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Range of the function , $f (x) = cot ^{-1}$ $\left( {{{\log }_{4/5}}\,\,(5\,{x^2}\,\, - \,\,8\,x\,\, + \,\,4)\,} \right)$ is :
$(0 , \pi )$
$\left[ {\frac{\pi }{4}\,\,,\,\,\pi } \right)$
$\left( {0\,\,,\,\,\frac{\pi }{4}} \right]$
$\left( {0\,\,,\,\,\frac{\pi }{2}} \right)$
Solution
Let $g(x)=\left(5 x^{2}-8 x+4\right)$
$=5\left(x^{2}-\frac{8}{5} x+\frac{4}{5}\right)$
$=\left(5\left(x-\frac{4}{5}\right)^{2}+\frac{4}{5}\right)$
Thus, $R_{g}=\left[\frac{4}{5}, \infty\right)$
$\frac{4}{5} \leq g(x)<\infty$
$\log _{5 / 4}\left(\frac{4}{5}\right) \leq \log _{5 / 4}(g(x))<\log _{5 / 4}(\infty)$
$-1 \leq \log _{5 / 4}(g(x))<\infty$
$\tan ^{-1}(-1) \leq \tan ^{-1}\left(\log _{5 / 4}(g(x))\right)<\tan ^{-1}(\infty)$
$-\frac{\pi}{4} \leq \tan ^{-1}\left(\log _{5 / 4}((g(x)))<\frac{\pi}{2}\right.$
$-\frac{\pi}{4} \leq f(x)<\frac{\pi}{2}$
Thus, $\quad R_{f}=\left[-\frac{\pi}{4}, \frac{\pi}{2}\right)$
