If alpha leq 2 sin^{-1} x + cos^{-1} x leq be | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(C) Given the expression $f(x) = 2 \sin^{-1} x + \cos^{-1} x$. We can rewrite this as $f(x) = \sin^{-1} x + (\sin^{-1} x + \cos^{-1} x)$. Since we kno

Vedclass · Accepted Answer

$\alpha = 0, \beta = \pi$

Vedclass · Answer

(C) Given the expression $f(x) = 2 \sin^{-1} x + \cos^{-1} x$. We can rewrite this as $f(x) = \sin^{-1} x + (\sin^{-1} x + \cos^{-1} x)$. Since we know the identity $\sin^{-1} x + \cos^{-1} x = \frac{\pi}{2}$ for $x \in [-1, 1]$,the expression becomes $f(x) = \sin^{-1} x + \frac{\pi}{2}$. The range of $\sin^{-1} x$ is $[-\frac{\pi}{2}, \frac{\pi}{2}]$. Adding $\frac{\pi}{2}$ to all parts of the inequality,we get $-\frac{\pi}{2} + \frac{\pi}{2} \leq \sin^{-1} x + \frac{\pi}{2} \leq \frac{\pi}{2} + \frac{\pi}{2}$. This simplifies to $0 \leq 2 \sin^{-1} x + \cos^{-1} x \leq \pi$. Comparing this with $\alpha \leq 2 \sin^{-1} x + \cos^{-1} x \leq \beta$,we find $\alpha = 0$ and $\beta = \pi$.

If $\alpha \leq 2 \sin^{-1} x + \cos^{-1} x \leq \beta$,then

Similar Questions

If $\sec^{-1} x = \csc^{-1} y$,then $\cos^{-1} \frac{1}{x} + \cos^{-1} \frac{1}{y} = $

If $\tan ^{-1} 2 x+\tan ^{-1} 3 x=\frac{\pi}{4}$,then the value of $x$ is

$2 \pi - \left(\sin ^{-1} \frac{4}{5} + \sin ^{-1} \frac{5}{13} + \sin ^{-1} \frac{16}{65}\right)$ is equal to

If $(\tan^{-1} x)^2 + (\cot^{-1} x)^2 = \frac{5\pi^2}{8}$, then the value of $x$ is equal to...

$\sin^{-1}x + \sin^{-1}\frac{1}{x} + \cos^{-1}x + \cos^{-1}\frac{1}{x} = $

Vedclass Test Series

Exam Paper Generator

Online Exam Module