Question

Let $f\left(x\right)=2{\tan }^{-1}x$ and $g\left(x\right)=x+2.$ Then the number of integer(s) satisfying $\left(\right(f\circ g\left)\right(x){)}^2-5(f\circ g\left)\right(x)+4>0$ where $x\in (-10,10)$ is

Answer 1

$f\left(x\right)=2{\tan }^{-1}x$
$\Rightarrow f\left(g\left(x\right)\right)=2{\tan }^{-1}(x+2)$
$(f\left(g\left(x\right)\right){)}^2-5f\left(g\left(x\right)\right)+4>0$
$\Rightarrow \left(f\right(g\left(x\right))-1)\left(f\right(g\left(x\right))-4)>0$
$\Rightarrow f\left(g\left(x\right)\right)<1$ or $f\left(g\left(x\right)\right)>4$
$\Rightarrow {\tan }^{-1}(x+2)<\frac{1}{2}$
or ${\tan }^{-1}(x+2)>2$ (not possible)
$\Rightarrow {\tan }^{-1}(x+2)<\frac{1}{2}$
$\Rightarrow x+2<\tan \left(\frac{1}{2}\right)$
$\Rightarrow x\in (-\infty ,\tan \left(\frac{1}{2}\right)-2)$

As $x\in (-10,10),$
$x\in (-10,\tan \left(\frac{1}{2}\right)-2)$
As $\frac{1}{2}<\frac{\pi }{6}$
$\Rightarrow \tan \frac{1}{2}<\frac{1}{\sqrt{3}}$
$\Rightarrow \tan \frac{1}{2}-2<\frac{1}{\sqrt{3}}-2\approx -1.4$
Hence, integers in the range are $-9,-8,-7,-6,-5,-4,-3,-2$
i.e., $8$ integers