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Let $\mathrm{f}(x)=\log (\sin x), 0 < x < \pi$ and $\mathrm{g}(x)=\sin ^{-1}\left(\mathrm{e}^{-x}\right), x \geq 0$.
If $\alpha$ is a positive real number such that $a=(f \circ g)^{\prime}(\alpha)$ and $b=(f \circ g)(\alpha)$, then
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If $\alpha$ is a positive real number such that $a=(f \circ g)^{\prime}(\alpha)$ and $b=(f \circ g)(\alpha)$, then
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Verified Answer
The correct answer is:
$a \alpha^2-b \alpha-a=1$
$\begin{array}{ll} & \mathrm{f}(x)=\log (\sin x), 0 < x < \pi \text { and } \\ & \mathrm{g}(x)=\sin ^{-1}\left(\mathrm{e}^{-x}\right), x \geq 0 \\ \therefore \quad & \left(\text { fog) }(x)=\log \left[\sin \left(\sin ^{-1} \mathrm{e}^{-x}\right)\right]=\log \left(\mathrm{e}^{-x}\right)=-x\right. \\ \therefore \quad & (\text { fog })^{\prime}(x)=-1 \\ \therefore \quad & \mathrm{a}=(\text { fog })^{\prime}(\alpha)=-1 \text { and } \mathrm{b}=(\mathrm{fog})(\alpha)=-\alpha \\ & \text { These values satisfy only option (B). } \\ \therefore \quad & \text { Option (B) is correct. }\end{array}$
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