Question

Let $\left\{y\right\}$ and $\left[y\right]$ denote fractional part function and greatest integer function respectively.
If $f\left(x\right)={\sin }^{-1}\{[3x+2]-\{3x+(x-\left\{2x\right\}\left)\right\}\}$ for $x\in (0,\frac{\pi }{12})$ and $(g\circ f)\left(x\right)=x$ for all $x\in (0,\frac{\pi }{12}),$ then $g^{\prime }\left(\frac{\pi }{6}\right)$ is equal to

• A. $\frac{\sqrt{3}}{8}$
• B. $\frac{-1}{4}$
• C. $\frac{1}{8}$
• D. $\frac{-\sqrt{3}}{4}$

Answer 1

The correct option is D $\frac{-\sqrt{3}}{4}$

$f\left(x\right)={\sin }^{-1}\left\{\right[3x+2]-\{3x+(x-\left\{2x\right\}\left)\right\}\}$
Since $x\in (0,\frac{\pi }{12}),$
$\Rightarrow 2x\in (0,\frac{\pi }{6})$ and $3x\in (0,\frac{\pi }{4})$
$\therefore \left\{2x\right\}=2x$ and $\left\{3x\right\}=3x$

We know $\{x+n\}=\left\{x\right\}$ for $n\in Z$
$\therefore f\left(x\right)={\sin }^{-1}\{-\{3x+(x-2x\left)\right\}\}f\left(x\right)={\sin }^{-1}\{-\{3x-x\}\}={\sin }^{-1}\{-2x\}={\sin }^{-1}(1-\{2x\left\}\right)={\sin }^{-1}(1-2x)$

Given that $(g\circ f)\left(x\right)=x$ for all $x\in (0,\frac{\pi }{12})$
It means $g\left(x\right)=f^{-1}\left(x\right)$
$f\left(x\right)={\sin }^{-1}(1-2x)$
Let $y={\sin }^{-1}(1-2x)$
Then $x=\frac{1-\sin y}{2}=f^{-1}\left(y\right)$
$\Rightarrow g\left(x\right)=f^{-1}\left(x\right)=\frac{1-\sin x}{2}\Rightarrow g^{\prime }\left(x\right)=-\frac{\cos x}{2}\Rightarrow g^{\prime }\left(\frac{\pi }{6}\right)=-\frac{\sqrt{3}}{4}$