Question

The number of integral value(s) in the range of the expression $\frac{6\tan x+4-4{\tan }^{2}x}{1+{\tan }^{2}x}$ is

Answer 1

$\frac{6\tan x+4-4{\tan }^{2}x}{1+{\tan }^{2}x}=3\left(\frac{2\tan x}{1+{\tan }^{2}x}\right)+4\left(\frac{1-{\tan }^{2}x}{1+{\tan }^{2}x}\right)=3\sin 2x+4\cos 2x$

Now,
$3\sin 2x+4\cos 2x\in [-\sqrt{3^2+4^2},\sqrt{3^2+4^2}]\Rightarrow 3\sin 2x+4\cos 2x\in [-5,5]$

Hence, the number of integral values in the range is $11.$


Alternate Solution:
Let $t=\tan x\Rightarrow t\in R$, so
$y=\frac{6\tan x+4-4{\tan }^{2}x}{1+{\tan }^{2}x}\Rightarrow y=\frac{6t+4-4t^2}{1+t^2}\Rightarrow t^2(y+4)-6t+(y-4)=0$
$\because t\in R$
$\Delta \ge 0\Rightarrow 36-4(y^2-16)\ge 0\Rightarrow y^2\le 25\Rightarrow y\in [-5,5]$

Hence, the number of integral values in the range is $11.$