Question

The maximum value of $\sin (x+\pi /6)+\cos (x+\pi /6)$ in the interval $(0,\pi /2)$ is attained at

• A. $\frac{\pi }{12}$
• B. $\frac{\pi }{6}$
• C. $\frac{\pi }{4}$
• D. $\frac{\pi }{3}$

Answer 1

The correct option is A $\frac{\pi }{12}$

The given equation is:

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

Differentiating once w.r.t to x and equating it to zero to get the extremum points we get,

$\Rightarrow f^{\prime }\left(x\right)=\cos (x+\frac{\pi }{6})-\sin (x+\frac{\pi }{6})$

Equating $f^{\prime }\left(x\right)=0$ we get,

$\Rightarrow \cos (x+\frac{\pi }{6})-\sin (x+\frac{\pi }{6})=0$

$\Rightarrow \cos (x+\frac{\pi }{6})=\sin (x+\frac{\pi }{6})$

$\Rightarrow x+\frac{\pi }{6}=\frac{\pi }{4}$

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

Again differentiating to perform double derivative test and check whether the extremum gives a maximum or minimum

$\Rightarrow f^{\prime \prime }\left(x\right)=-\sin (x+\frac{\pi }{6})-\cos (x+\frac{\pi }{6})$

$\Rightarrow f^{\prime \prime }\left(\frac{\pi }{12}\right)=-\sin (\frac{\pi }{12}+\frac{\pi }{6})-\cos (\frac{\pi }{12}+\frac{\pi }{6})$

$\Rightarrow f^{\prime \prime }\left(\frac{\pi }{12}\right)=-\sin \frac{\pi }{4}-\cos \frac{\pi }{4}$

$\Rightarrow f^{\prime \prime }\left(\frac{\pi }{12}\right)=-\sqrt{2}$

$\Rightarrow f^{\prime \prime }\left(\frac{\pi }{12}\right)<0$.

Hence the point $x=\frac{\pi }{12}$ is a point of maxima. .....Answer