Question

The maximum value of $\sin \left(x+\frac{\mathrm{\pi }}{6}\right)+\cos \left(x+\frac{\mathrm{\pi }}{6}\right)$ in the interval $\left(0,\frac{\mathrm{\pi }}{2}\right)$ is attained at

• A. $x=\frac{\mathrm{\pi }}{12}$
• B. $x=\frac{\mathrm{\pi }}{6}$
• C. $x=\frac{\mathrm{\pi }}{3}$
• D. $x=\frac{\mathrm{\pi }}{2}$

Answer 1

The correct option is A

$x=\frac{\mathrm{\pi }}{12}$

Explanation for the correct option:

$\sin \left(x+\frac{\mathrm{\pi }}{6}\right)+\cos \left(x+\frac{\mathrm{\pi }}{6}\right)$

Multiplying and dividing by $\sqrt{2},$

$$\begin{gathered} =\left(\sqrt{2}\right)\left[\left(\frac{1}{\sqrt{2}}\right)\sin \left(x+\frac{\mathrm{\pi }}{6}\right)+\left(\frac{1}{\sqrt{2}}\right)\cos \left(x+\frac{\mathrm{\pi }}{6}\right)\right] \\ =\left(\sqrt{2}\right)\left[\cos \frac{\mathrm{\pi }}{4}\sin \left(x+\frac{\mathrm{\pi }}{6}\right)+\sin \frac{\mathrm{\pi }}{4}\cos \left(x+\frac{\mathrm{\pi }}{6}\right)\right] \\ =\sqrt{2}\sin \left(x+\frac{\mathrm{\pi }}{6}+\frac{\mathrm{\pi }}{4}\right)\left[\because \mathrm{cosAsinB}+\mathrm{sinAcosB}=\sin (A+B)\right] \end{gathered}$$

The expression has a a maximum value when $x+\frac{\mathrm{\pi }}{6}+\frac{\mathrm{\pi }}{4}=\frac{\mathrm{\pi }}{2}$ and the maximum value is $\sqrt{2}.$

Thus, $x=\frac{\mathrm{\pi }}{2}-\frac{\mathrm{\pi }}{6}-\frac{\mathrm{\pi }}{4}=\frac{\mathrm{\pi }}{12}$

Hence the correct option is option(A)