Question

The value of $a$ for which the function $f\left(x\right)=a\sin x+\left(\frac{1}{3}\right)\sin 3x$ has an extremum at $x=\frac{\pi }{3}$ is

• A. $1$
• B. $-1$
• C. $0$
• D. $2$

Answer 1

Answer: (D)

$2$

The given information in the question is:

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

An extremum is calculated from the derivative of the function about a point where the derivative is equal to 0.

So we calculate the derivative and equate it to 0.

$\Rightarrow f^{\prime }\left(x\right)=a\cos x+\cos 3x$

Now equating $f^{\prime }\left(x\right)$ to zero.

$\Rightarrow f^{\prime }\left(x\right)=0$

$\Rightarrow a\cos x+\cos 3x=0$

Now it is given that the extremum is at $x=\frac{\pi }{3}$

so substituting the value of extremum in the derivative equation which is its solution, we get

$\Rightarrow a\cos \frac{\pi }{3}+\cos \pi =0$

We know the value of $\cos \frac{\pi }{3}=\frac{1}{2}$ and $\cos \pi =-1$. Substituting these values we get,

$\Rightarrow \frac{a}{2}-1=0$

$\Rightarrow a=2$ .....Answer{Option(B)}

For the function $f\left(x\right)$ to have an the extremum at the required point the value of $a=2$.