Question Error Report

Thank you for reporting, we will resolve it shortly

Back to Question

Q. Let $f(x)=\sqrt{\left(\sin ^{-1} x-\cos ^{-1} x\right)}, \quad g(x)=\sqrt{\left(\tan ^{-1} x-\cot ^{-1} x\right)}$ and $h(x)=$ $\sqrt{\left(\sec ^{-1} x -\operatorname{cosec}^{-1} x \right)}$ then correct statement(s) is(are)

Inverse Trigonometric Functions

Solution:

$ \sin ^{-1}-\cos ^{-1} x=2 \sin ^{-1} x-\pi / 2$
$f(x): \sqrt{\left(2 \sin ^{-1} x-\pi / 2\right)}$ so $2 \sin ^{-1} x-\pi / 2 \geq 0 \Rightarrow \sin ^{-1} x \geq \pi / 4$
so Domain of, $f(x): \frac{1}{\sqrt{2}} \leq x \leq 1 $ similarly $g(x)=\sqrt{\left(2 \tan ^{-1} x-\frac{\pi}{2}\right)}, \tan ^{-1} x \geq \frac{\pi}{4}$
so Domain of, $g(x): 1 \leq x<\infty$ similarly $h(x)=\sqrt{\left(2 \sec ^{-1} x-\frac{\pi}{2}\right)}, \sec ^{-1} x \geq \frac{\pi}{4}$
so Domain of, $h(x):(-\infty,-1] \cup[\sqrt{2}, \infty)$ so Domain of $f(x)+g(x)$ is 1 only.
for $g(x)+h(x)$ Domain will be $[\sqrt{2}, \infty)$,
for $h(x)+f(x)$, Domain will be null set
for $g(x)+h(x)+f(x)$, Domain will be null set .