Question

If $y={\text{Cosec}}^{-1}(x)$ and $\frac{dy}{dx}=\frac{-1}{|x|\sqrt{x^2-1}}$, then

• A. $y\in \left(-\frac{\pi }{2},0\right)$
• B. $y\in \left(-\frac{\pi }{2},2\pi \right)$
• C. $y\in \left(-\frac{\pi }{2},0\right)\cup \left(0,\frac{\pi }{2}\right)$
• D. $y\in R$

Answer 1

Answer: (C)

It is given that, $y={\text{cosec}}^{-1}x$
and $\frac{dy}{dx}=\frac{-1}{|x|\sqrt{x^2-1}}$
$\because$ Domain of ${\text{cosec}}^{-1}x$ is $\left[-\frac{\pi }{2},\frac{\pi }{2}\right]-\{0\}$
and the $\frac{dy}{dx}$ is define for $x\in \left(-\frac{\pi }{2},0\right)\cup \left(0,\frac{\pi }{2}\right)$.
Because for derivatives we should exclude the end points of a internal.