Question

If $f\left(x\right)=\sin x+\sin x\cos \frac{x}{2}+\sin x{\cos }^2\frac{x}{2}+\cdots$ and $g\left(x\right)=f\left(x\right)\cdot \sin \frac{x}{4}-\cos \frac{x}{4},$ then

• A. number of solutions of $g\left(x\right)=1$ in $[0,4\pi ]$ is $1$
• B. number of solutions of $g\left(x\right)=1$ in $[0,4\pi ]$ is $2$
• C. number of solutions of $f\left(x\right)\sin \frac{x}{4}=2$ in $[0,8\pi ]$ is $2$
• D. number of solutions of $f\left(x\right)\sin \frac{x}{4}=2$ in $[0,8\pi ]$ is $0$

Answer 1

Answer: (B), (D)

The correct options are
B number of solutions of $g\left(x\right)=1$ in $[0,4\pi ]$ is $2$
D number of solutions of $f\left(x\right)\sin \frac{x}{4}=2$ in $[0,8\pi ]$ is $0$

Case $1:$ When $\sin x\ne 0$
$f\left(x\right)=\frac{\sin x}{1-\cos \frac{x}{2}}$
$=\frac{2\sin \frac{x}{2}\cos \frac{x}{2}}{2{\sin }^2\frac{x}{4}}$
$=\frac{2\cos \frac{x}{4}\cos \frac{x}{2}}{\sin \frac{x}{4}}$
$=\frac{\cos \frac{3x}{4}+\cos \frac{x}{4}}{\sin \frac{x}{4}}$

$\therefore g\left(x\right)=\cos \frac{3x}{4}$

If $g\left(x\right)=1\Rightarrow \cos \frac{3x}{4}=1$
$\Rightarrow \frac{3x}{4}=2n\pi ,n\in Z$
$\Rightarrow x=\frac{8n\pi }{3},n\in Z$
In $[0,4\pi ],$ only $x=\frac{8\pi }{3}$ is possible as $\sin x\ne 0$

Case $2$: When $\sin x=0$, then
$f\left(x\right)=0$
$g\left(x\right)=f\left(x\right)\cdot \sin \frac{x}{4}-\cos \frac{x}{4}=-\cos \frac{x}{4}$

If $g\left(x\right)=1\Rightarrow -\cos \frac{x}{4}=1$
$\Rightarrow \cos \frac{x}{4}=-1$
$\Rightarrow \frac{x}{4}=(2n+1)\pi ,n\in Z$
$\Rightarrow x=4(2n+1)\pi ,n\in Z$
In $[0,4\pi ],$ only $x=4\pi$ is possible.

So, there are only $2$ solutions of $g\left(x\right)=1$ in $[0,4\pi ]$

If $f\left(x\right)\sin \frac{x}{4}=2$, then
$\Rightarrow 2\cos \frac{x}{4}\cdot \cos \frac{x}{2}=2$
$\Rightarrow \cos \frac{x}{4}\cdot \cos \frac{x}{2}=1$
$\Rightarrow \cos \frac{x}{4}=1\text{and}\cos \frac{x}{2}=1$
$\Rightarrow x=8n\pi ,n\in Z$
which is not possible.
So, no such $x$ exists in $[0,8\pi ]$ for $f\left(x\right)\sin \frac{x}{4}=2$