Question

Let \(f(x)=\log_e(\sin x)\), \((0\lt x\lt\pi)\) and \(g(x)=\sin^{-1}(e^{-x})\), \((x\ge 0)\). If \(\alpha\) is a positive real number such that \(a=(f\circ g)^{\prime}(\alpha)\) and \(b=(f\circ g)(\alpha)\), then :

• A. \(a\alpha^2+b\alpha-a=-2\alpha^2\)
• B. \(a\alpha^2+b\alpha+a=0\)
• C. \(a\alpha^2-b\alpha-a=0\)
• D. \(a\alpha^2-b\alpha-a=1\)

Hint:

Simplify \(f(g(x))=\ln\left(\sin(\sin^{-1}e^{-x})\right)=-x\) first.

Answer 1

The Correct Option is D

\((f\circ g)(x)=\log_e\left(\sin(\sin^{-1}e^{-x})\right)=\log_e(e^{-x})=-x\) (valid since \(0\lt e^{-x}\le1\)).
So \(b=(f\circ g)(\alpha)=-\alpha\) and \(a=(f\circ g)^{\prime}(\alpha)=-1\).
Test the options: \(a\alpha^2-b\alpha-a=-\alpha^2+\alpha^2+1=1\). This matches option (4).
Others: (1) LHS \(=-\alpha^2-\alpha^2+1=1-2\alpha^2\ne-2\alpha^2\); (2) \(-\alpha^2-\alpha^2-1\ne0\); (3) LHS \(=1\ne0\).
Answer: (4) \(a\alpha^2-b\alpha-a=1\)