Question

The function $f\left(x\right)=a\sin \left(x\right)+\frac{1}{3}\sin \left(3x\right)$ has maximum value at $x=\frac{\pi }{3}$. The value of $a$ is:

• A. $3$
• B. $\frac{1}{3}$
• C. $2$
• D. $\frac{1}{2}$

Answer 1

The correct option is C

$2$

Explanation for correct option

Given function $f\left(x\right)=a\sin \left(x\right)+\frac{1}{3}\sin \left(3x\right)$

Then,

$$\begin{gathered} f\text{'}\left(x\right)=a\cos \left(x\right)+\frac{1}{3}\left(\cos \left(3x\right)\right)·3 \\ =a\cos \left(x\right)+\cos \left(3x\right) \end{gathered}$$

We know that the maximum value or global maxima of any function occurs at the critical point which is given as $f\text{'}\left(x\right)=0$.

Also, it is given that the maximum value of $f\left(x\right)=a\sin \left(x\right)+\frac{1}{3}\sin \left(3x\right)$ occur at $x=\frac{\mathrm{\pi }}{3}$

So,

$$\begin{gathered} f\text{'}\left(\frac{\mathrm{\pi }}{3}\right)=0 \\ \Rightarrow a\cos \left(\frac{\mathrm{\pi }}{3}\right)+\cos \left(3.\frac{\mathrm{\pi }}{3}\right)=0 \\ \Rightarrow \frac{a}{2}-1=0 \\ \Rightarrow \frac{a}{2}=1 \\ \Rightarrow a=2 \end{gathered}$$

Hence, the correct option is $\left(\mathrm{C}\right)$.