Let f(x) = log_e(sin x) for 0

Question

(C) Given $f(x) = \log_e(\sin x)$ and $g(x) = \sin^{-1}(e^{-x})$.First,find the composite function $(fog)(x) = f(g(x)) = \log_e(\sin(\sin^{-1}(e^{-x})

Vedclass · Accepted Answer

$a\alpha^2 - b\alpha - a = 1$

Vedclass · Answer

(C) Given $f(x) = \log_e(\sin x)$ and $g(x) = \sin^{-1}(e^{-x})$.First,find the composite function $(fog)(x) = f(g(x)) = \log_e(\sin(\sin^{-1}(e^{-x})))$.Since $\sin(\sin^{-1}(u)) = u$,we have $(fog)(x) = \log_e(e^{-x}) = -x$.Given $b = (fog)(\alpha)$,we have $b = -\alpha$.Next,find the derivative $(fog)'(x) = \frac{d}{dx}(-x) = -1$.Given $a = (fog)'(\alpha)$,we have $a = -1$.Now,substitute $a = -1$ and $b = -\alpha$ into the options.Checking option $B$: $a\alpha^2 - b\alpha - a = (-1)\alpha^2 - (-\alpha)\alpha - (-1) = -\alpha^2 + \alpha^2 + 1 = 1 
eq 0$.Checking option $D$: $a\alpha^2 + b\alpha + a = (-1)\alpha^2 + (-\alpha)\alpha + (-1) = -\alpha^2 - \alpha^2 - 1 = -2\alpha^2 - 1 
eq 0$.Re-evaluating the expression $a\alpha^2 - b\alpha - a$: Since $a = -1$ and $b = -\alpha$,then $a\alpha^2 - b\alpha - a = (-1)\alpha^2 - (-\alpha)\alpha - (-1) = -\alpha^2 + \alpha^2 + 1 = 1$.Thus,$a\alpha^2 - b\alpha - a = 1$ is correct,which corresponds to option $C$.

Let $f(x) = \log_e(\sin x)$ for $0 < x < \pi$ and $g(x) = \sin^{-1}(e^{-x})$ for $x \ge 0$. If $\alpha$ is a positive real number such that $a = (fog)'(\alpha)$ and $b = (fog)(\alpha)$,then which of the following is true?

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