Question

If $f\left(x\right)=2(7\cos x+24\sin x)(7\sin x-24\cos x)$for every $xϵR$ then maximum value of $\left(f\right(x){)}^{\frac{1}{4}}$ is

Answer 1

$f\left(x\right)=2(7\cos x+24\sin x)(7\sin x-24\cos x)f(x)=2(49\sin x\cos x-168{\cos }^{2}x+168{\sin }^{2}x-576\sin x\cos x\left)f\right(x)=-527\sin 2x+336\cos 2x[Note:-\sin 2x=2\sin x\cos xand\cos 2x={\cos }^{2}x-{\sin }^{2}x]Rangeofa\sin x+b\cos xis[-\sqrt{a^2+b^2},\sqrt{a^2+b^2}]Somaximumvalueoff(x)=\sqrt{{527}^2+{336}^2}=625So{f\left(x\right)}^{\frac{1}{4}}={625}^{\frac{1}{4}}=5$