જો વિધેય f(mathrm{x})=frac{cos ^{-1} sqrt{x^{ | vedclass

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Question

$0 \leq x^{2}-x+1 \leq 1$

$\Rightarrow x^{2}-x \leq 0$

$\Rightarrow x \in[0,1]$

$	ext { Also, } 0\,<\,\sin ^{-1}\left(\frac{2 x-1}{2}ig

Vedclass · Accepted Answer

$\frac{3}{2}$

Vedclass · Answer

$0 \leq x^{2}-x+1 \leq 1$

$\Rightarrow x^{2}-x \leq 0$

$\Rightarrow x \in[0,1]$

$	ext { Also, } 0\,<\,\sin ^{-1}\left(\frac{2 x-1}{2}ight) \leq \frac{\pi}{2}$

$\Rightarrow 0\,<\,\frac{2 x-1}{2} \leq 1$

$\Rightarrow 0\,<\,2 x-1 \leq 2$

$1\,<\,2 x \leq 3$

$\frac{1}{2}\,<\,x \leq \frac{3}{2}$

Taking intersection

$x \in\left(\frac{1}{2}, 1ight]$

$\Rightarrow \alpha=\frac{1}{2}, \beta=1$

$\Rightarrow \alpha=\frac{1}{2}, \beta=1$

જો વિધેય $f(\mathrm{x})=\frac{\cos ^{-1} \sqrt{x^{2}-x+1}}{\sqrt{\sin ^{-1}\left(\frac{2 x-1}{2}\right)}}$ નો પ્રદેશ $(\alpha, \beta]$ હોય તો $\alpha+\beta$ ની કિમંત મેળવો.

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