Question

The maximum value of function $f\left(x\right)=\sin x(1+\cos x),x\in R$ is:

• A. $\frac{3^{\frac{3}{2}}}{4}$
• B. $\frac{3^{\frac{5}{3}}}{4}$
• C. $\frac{3}{2}$
• D. $\frac{3^{\frac{7}{5}}}{2}$

Answer 1

The correct option is A $\frac{3^{\frac{3}{2}}}{4}$

Let $f\left(x\right)=\sin x(1+\cos x)$.

Now, $f^{\prime }\left(x\right)=\cos x(1+\cos x)-{\sin }^{2}x=\cos x+{\cos }^{2}x-(1-{\cos }^{2}x)$

$=2{\cos }^{2}x+\cos x-1$.

Again $f^{\prime \prime }\left(x\right)=-4\cos x\sin x-\sin x$.

For, maximum value of $f\left(x\right)$ we must have $f^{\prime }\left(x\right)=0$.

or, $2{\cos }^{2}x+\cos x-1=0$

or, $\cos x=\frac{-1\pm \sqrt{1+4\times 2}}{4}$

or, $\cos x=-1,\frac{1}{2}\Rightarrow \sin x=0,\frac{\sqrt{3}}{2}$ respectively.

For, $\cos x=\frac{1}{2}$ and $\sin x=\frac{\sqrt{3}}{2}$ we have $f^{\prime \prime }\left(x\right)<0$.

This values of $\cos x$ and $\sin x$ gives the maximum value of $f\left(x\right)$.

$\therefore$ Maximum value of $f\left(x\right)$ is $\frac{\sqrt{3}}{2}\times \frac{3}{2}=\frac{3^{\frac{3}{2}}}{4}$