Question

For the function $f\left(x\right)=\frac{{\tan }^{2}x-{\cot }^{2}x+1}{{\tan }^{2}x+{\cot }^{2}x-1}$ , which of the following statement(s) is/are correct

• A. Global maximum value of $f\left(x\right)$ is $\frac{5}{3}$
• B. Global minimum value of $f\left(x\right)$ is $-1$
• C. Global maximum value of $f\left(x\right)$ does not exist
• D. Global minimum value of $f\left(x\right)$ does not exist

Answer 1

Answer: (A), (D)

The correct options are
A Global maximum value of $f\left(x\right)$ is $\frac{5}{3}$
D Global minimum value of $f\left(x\right)$ does not exist

$f\left(x\right)=\frac{ta{n}^{2}x-co{t}^{2}x+1}{ta{n}^{2}x+co{t}^{2}x-1}$
$f\left(x\right)=\frac{{\tan }^{4}x+{\tan }^{2}x-1}{{\tan }^{4}x-{\tan }^{2}x+1}$
Put $tanx=t$
Then , $g\left(t\right)=\frac{t^4+t^2-1}{t^4-t^2+1}$
$g^{\prime }\left(t\right)=\frac{-4t^5-8t^3}{(t^4-t^2+1{)}^2}$
For maxima or minima,
$g^{\prime }\left(t\right)=0$
$\Rightarrow t=0$(not possible), $t^2=2$
$g^{\prime \prime }\left(t\right)<0$ at $t^2=2$
So, g(t) has a maximum at $t^2=2$
So, $g\left(2\right)=\frac{5}{3}$
Hence, f(x) has a maximum value $\frac{5}{3}$ and there is no value at which minimum of f(x) exists.