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We are given that a twice differentiable function, $f:\Bbb R\rightarrow[-2,2]$ satisfies the condition $$(f(0))^2 + (f'(0))^2 = 85$$

We are asked if exists a value of $x$, say $α \in (-4,4)$, for which $f(α) + f''(α) = 0$ and $f'(α) \neq0$.

(Please note we are merely asked about the existence of such value and not the value itself)

My attempt:

Suppose I take a function $$p(x) = (f(x))^2 + (f'(x))^2$$ Taking the derivative I obtained, $$p'(x) = 2f(x)f'(x) + 2f'(x)f''(x)$$ Which can be rewritten as $$p'(x) = 2f'(x)[f'(x)+ f''(x)]$$

So if I can somehow prove that p'(x) = 0 at some point and f'(x) is not zero at that point, I know that there is a value of α. But not sure how to proceed any further. Can I apply LMVT somewhere? Have I missed something? Any help would be appreciated!

Scilife
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  • I dont think such a function exists. – Archis Welankar Apr 11 '20 at 06:39
  • This question has appeared in a popular examination, the JEE twice - once in 2005 and again in 2018. I don't believe it is incorrect. EDIT: I made an error in typing the question. Please see it now. – Scilife Apr 11 '20 at 07:06
  • @Hetebrij I didn't see that before. But I still didnt understand the solution. It seems a bit beyond what I have learnt till now. Can it be solved using elementary techniques like LMVT, rolles theorem, etc – Scilife Apr 11 '20 at 07:50
  • @Hetebrij why did you remove it? I clearly mentioned that it did not resolve my problem. Thats why I asked this. You are forcing me to "ask a new one" but I already have!! – Scilife Apr 11 '20 at 09:38
  • @Scilife I didn't remove the question all by myself, but try asking about the specific part of the solution you don't understand. – Hetebrij Apr 11 '20 at 11:00
  • I was looking for a different solution. One involving elementary techniques. The solution given on that answer uses techniques unknown to me. This is from an examination for undergrad students. – Scilife Apr 11 '20 at 12:42
  • Anyways I figured it out myself. Might post my answer on the other question. – Scilife Apr 11 '20 at 12:43

1 Answers1

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You can try using the function $\sin(\sqrt{85}x)$ as $f(x)$ since it satisfies the mentioned functional equation. The problem becomes easy to solve from this point on and we can see that $f(x)+f''(x)=-84\sin(\sqrt{85}x)$, which is zero when $x = \frac{n\pi}{\sqrt{85}}$. At these values of x, $f'(x) \ne 0$ and hence, $\alpha =\frac{\pi}{\sqrt{85}}$ exists satisfying the given conditions.

Aniruddha Deb
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