Inverse Circular Functions 2 Question 1
1. Let $f(x)=\log _e(\sin x),(0<x<\pi)$ and $g(x)=\sin ^{-1}\left(e^{-x}\right)$, $(x \geq 0)$. If $\alpha$ is a positive real number such that $a=(f \circ g)^{\prime}(\alpha)$ and $b=(f \circ g)(\alpha)$, then (2019 Main, 10 April II)
(a) $a \alpha^{2}-b \alpha-a=0$
(b) $a \alpha^{2}-b \alpha-a=1$
(c) $a \alpha^{2}+b \alpha-a=-2 \alpha^{2}$
(d) $a \alpha^{2}+b \alpha+a=0$
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Answer:
Correct Answer: 1. (b)
Solution:
- Given functions, $f(x)=\log _e(\sin x), \quad(0<x<\pi)$ and $g(x)=\sin ^{-1}\left(e^{-x}\right), x \geq 0$. Now, $f \circ g(x)=f(g(x))=f\left(\sin ^{-1}\left(e^{-x}\right)\right)$
$=\log _e\left(\sin \left(\sin ^{-1}\left(e^{-x}\right)\right)\right)$
$=\log _e\left(e^{-x}\right)$
$\because \sin (\sin^{-1} x)=x$,
if $x \in[-1,1] $
$=-x$
and
$ (f \circ g)^{\prime}(x)=\frac{d}{d x}(-x)=-1 $
According to the question, $\because a=(f \circ g)^{\prime}(\alpha)=-1$ [from Eq. (ii)] [from Eq. (i)]
and $\quad b=(f \circ g)(\alpha)=-(\alpha)$
for a positive real value ’ $\alpha$ ‘.
Since, the value of $a=-1$ and $b=-\alpha$, satisfy the quadratic equation (from the given options)
$ a \alpha^{2}-b \alpha-a=1 $
