The range of sin^{-1}left(frac{1+x^2}{2+x^2}r | vedclass

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

Question

(D) Let $f(x) = \sin^{-1}\left(\frac{1+x^2}{2+x^2}\right)$.We can rewrite the argument as $\frac{1+x^2}{2+x^2} = \frac{2+x^2-1}{2+x^2} = 1 - \frac{1}{

Vedclass · Accepted Answer

$[ \frac{\pi}{6}, \frac{\pi}{2} )$

Vedclass · Answer

(D) Let $f(x) = \sin^{-1}\left(\frac{1+x^2}{2+x^2}\right)$.We can rewrite the argument as $\frac{1+x^2}{2+x^2} = \frac{2+x^2-1}{2+x^2} = 1 - \frac{1}{2+x^2}$.Since $x^2 \ge 0$,we have $2+x^2 \ge 2$,which implies $0 Multiplying by $-1$,we get $-\frac{1}{2} \le -\frac{1}{2+x^2} Adding $1$ to all parts,we get $1 - \frac{1}{2} \le 1 - \frac{1}{2+x^2} Since the function $\sin^{-1}(u)$ is strictly increasing,the range is $[\sin^{-1}(\frac{1}{2}), \sin^{-1}(1))$.Therefore,the range is $[\frac{\pi}{6}, \frac{\pi}{2})$.

The range of $\sin^{-1}\left(\frac{1+x^2}{2+x^2}\right)$ is:

Similar Questions

The range of $\operatorname{Sin}^{-1} x + \operatorname{Cos}^{-1} x + \operatorname{Tan}^{-1} x$ is

If $y = \tan^{-1} x$,then . . . . . . .

If $f(x) = \sqrt{2x - 1} + 5 \cos^{-1}\left(\frac{2x - 1}{3}\right)$,then the domain of the function $f(x)$ is

The number of real roots of the equation $\tan ^{-1} \sqrt{x(x+1)}+\sin ^{-1} \sqrt{x^{2}+x+1}=\frac{\pi}{4}$ is:

Let $[\cdot]$ denote the greatest integer function. If the domain of the function $f(x) = \cos^{-1} \left( \frac{4x+2[x]}{3} \right)$ is $[\alpha, \beta]$,then $12(\alpha + \beta)$ is equal to:

Vedclass Test Series

Exam Paper Generator

Online Exam Module