4
$\begingroup$

A problem from IIT 1981

Let $f(x)$ be a function defined for all x∈R such that $f(x+y)=f(x)⋅f(y)$ for all x,y∈R. Let $f′(0)=3$ and $f(5)=2$. Find the value of $f′(5)$.

The author of http://paramanands.blogspot.in/2010/08/two-problems-from-iit-jee.html#.ViuP3X4rLIU has pointed out that in no way f(x) can be an exponential function.But my question is any such function possible which satisfies the above conditions?Or is the question truly wrong?

$\endgroup$
  • $\begingroup$ The question is probably hoping you'll write $f'(y)=f'(0)f(y)$ which follows from the functional equation. However, the question is entirely wrong since, as you note, $f'(0)=3$ implies $f(5)=e^{15}$ and could well claim $f'(5)=3e^{15}$. This also follows from the givens (as does any other answer). $\endgroup$ – Milo Brandt Oct 24 '15 at 14:40
  • $\begingroup$ @MiloBrandt could you explain how you wrote $f'(0)=3$ then $f(5)=e^{15}$ ? $\endgroup$ – user220382 Oct 24 '15 at 14:47
  • $\begingroup$ The only solution to $f(x+y)=f(x)f(y)$ with $f'(0)=a$ is $f(x)=e^{ax}$ - in particular note that this follows from seeing that $f'(y)=f'(0)f(y)=af(y)$. Here $a$ is three. $\endgroup$ – Milo Brandt Oct 24 '15 at 14:50
  • $\begingroup$ @MiloBrandt got you! :-) Thanks! $\endgroup$ – user220382 Oct 24 '15 at 14:52
  • 1
    $\begingroup$ Good to see some discussion on MSE based on my blog post. BTW I "did not mention" that $f(x)$ can't be an exponential function, rather I mentioned that a function which satisfies all the conditions of the problem simultaneously can not exist. $\endgroup$ – Paramanand Singh Oct 26 '15 at 4:20
4
$\begingroup$

you can prove, as done in the paper that if $f(x+y)=f(x)f(y)$ (and is derivable in $0$), then $f'(x)=f'(0)f(x)$, here $f'(x)=3f(x)$ and $f(0)=1$. the only solution of this equation is $f(x) = e^{3x}$, and so we cannot have f(5)=2.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy