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 $\mathrm{a}=(\mathrm{fog})^{\prime}(\alpha)$ and $\mathrm{b}=(\mathrm{fog})(\alpha)$, then

(C) $\mathrm{a} \alpha^{2}-\mathrm{b} \alpha-\mathrm{a}=1$

(A) $a \alpha^{2}+b \alpha-a=2 \alpha^{2}$

(B) $\mathrm{a} \alpha^{2}-\mathrm{b} \alpha-\mathrm{a}=0$

(D) $a \alpha^{2}+b \alpha+a=0$

JEE Main · 2019 · 10 Apr 2019 · Shift 2 · Differentiation

Let

f (x) = lo g_{e} (sin x), (0 < x < π)

and

g (x) = sin^{- 1} (e^{- x}), (x \geq 0)

. If

α

is a positive real number such that

a = (fog)^{'} (α)

and

b = (fog) (α)

, then

Answer: (C)

Compose the two functions:

(f \circ g) (x) = ln (sin (sin^{- 1} (e^{- x}))) = ln (e^{- x}) = - x

. So at the point

α

the value is

b = (f \circ g) (α) = - α

, and since the composite is simply

- x

its derivative is

- 1

everywhere, giving

a = (f \circ g)^{'} (α) = - 1

. Substituting

a = - 1

and

b = - α

into

a α^{2} - b α - a

gives

- α^{2} - (- α) α - (- 1) = - α^{2} + α^{2} + 1 = 1

, so the relation that holds is

a α^{2} - b α - a = 1

JEE Main · 2019 · 10 Apr 2019 · Shift 2 · Single correct option