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Q. Which of the following option is/are correct?

Relations and Functions - Part 2

Solution:

(A)$\tan ^{-1}\left(x^2\right)=\cot ^{-1}\left(\frac{1}{x^2}\right)$ is obvious true for all $x \in R-\{0\}$
(B)$\cos ^{-1}\left(\frac{1-x^2}{1+x^2}\right) \in[0, \pi)$ for all $x \in R \Rightarrow 2 \cos ^{-1}\left(\frac{1-x^2}{1+x^2}\right) \neq 2 \pi$
(C) Domain of $f ( x )$ is $\{ \pm 1\}$ range of $f(x)$ is $\{0\}$
$\therefore f ( x )=0 \forall x \in \text { domain of } f ( x )$
(D) $\tan ^2 x+\cot ^2 x \geq 2$ for all $x \in R-\left(\frac{n \pi}{2}\right) n \in I$ $\Rightarrow $
(D) is incorrect ( $f$ is not defined for any $x$. Domain is $\phi$ )