Question

If $f\left(x\right)=ln\left(\sin x\right)$ and $g\left(x\right)={\sin }^{-1}\left(e^{-x}\right)$ for all $x\in (0,\pi )$ and $f\left(g\right(\alpha \left)\right)=b$ and $\left(f\right(g\left(x\right){)}^{\prime }$ at $x=\alpha$ is a then which is true?

• A. $a{\alpha }^2-b\alpha +1=a$
• B. $a{\alpha }^2-b\alpha =-a$
• C. $a{\alpha }^2-b\alpha +1=-a$
• D. $a{\alpha }^2-b\alpha -1=-a$

Answer 1

Answer: (C)

$a{\alpha }^2-b\alpha +1=-a$

$f\left(g\right(x\left)\right)=ln\left(\sin \right({\sin }^{-1}\left(e^{-x}\right)\left)\right)$
$=ln\left(e^{-x}\right)=-x$
$\left(f\right(g\left(x\right){)}^{\prime }=-1$
Now $f\left(g\right(\alpha )=-\alpha =b$
and $f\left(g\right(x){)}^{\prime }$ at $x=\alpha$ is $-1=a$
Now $a{\alpha }^2-b\alpha +1=-{\alpha }^2-(-\alpha )\alpha +1=1=-a$.