Question

Which of the following option is/are correct?

• A. ${\tan }^{-1}\left(x^2\right)={\cot }^{-1}\left(\frac{1}{x^2}\right)$ is true for all $x\in R-\{0\}$
• B. $2{\cos }^{-1}\left(\frac{1-x^2}{1+x^2}\right)=2\pi \Rightarrow$ no real solution
• C. If $f(x)={\cos }^{-1}|x|+{\sec }^{-1}|x|$ then $f(x)$ is even as well as odd.
• D. If $f(x)={\sin }^{-1}\left({\tan }^{2}x+{\cot }^{2}x\right)+{\mathrm{cosec}}^{-1}\left({\sin }^{2}x+{\mathrm{cosec}}^{2}x\right)$ then domain of $f(x)$ is $R-\left\{\frac{n\pi }{2}\right\}n\in I$.

Answer 1

Answer: (A), (B), (C)

(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 $x\in R\Rightarrow 2{\cos }^{-1}\left(\frac{1-x^2}{1+x^2}\right)\ne 2\pi$
(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) ${\tan }^{2}x+{\cot }^{2}x\ge 2$ 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 $\varphi$ )