Find the range of $f(x)=\cos^{-1}(\frac{\sqrt{1+2x^2}}{1+x^2})$

$[0,\frac{π}{2})$

$f(x)=\cos^{-1}(\frac{\sqrt{1+2x^2}}{1+x^2})$

$=\sin^{-1}\left(\sqrt{1-\frac{1+2x^2}{(1+x^2)^2}}\right)$  $(∵\cos^{-1}x=\sin^{-1}\sqrt{1-x^2})$

$=\sin^{-1}\left(\sqrt{\frac{x^4}{(1+x^2)^2}}\right)$

$=\sin^{-1}(\frac{x^2}{1+x^2})$

$=\sin^{-1}(1-\frac{1}{1+x^2})$

Now, $(1-\frac{1}{1+x^2})∈[0,1)$

$∴f(x)∈[0,\frac{π}{2})$