Q60 / 90 · JEE Main 2019 (11 Jan Shift 1) · Mathematics

Let $f_{k}(x)=\frac{1}{k}\left(\sin ^{k} x+\cos ^{k} x\right)$ for $\mathrm{k}=1,2,3, \ldots$ Then for all $\mathrm{x} \in \mathrm{R},$ the value of $f_{4}(x)-f_{6}(x)$ is equal to :

  • Correct answer
    A
    $\frac{1}{12}$
  • B
    $\frac{1}{4}$
  • C
    $\frac{-1}{12}$
  • D
    $\frac{5}{12}$

Solution

$f_{k}(x)=\frac{1}{k}\left(\sin ^{k} x+\cos ^{k} x\right)$ $f_{4}(x)=\frac{1}{4}\left[\sin ^{4} x+\cos ^{4} x\right]$ $=\frac{1}{4}\left[\left(\sin ^{2} x+\cos ^{2} x\right)^{2}-\frac{(\sin 2 x)^{2}}{2}\right]$ $=\frac{1}{4}\left[1-\frac{(\sin 2 x)^{2}}{2}\right]$ $f_{6}(x)=\frac{1}{6}\left[\sin ^{6} x+\cos ^{6} x\right]$ $=\frac{1}{6}\left[\left(\sin ^{2} x+\left(\cos ^{2} x\right)-\frac{3}{4}\left(\sin ^{2} x\right)^{2}\right]\right.$ $=\frac{1}{6}\left[1-\frac{3}{4}(\sin 2 x)^{2}\right]$ Now $f_{4}(x)-f_{(6)}(x)=\frac{1}{4}-\frac{1}{6}-\frac{(\sin 2 x)^{2}}{8}+\frac{1}{8}(\sin 2 x)^{2}$ $=\frac{1}{12}$

Hence, the correct answer is option: 'A'.

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