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

Question

Let f(x) = loge( sin x), (0 < x < π) and g(x) = sin-1(e-x), (x ge 0). If α is a positive real number such that a = (fog)'(α) and b = (fog)(α), then : (A) a α2 - b α - a = 0 (B) a α2 - b α - a = - 2 α2 (C) a α2 + b α + a = 0 (D) a α2 - b α - a = 1

Tardigrade Admin · Accepted Answer

fog(x) = (-x) ⇒ (fg(α)) = -a = b (fg(x))' = - 1 ⇒ (fg(α))' = -1 = a

Q. 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 $a = (f o g)^{'} (α)$ and $b = (f o g) (α)$ , then :

$a α^{2} - b α - a = 0$

$a α^{2} - b α - a = - 2 α^{2}$

$a α^{2} + b α + a = 0$

$a α^{2} - b α - a = 1$

$f o g (x) = (- x) \Rightarrow (f g (α)) = - a = b$
$(f g (x))^{'} = - 1 \Rightarrow (f g (α))^{'} = - 1 = a$

Q. Let f(x)=loge​(sinx),(0<x<π) and g(x)=sin−1(e−x),(x≥0). If α is a positive real number such that a=(fog)′(α) and b=(fog)(α), then :

aα2−bα−a=0

aα2−bα−a=−2α2

aα2+bα+a=0

aα2−bα−a=1