A calculus problem by Aakhyat Singh

Calculus Level 3

If e f ( x ) = 10 + x 10 − x e^{f(x)}=\dfrac {10+x}{10-x} , for − 10 < x < 10 -10<x<10 and f ( x ) = k f ( 200 x x 2 + 100 ) f(x)=kf \left(\dfrac {200x}{x^2+100} \right) , what is 10 k 10k ?


The answer is 5.

Aakhyat Singh by Aakhyat Singh , 20, India

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1 solution

Tom Engelsman
Aug 15, 2017

If e f ( x ) = 10 + x 10 − x e^{f(x)} = \frac{10+x}{10-x} for x ∈ ( − 10 , 10 ) , x \in (-10, 10), then f ( x ) = l n ( 10 + x 10 − x ) . f(x) = ln(\frac{10+x}{10-x}). Now if f ( x ) = k ⋅ f ( 200 x x 2 + 100 ) f(x) = k \cdot f(\frac{200x}{x^2 + 100}) , then we obtain:

l n ( 10 + x 10 − x ) = k ⋅ l n ( 10 + 200 x x 2 + 100 10 − 200 x x 2 + 100 ) ; ln(\frac{10+x}{10-x}) = k \cdot ln(\frac{10 + \frac{200x}{x^2+100}}{10 - \frac{200x}{x^2 + 100}});

or l n ( 10 + x 10 − x ) = l n ( 100 + 20 x + x 2 100 − 20 x + x 2 ) k ; ln(\frac{10+x}{10-x}) = ln(\frac{100 + 20x + x^2}{100 - 20x + x^2})^{k};

or l n ( 10 + x 10 − x ) = l n ( 10 + x 10 − x ) 2 k ln(\frac{10+x}{10-x}) = ln(\frac{10 + x}{10 - x})^{2k} .

We finally arrive at 2 k = 1 ⇒ k = 1 2 2k = 1 \Rightarrow k = \frac{1}{2} , and 10 k = 5 . 10k = \boxed{5}.

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