Mother of All Calculus!

Calculus Level 4

If a positive real valued continuously differentiable functions f f on the real line such that for all x x

f 2 ( x ) = ∫ 0 x ( ( f ( t ) ) 2 + ( f ′ ( t ) ) 2 ) d t + e 2 \large\ f^{ 2 }\left( x \right) = \int _{ 0 }^{ x }{ \left( { \left( f\left( t \right) \right) }^{ 2 } + { \left( f'\left( t \right) \right) }^{ 2 } \right) dt } + { e }^{ 2 }

is satisfied.

Then find the value of f ( − 1 ) \large\ f(-1) .


The answer is 1.

Priyanshu Mishra by Priyanshu Mishra , 20, India

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1 solution

Zach Abueg
Aug 21, 2017

( f ( x ) ) 2 = ∫ 0 x ( ( f ( t ) ) 2 + ( f ′ ( t ) ) 2 ) d t + e 2 Differentiate both sides 2 f ( x ) f ′ ( x ) = ( f ( x ) ) 2 + ( f ′ ( x ) ) 2 a 2 − 2 a b + b 2 = ( a − b ) 2 ( f ( x ) − f ′ ( x ) ) 2 = 0 f ( x ) = f ′ ( x ) ⟹ f ( x ) = C e x f ( 0 ) 2 = e 2 ⟹ f ( 0 ) = e ⟹ C = e f ( x ) = e x + 1 \displaystyle \begin{aligned} \big(f(x)\big)^2 & = \int_0^x \bigg(\big(f(t)\big)^2 + \big(f'(t)\big)^2\bigg) \ dt + e^2 & \small \color{#3D99F6} \text{Differentiate both sides} \\ 2f(x)f'(x) & = \left(f(x)\right)^2 + \left(f'(x)\right)^2 & \small \color{#3D99F6} a^2 - 2ab + b^2 = (a - b)^2 \\ \left(f(x) - f'(x)\right)^2 & = 0 \\ f(x) & = f'(x) \\ \implies f(x) & = Ce^x & \small \color{#3D99F6} f(0)^2 = e^2 \implies f(0) = e \implies C = e \\ f(x) & = e^{x + 1} \end{aligned}

⟹ f ( − 1 ) = 1 \implies f(-1) = \boxed{1}

6 Helpful 0 Interesting 0 Brilliant 0 Confused

Simply great solution to a bad problem!.

Priyanshu Mishra - 3 years, 9 months ago

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Thanks :) But what makes it a bad problem?

Zach Abueg - 3 years, 9 months ago

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You mean bad looking right ?!!

Arghyadeep Chatterjee - 3 years, 1 month ago

This is a good problem. Why do you say that?

Milly Choochoo - 3 years, 9 months ago

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