Logarithmic Earthquake!

Algebra Level 5

Suppose $f$ and $g$ are differentiable functions such that

$\large\ xg\left( f\left( x \right) \right)f^{ ' }\left( g\left( x \right) \right)g^{ ' }\left( x \right) = f\left( g\left( x \right) \right)g^{ ' }\left( f\left( x \right) \right)f^{ ' }\left( x \right)$

$\forall x \in \Re$ and

$\large\ \int _{ 0 }^{ a }{ f\left( g\left( x \right) \right)dx } = \frac { 1 - { e }^{ -2a } }{ 2 } \forall a \in \Re$ .

Given that $g(f(0)) = 1$ and if the value of $g(f(4)) = {e} ^ {-4k}$ , where $k$ is a natural number, then find $k$ .

The answer is 4.

1 solution

Mark Hennings
Dec 4, 2017

The differential equation can be written as $\frac{d}{dx}\left[\ln\big(g(f(x))\big)\right] \; = \; x\frac{d}{dx}\left[\ln\big(f(g(x))\big)\right]$ and, differentiating the integral formula, we know that $f(g(x)) = e^{-2x}$ , and so $\frac{d}{dx}\left[\ln\big(g(f(x))\big)\right] \; = \; -2x$ so that $\ln\big(g(f(x))\big) = -x^2$ , and hence $g(f(x)) = e^{-x^2}$ . We see that we have $f(x) = x^2$ and $g(x) = e^{-x}$ .

Thus $g(f(4)) = e^{-16}$ ,which makes $k = \boxed{4}$ .

Wow!

I was stuck there integrating the first step and then applying it somewhere, but actually i didn't need to integrate that.

Rightly Judged.

Priyanshu Mishra - 3 years, 6 months ago

Logarithmic Earthquake!

The answer is 4.

1 solution

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