Question

Suppose f(x) and g(x) are differentiable functions s.t. $xg(f\left(x\right))f^{\prime }\left(g\right(x\left)\right)g^{\prime }\left(x\right)=f\left(g\right(x\left)\right)g^{\prime }\left(f\right(x\left)\right)f^{\prime }\left(x\right)\forall xϵR$ and ${\int }_0^{a}f\left(g\right(x\left)\right)dx$ =$\frac{1-e^{-2a}}{2}$$\forall aϵR$ . Given that g(f(0)) = 1, if value of g (f(4)) = $e^{-4k}$ , Where $kϵN$ then k is equal to

Answer 1

$xg\left(f\right(x\left)\right)f^{\prime }\left(g\right(x\left)\right)g^{\prime }\left(x\right)=f\left(g\right(x\left)\right)g^{\prime }\left(f\right(x\left)\right)f^{\prime }\left(x\right)$
x \dfrac {f'(g(x) g'(x) }{f(g(x) }=\dfrac{ g'(f(x) f(x)}{g(f(x)) }
integrating the equation , as we know integration of $ \dfrac {f'(g(x) g'(x) }{f(g(x) } =lnf(g(x)) $andintegrationof$ \dfrac {g'(f(x) g'(x) }{g(f(x) }= ln g(f(x)$
after integration we get
xln f(g(x)) - integration of (ln f(g(x)) dx =lng(f(x))........$1$ equation
integration from zero to a of f(g(x))dx = \dfrac {(1-e^-2a }{ 2}
now differntiating this equation
f(g(x))=e^-2a...........equation $2$
now putting equation 2 in equation 1 we get
xln(e^-2x)- integration ln(e^-2x) dx = ln(g(f( x))
-2x \times x -x^2 = ln g(f(X))
lng(f(X))= -x^2
g(f(x))= e^(-x^2)
g(f(4)) = e^(-4 *4)
so k= 4