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I have this question in a exam in which you have to choose only one of the possible answers: What would be correct?

Let be $ f,g : \mathbb{R} \rightarrow \mathbb{R} $ two derivable functions such that $ f'(x)=g(x),g'(x)=f(x),f(0)=1,g(0)=0 $. And let be the function $F(x)=f^2(x)+g^2(x)$, then:

a) $F(x)=1-\sin(x)$.

b) $F$ is the constant function with value 1.

c) None of the other two answers.

Frankly I don't even know where to start and the possible relationship between the conditions and possible responses.

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I wonder whether the condition wasn't supposed to be $f'(x)=-g(x)$. David's method still applies, but the answer is different. – Gerry Myerson Jan 20 '12 at 22:50
I just noticed: Is this an exam that's yet to be turned in? – David Mitra Jan 20 '12 at 23:11
It is a previous exam for public use to train students for next one. – jneira Jan 20 '12 at 23:16
up vote 1 down vote accepted

The first two conditions give $f''=f$. The initial conditions then give $$f(x)=\cosh x,\quad\text{and}\quad g(x)=\sinh x,$$ where $\cosh $ is the hyperbolic cosine function $\cosh x={e^x+e^{-x}\over 2}$ and $\sinh$ is the hyperbolic sine function $\sinh x={e^x-e^{-x}\over 2}$.

One has the identity $\cosh^2 x+\sinh^2 x=\cosh(2x) $.

Per Gerry's comment:

If the first condition were $f'(x)=-g(x)$, then you could argue in a manner similar to the above; or note that $$ F'(x)=2f(x)f'(x)+2g(x)g'(x)=-2f(x)g(x)+2g(x)f(x)=0. $$ Thus $F$ is a constant function. From the initial conditions, $F$ must be identically 1.

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But $f'$ may not be derivable. So, this seems a bit of stretching. Can you explain this? – user21436 Jan 20 '12 at 22:35
@Kannappan Sampath $f'=g$ is given, and $g$ is differentiable. – David Mitra Jan 20 '12 at 22:39
Oh sorry, I have been missing that! +1 for your answer! – user21436 Jan 20 '12 at 22:43
Sorry if i missed something: i see some changes in the answer and now i dont know what is the answer: a?? – jneira Jan 20 '12 at 23:05
@jneira It's "c)", since $\cosh(2x)$ is not identically 1 nor identically $1-\sin x$ (I edited the answer to leave something for you to think about) – David Mitra Jan 20 '12 at 23:09

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