Question

The maximum value of $f\left(x\right)=2\sin x+\cos 2x,0\le x\le \frac{\mathrm{\pi }}{2}$ occurs at $x$is

• A. $0$
• B. $\frac{\mathrm{\pi }}{6}$
• C. $\frac{\mathrm{\pi }}{2}$
• D. None of these

Answer 1

The correct option is B

$\frac{\mathrm{\pi }}{6}$

Explanation for the correct options:

Solve for the maximum value of $f\left(x\right)=2\sin x+\cos 2x,0\le x\le \frac{\mathrm{\pi }}{2}$ occurs at $x$is

$f\left(x\right)=2\sin x+\cos 2x$

On differentiating w.r.t $x,$ we get

$f\text{'}\left(x\right)=2\cos x-2\sin 2x$

Put $f\text{'}\left(x\right)=0$

$$\begin{gathered} 2\cos x-4\sin x\cos x=0\mathbf{[}\mathbf{\because }\mathbf{sin}\mathbf{2}\mathbf{x}\mathbf{=}\mathbf{2}\mathbf{sinx}\mathbf{}\mathbf{cosx}\mathbf{]} \\ 2\cos x(1-2\sin x)=0 \\ \cos x=0,-2\sin x=-1 \\ \cos x=0,\sin x=\frac{1}{2} \\ x=\frac{\mathrm{\pi }}{2},x=\frac{\mathrm{\pi }}{6} \\ f\text{'}\text{'}\left(x\right)=-2\sin x-4\cos 2x \end{gathered}$$

At $x=\frac{\mathrm{\pi }}{2},f\text{'}\text{'}\left(\frac{\mathrm{\pi }}{2}\right)=-2\sin \left(\frac{\mathrm{\pi }}{2}\right)-4\cos 2\left(\frac{\mathrm{\pi }}{2}\right)$

$=2>0;$ (minima)

At $x=\frac{\mathrm{\pi }}{6},f\text{'}\text{'}\left(\frac{\mathrm{\pi }}{6}\right)=-2\sin \left(\frac{\mathrm{\pi }}{6}\right)-4\cos 2\left(\frac{\mathrm{\pi }}{6}\right)$

$=-3<0;$ (maxima)

Hence the maximum value of $f\left(x\right)=2\sin x+\cos 2x,0\le x\le \frac{\mathrm{\pi }}{2}$ occurs at $x=\frac{\mathrm{\pi }}{6}.$

Hence, the correct option is option (B)