Question

The maximum value of $\sin \left( x+\dfrac{\pi }{6} \right)+\cos \left( x+\dfrac{\pi }{6} \right)$ in the interval $\left( 0,\dfrac{\pi }{2} \right)$ is attained at
(a) $\dfrac{\pi }{12}$
(b) $\dfrac{\pi }{6}$
(c) $\dfrac{\pi }{3}$
(d) $\dfrac{\pi }{2}$

Answer 1

We start solving the problem by assuming $f\left( x \right)=\sin \left( x+\dfrac{\pi }{6} \right)+\cos \left( x+\dfrac{\pi }{6} \right)$. We then make use of the property $\sin \left( \dfrac{\pi }{2}-x \right)=\cos x$ for the cosine function present in the problem. We then make use of sum to product formula $\sin A+\sin B=2\sin \left( \dfrac{A+B}{2} \right)\cos \left( \dfrac{A-B}{2} \right)$ to proceed through the problem. We then make the necessary calculations to simplify the function. We then make use of the property that the function $\cos \theta$ is maximum if the value of $\theta$ is 0 to get the required value of ‘x’.

Complete step by step answer:
According to the problem, we are asked to find the value of ‘x’ at which the maximum value of the function $\sin \left( x+\dfrac{\pi }{6} \right)+\cos \left( x+\dfrac{\pi }{6} \right)$ is attained in the interval $\left( 0,\dfrac{\pi }{2} \right)$.
Let us assume $f\left( x \right)=\sin \left( x+\dfrac{\pi }{6} \right)+\cos \left( x+\dfrac{\pi }{6} \right)$.
We know that $\sin \left( \dfrac{\pi }{2}-x \right)=\cos x$.
$\Rightarrow f\left( x \right)=\sin \left( x+\dfrac{\pi }{6} \right)+\sin \left( \dfrac{\pi }{2}-\left( x+\dfrac{\pi }{6} \right) \right)$.
$\Rightarrow f\left( x \right)=\sin \left( x+\dfrac{\pi }{6} \right)+\sin \left( \dfrac{\pi }{3}-x \right)$.
We know that $\sin A+\sin B=2\sin \left( \dfrac{A+B}{2} \right)\cos \left( \dfrac{A-B}{2} \right)$.
$\Rightarrow f\left( x \right)=2\sin \left( \dfrac{x+\dfrac{\pi }{6}+\dfrac{\pi }{3}-x}{2} \right)\cos \left( \dfrac{x+\dfrac{\pi }{6}-\dfrac{\pi }{3}+x}{2} \right)$.
$\Rightarrow f\left( x \right)=2\sin \left( \dfrac{\dfrac{2\pi +\pi }{6}}{2} \right)\cos \left( \dfrac{2x+\dfrac{\pi -2\pi }{6}}{2} \right)$.
$\Rightarrow f\left( x \right)=2\sin \left( \dfrac{\dfrac{3\pi }{6}}{2} \right)\cos \left( \dfrac{2x+\left( \dfrac{-\pi }{6} \right)}{2} \right)$.
$\Rightarrow f\left( x \right)=2\sin \left( \dfrac{\pi }{4} \right)\cos \left( x-\dfrac{\pi }{12} \right)$.
$\Rightarrow f\left( x \right)=2\left( \dfrac{1}{\sqrt{2}} \right)\cos \left( x-\dfrac{\pi }{12} \right)$.
$\Rightarrow f\left( x \right)=\sqrt{2}\cos \left( x-\dfrac{\pi }{12} \right)$ ---(1).
From equation (1), we can see that the function $f\left( x \right)$ will be maximum if the function $\cos \left( x+\dfrac{-\pi }{12} \right)$ is maximum.
We know that the function $\cos \theta$ is maximum if the value of $\theta$ is 0.
So, the function $f\left( x \right)$ is maximum if $x-\dfrac{\pi }{12}=0$.
$\Rightarrow x=\dfrac{\pi }{12}$.
So, we have found that the function $\sin \left( x+\dfrac{\pi }{6} \right)+\cos \left( x+\dfrac{\pi }{6} \right)$ will attain maximum at $x=\dfrac{\pi }{12}$.

So, the correct answer is “Option a”.

Note: We can see that the given problem contains a huge amount of calculation, so we need to perform each step carefully. We can also solve this problem as shown below:
We have $f\left( x \right)=\sin \left( x+\dfrac{\pi }{6} \right)+\cos \left( x+\dfrac{\pi }{6} \right)$.
Let us differentiate w.r.t x on both sides.
So, we get ${{f}^{'}}\left( x \right)=\dfrac{d}{dx}\left( \sin \left( x+\dfrac{\pi }{6} \right)+\cos \left( x+\dfrac{\pi }{6} \right) \right)$.
$\Rightarrow {{f}^{'}}\left( x \right)=\dfrac{d}{dx}\left( \sin \left( x+\dfrac{\pi }{6} \right) \right)+\dfrac{d}{dx}\left( \cos \left( x+\dfrac{\pi }{6} \right) \right)$.
$\Rightarrow {{f}^{'}}\left( x \right)=\cos \left( x+\dfrac{\pi }{6} \right)-\sin \left( x+\dfrac{\pi }{6} \right)$ ---(1).
Now, let us find the value of x in the interval $\left( 0,\dfrac{\pi }{2} \right)$ for which we get ${{f}^{'}}\left( x \right)=0$.
$\Rightarrow \cos \left( x+\dfrac{\pi }{6} \right)-\sin \left( x+\dfrac{\pi }{6} \right)=0$.
$\Rightarrow \cos \left( x+\dfrac{\pi }{6} \right)=\sin \left( x+\dfrac{\pi }{6} \right)$. We know that this will be possible if $x+\dfrac{\pi }{6}=\dfrac{\pi }{4}$.
$\Rightarrow x=\dfrac{\pi }{4}-\dfrac{\pi }{6}$.
$\Rightarrow x=\dfrac{3\pi -2\pi }{12}=\dfrac{\pi }{12}$ ---(2).
Now, let us differentiate equation (1) again.
So, we get ${{f}^{''}}\left( x \right)=\dfrac{d}{dx}\left( \cos \left( x+\dfrac{\pi }{6} \right)-\sin \left( x+\dfrac{\pi }{6} \right) \right)$.
$\Rightarrow {{f}^{''}}\left( x \right)=\dfrac{d}{dx}\left( \cos \left( x+\dfrac{\pi }{6} \right) \right)-\dfrac{d}{dx}\left( \sin \left( x+\dfrac{\pi }{6} \right) \right)$.
$\Rightarrow {{f}^{''}}\left( x \right)=-\sin \left( x+\dfrac{\pi }{6} \right)-\cos \left( x+\dfrac{\pi }{6} \right)$ ---(3).
Let us substitute the value of ‘x’ obtained from equation (2) in equation (3).
So, we get ${{f}^{''}}\left( \dfrac{\pi }{12} \right)=-\sin \left( \dfrac{\pi }{12}+\dfrac{\pi }{6} \right)-\cos \left( \dfrac{\pi }{12}+\dfrac{\pi }{6} \right)$.
$\Rightarrow {{f}^{''}}\left( \dfrac{\pi }{12} \right)=-\sin \left( \dfrac{\pi }{4} \right)-\cos \left( \dfrac{\pi }{4} \right)$.
$\Rightarrow {{f}^{''}}\left( \dfrac{\pi }{12} \right)=-\dfrac{1}{\sqrt{2}}-\dfrac{1}{\sqrt{2}}$.
$\Rightarrow {{f}^{''}}\left( \dfrac{\pi }{12} \right)=-\sqrt{2}<0$. Which suggests there is a maximum at $x=\dfrac{\pi }{12}$.