}x\cos^{2}x\left(\sin^{2}x + \cos^{2}x\right)\right]

= \dfrac{1}{6}\left[1 - 3\sin^{2}x\cos^{2}x\left(1\right)\right]

{f}_{4}\left(x\right) - {f}_{6}\left(x\right) = \dfrac{1}{4} - \dfrac{1}{6}

= \dfrac{1}{12}

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