Range of the function , f (x) = cot ^{-1} lef | vedclass

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Question

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

Vedclass · Accepted Answer

$\left[ {\frac{\pi }{4}\,\,,\,\,\pi } \right)$

Vedclass · Answer

Let $g(x)=\left(5 x^{2}-8 x+4ight)$

$=5\left(x^{2}-\frac{8}{5} x+\frac{4}{5}ight)$

$=\left(5\left(x-\frac{4}{5}ight)^{2}+\frac{4}{5}ight)$

Thus, $R_{g}=\left[\frac{4}{5}, \inftyight)$

$\frac{4}{5} \leq g(x)<\infty$

$\log _{5 / 4}\left(\frac{4}{5}ight) \leq \log _{5 / 4}(g(x))<\log _{5 / 4}(\infty)$

$-1 \leq \log _{5 / 4}(g(x))<\infty$

$	an ^{-1}(-1) \leq 	an ^{-1}\left(\log _{5 / 4}(g(x))ight)<	an ^{-1}(\infty)$

$-\frac{\pi}{4} \leq 	an ^{-1}\left(\log _{5 / 4}((g(x)))<\frac{\pi}{2}ight.$

$-\frac{\pi}{4} \leq f(x)<\frac{\pi}{2}$

Thus, $\quad R_{f}=\left[-\frac{\pi}{4}, \frac{\pi}{2}ight)$

Range of the function , $f (x) = cot ^{-1}$ $\left( {{{\log }_{4/5}}\,\,(5\,{x^2}\,\, - \,\,8\,x\,\, + \,\,4)\,} \right)$ is :

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