Question

Let $f\left(x\right)=\frac{\left[x\sin a\right]}{\left[x+1\right]},x\ne -1,f\ne 1,f\left(x\right)\ne 0.$ Then for what value of $\sin a$is $f\left(f\right(x\left)\right)=x$?

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

Answer 1

The correct option is D

$-1$

Explanation for the correct answer

Given that: $f\left(x\right)=\frac{\left[x\sin a\right]}{\left[x+1\right]},x\ne -1,f\ne 1,f\left(x\right)\ne 0.$

Now, calculating$f\left(fx\right))$:

$$\begin{gathered} \Rightarrow f\left(fx\right)=\frac{\left[f\left(x\right)\sin a\right]}{f\left(x\right)+1} \\ \Rightarrow f\left(fx\right)=\frac{{\displaystyle \frac{x\sin a}{x+1}}.\sin a}{{\displaystyle \frac{x\sin a}{x+1}}+1} \\ \Rightarrow f\left(fx\right)=\frac{x{\sin }^{2}a}{x\sin a+x+1} \\ \Rightarrow x=\frac{x{\sin }^{2}a}{x\sin a+x+1}\left[\mathrm{Given}\mathrm{that}:f\left(fx\right)=x\right] \\ \Rightarrow x\sin a+x+1={\sin }^{2}a \\ \Rightarrow {\sin }^{2}a-x\sin a-x-1=0 \\ \Rightarrow {\sin }^{2}a+\sin a-x\sin a-x-1-\sin a=0 \\ \Rightarrow \sin a(\sin a-1-x)+1(\sin a-x-1)=0 \\ \Rightarrow (\sin a+1)(\sin a-x-1)=0 \\ \Rightarrow \sin a=-1or(x+1) \end{gathered}$$

Hence, the correct answer is Option (D).