Question

if $f\left(x\right)=\sqrt{x}$, $g\left(x\right)=\frac{3x^2-x^6}{1-3x^4}$ and $h\left(x\right)=\tan x,$ then $(g\circ f\circ h)\left(\frac{\pi }{3}\right)$ is

• A. $0$
• B. $1$
• C. $-1$
• D. $2$

Answer 1

The correct option is A $0$

First find $f\left(h\right(x\left)\right)$
$f\left(h\right(x\left)\right)=\sqrt{\tan x}$
Then, $(g\circ f\circ h)\left(x\right)=g\left(f\right(h\left(x\right)\left)\right)=\frac{3(\sqrt{\tan x}{)}^2-(\sqrt{\tan x}{)}^6}{1-3(\sqrt{\tan x}{)}^4}$
$\Rightarrow g\left(f\right(h\left(x\right)\left)\right)=\frac{3\tan x-(\tan x{)}^3}{1-3(\tan x{)}^2}$
$\Rightarrow g\left(f\right(h\left(x\right)\left)\right)=\tan 3x$

$\therefore (g\circ f\circ h)\left(\frac{\pi }{3}\right)=\tan 3\left(\frac{\pi }{3}\right)=\tan \pi =0$