Question

Let $f:(0,\infty )\to R$ be defined as $f\left(x\right)={\tan }^{-1}\left({\log }_2\right(x\left)\right)$ and $g:R\to R$ be defined as $g\left(x\right)=\frac{2x^2-3x+2}{2x^2+3x+2}$

If $f\circ g$ denotes the composition of $f$ and $g,$ then

• A. $f\circ g(\frac{-1}{x})=f\circ g(-x)$
• B. $f\circ g\left(x\right)$ is strictly increasing in the domain $(-\infty ,\infty )$
• C. $f\circ g\left(x\right)$ is many-one function
• D. Range of $g\left(x\right)$ is $[\frac{1}{7},7]$

Answer 1

Answer: (A), (C), (D)

The correct options are
A $f\circ g(\frac{-1}{x})=f\circ g(-x)$
C $f\circ g\left(x\right)$ is many-one function
D Range of $g\left(x\right)$ is $[\frac{1}{7},7]$

$f\circ g\left(x\right)={\tan }^{-1}\left({\log }_2\left(\frac{2x^2-3x+2}{2x^2+3x+2}\right)\right)$
Replacing $x$ by $\frac{-1}{x},$ we get
$f\circ g\left(\frac{-1}{x}\right)={\tan }^{-1}\left({\log }_2\left(\frac{2x^2+3x+2}{2x^2-3x+2}\right)\right)$

Replacing $x$ by $-x$, we get
$f\circ g\left(-x\right)={\tan }^{-1}\left({\log }_2\left(\frac{2x^2+3x+2}{2x^2-3x+2}\right)\right)$
$\Rightarrow f\circ g\left(\frac{-1}{x}\right)=f\circ g\left(-x\right)$

As we already proved that $f\circ g\left(\frac{-1}{x}\right)=f\circ g\left(-x\right)$, so it can't be strictly increasing in the given domain $(-\infty ,\infty )$.
So, $f\circ g\left(x\right)$ is many-one function.

Now, Let $y=\frac{2x^2-3x+2}{2x^2+3x+2}$
$\Rightarrow (2y-2)x^2+(3y+3)x+2y-2=0$
Now $D\ge 0$
$(3y+3{)}^2-4(2y-2\left)\right(2y-2)\ge 0$
$\Rightarrow 9y^2+18y+9-16y^2+32y-16\ge 0$
$\Rightarrow -7y^2+50y-7\ge 0$
$\Rightarrow 7y^2-50y+7\le 0$

$\Rightarrow y\in [\frac{1}{7},7]$