JEE Main 2019 — Mathematics Trigonometry

$f_{k}(x)=\frac{1}{k}\left(\sin ^{k} x+\cos ^{k} x\right)$fk​(x)=k1​(sinkx+coskx) $f_{4}(x)=\frac{1}{4}\left[\sin ^{4} x+\cos ^{4} x\right]$f4​(x)=41​[sin4x+cos4x] $=\frac{1}{4}\left[\left(\sin ^{2} x+\cos ^{2} x\right)^{2}-\frac{(\sin 2 x)^{2}}{2}\right]$=41​[(sin2x+cos2x)2−2(sin2x)2​] $=\frac{1}{4}\left[1-\frac{(\sin 2 x)^{2}}{2}\right]$=41​[1−2(sin2x)2​] $f_{6}(x)=\frac{1}{6}\left[\sin ^{6} x+\cos ^{6} x\right]$f6​(x)=61​[sin6x+cos6x] $=\frac{1}{6}\left[\left(\sin ^{2} x+\left(\cos ^{2} x\right)-\frac{3}{4}\left(\sin ^{2} x\right)^{2}\right]\right.$=61​[(sin2x+(cos2x)−43​(sin2x)2] $=\frac{1}{6}\left[1-\frac{3}{4}(\sin 2 x)^{2}\right]$=61​[1−43​(sin2x)2] Now $f_{4}(x)-f_{(6)}(x)=\frac{1}{4}-\frac{1}{6}-\frac{(\sin 2 x)^{2}}{8}+\frac{1}{8}(\sin 2 x)^{2}$f4​(x)−f(6)​(x)=41​−61​−8(sin2x)2​+81​(sin2x)2 $=\frac{1}{12}$=121​