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

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(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}{

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

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(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

If $A = \{x \in R : \sin^{-1}(\sqrt{x^2+x+1}) \in [-\frac{\pi}{2}, \frac{\pi}{2}]\}$ and $B = \{y \in R : y = \sin^{-1}(\sqrt{x^2+x+1}), x \in A\}$,then which of the following is true?

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