Q62 / 150 · MHT CET 2023 (11 May Shift 1) · Mathematics

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

  • A
    $a \alpha^2-b \alpha-a=0$
  • Correct answer
    B
    $a \alpha^2-b \alpha-a=1$
  • C
    $\mathrm{a} \alpha^2+\mathrm{b} \alpha-\mathrm{a}=-2 \alpha^2$
  • D
    $a \alpha^2+b \alpha+a=0$

Solution

$\begin{array}{ll} & \mathrm{f}(x)=\log (\sin x), 0 < x < \pi \text { and } \\ & \mathrm{g}(x)=\sin ^{-1}\left(\mathrm{e}^{-x}\right), x \geq 0 \\ \therefore \quad & \left(\text { fog) }(x)=\log \left[\sin \left(\sin ^{-1} \mathrm{e}^{-x}\right)\right]=\log \left(\mathrm{e}^{-x}\right)=-x\right. \\ \therefore \quad & (\text { fog })^{\prime}(x)=-1 \\ \therefore \quad & \mathrm{a}=(\text { fog })^{\prime}(\alpha)=-1 \text { and } \mathrm{b}=(\mathrm{fog})(\alpha)=-\alpha \\ & \text { These values satisfy only option (B). } \\ \therefore \quad & \text { Option (B) is correct. }\end{array}$

Hence, the correct answer is option: 'B'.

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