[JEE Main 2014] let f k x 1k sin kx cos kx for k 1 then f
Let $f_k(x)=\frac1k(\sin^kx+\cos^kx)$ for $k\ge1$. Then $f_4(x)-f_6(x)$ equals
(a) $\frac14$
(b) $\frac1{12}$
(c) $\frac16$
(d) $\frac13$
1 Answer
VAVidaara Admin
✓ Vidaara Team
✓ Accepted
· 2d ago
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Correct answer: (b) $\frac1{12}$
$f_4=\frac14(1-2s^2c^2)$, $f_6=\frac16(1-3s^2c^2)$ where $s^2c^2=\sin^2x\cos^2x$; the $\sin^2x\cos^2x$ terms cancel, leaving $\frac14-\frac16=\frac1{12}$.
JEE Main 2014 · Trigonometry — verified solution by the Vidaara Team.
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