Question

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

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

Answer 1

Answer: (A), (C)

Number of solution of $g\left(x\right)=1$ in $[0,4\pi ]$ is $1$
Number of solution of $f\left(x\right)\sin \frac{x}{4}=2$ in $[0,8\pi ]$ is $2$

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 I\Rightarrow x=\frac{8n\pi }{3}$
In $[0,4\pi ]$, only $x=\frac{8\pi }{3}$
Case-2: When $\sin x=0$,and $g\left(x\right)=1$ then $\cos \frac{x}{4}=-1$
$\Rightarrow$ $x=4\pi$
In $[0,4\pi ]$, only two solutions
If $f\left(x\right)\sin \frac{x}{4}=2\Rightarrow \cos \frac{3x}{4}+\cos \frac{x}{4}=2$
$\Rightarrow \cos \frac{x}{4}=1$ and $\cos \frac{3x}{4}=1$
$\Rightarrow$ Only two $x$ exists in $[0,8\pi ]$, that is $0$ and $8\pi$