Question

If for the function $f\left(x\right)={\cos }^{-1}\left(\frac{x^2}{\sqrt{1+x^2}}\right)$ the range is (a,b), then find sin(b).

Answer 1

Consider $f\left(\theta \right)=co{s}^{-1}\left(\theta \right)$

Then $\theta ϵ[-1,1]$

Similarly in the above case

$-1\le \frac{x^2}{\sqrt{1+x^2}}\le 1$

Now $\frac{x^2}{\sqrt{1+x^2}}$ will always be greater than 0.

Hence, $0\le \frac{x^2}{\sqrt{1+x^2}}\le 1$

$\frac{x^2}{\sqrt{1+x^2}}ϵ[0,1]$

Hence, ${\cos }^{-1}\left(\frac{x^2}{\sqrt{1+x^2}}\right)ϵ[0,\frac{\pi }{2}]$

$f\left(x\right)ϵ[0,\frac{\pi }{2}]$

Hence $a=0$ and $b=\frac{\pi }{2}$

Therefore $\sin b=\sin \left(\frac{\pi }{2}\right)$$=1$.