Let f(x)=log (sin x), 0

Question

(B) Given $f(x)=\log (\sin x)$ for $0 < x < \pi$ and $g(x)=\sin ^{-1}(e^{-x})$ for $x \geq 0$. First,we find the composition $(f \circ g)(x) = f(g(x))

Vedclass · Accepted Answer

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

Vedclass · Answer

(B) Given $f(x)=\log (\sin x)$ for $0 First,we find the composition $(f \circ g)(x) = f(g(x))$. $(f \circ g)(x) = \log(\sin(\sin^{-1}(e^{-x}))) = \log(e^{-x}) = -x$. Now,we find the derivative $(f \circ g)^{\prime}(x) = \frac{d}{dx}(-x) = -1$. Given $a = (f \circ g)^{\prime}(\alpha) = -1$ and $b = (f \circ g)(\alpha) = -\alpha$. Substitute these values into the options: For option $(B)$: $a \alpha^2 - b \alpha - a = (-1)\alpha^2 - (-\alpha)\alpha - (-1) = -\alpha^2 + \alpha^2 + 1 = 1$. Thus,$a \alpha^2 - b \alpha - a = 1$ is correct.

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

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