Q) For every twice differentiable function $f:\mathbb{R}\longrightarrow [-2,2] $ with $[f(0)]^2+[f'(0)]^2=85$ , which of the following statement(s) is(are) TRUE?

(A) There exists $r,s \in\mathbb{R}$ , where $r<s$ , such that f is one-one on the open interval $(r,s)$

(B) There exists $x_0 \in (-4,0)$ such that $|f'(x_0)|\leq 1$

(C) $lim_{x\to \infty}f(x)=1$

(D) There exists $\alpha \in (-4,4)$ such that $f(\alpha)+f"(\alpha)=0 and f'(\alpha)=0$

I have problem with option B. Here's the given solution- enter image description here

But I think this is wrong because $f(0)$ cannot be equal to $2$ as that gives us the value of $f'(0)=9$ (from the condition given in the question). If the slope is positive at $x=0$ and the function is achieving its highest value there then for the points in the right neighbourhood of $x=0$ the slope would have to abruptly change to $0$ otherwise the function would obtain values greater than $2$ which are not in its co-domain. But it cannot abruptly change either coz it's given to be a twice differentiable function.

And I took $f(x)=9$ only but of course similar argument can be made with the negative value and a similar argument can be made if they took $f(0)=-2$ and $f(-4)=2$ in the solution.

My question-

  1. Is the given solution wrong because of what I just said? Or did I go wrong somewhere? If not then can this be put into some more concrete words or if there's a theorem related to this?
  2. Can an alternate solution be proposed for option B?

I'm under confident about this because this is a JEE Advanced 2018 question and no objections were made against this that I'm aware of.

  • 1
    $\begingroup$ what do you mean "with $[f(0)]^2+[f'(0)]^2$"? $\endgroup$
    – robjohn
    May 21, 2021 at 18:22
  • $\begingroup$ Did you mean to place a condition/restriction on $[f(0)]^2+[f’(0)]^2$? $\endgroup$ May 21, 2021 at 18:23
  • $\begingroup$ Yeah, edited, thanks $\endgroup$
    – Robin
    May 21, 2021 at 18:30
  • $\begingroup$ Why is the range of $f$ $[-2,2]$? $\endgroup$
    – Vishu
    May 21, 2021 at 18:33
  • 1
    $\begingroup$ $f(0)$ doesn't have to be 2, they are just using the bounds to get a bound on $f'(x_0)$. Indeed it cannot be that $f(0)=2$, although it isn't exactly because $f'(0)$ would be $9$, it's because it would be either $9$ (in which case an "illegal" value would be attained to the right of zero) or $-9$ (in which case an "illegal" value would be attained to the left of zero). Now you might object that this bound seems unnecessarily suboptimal, but at the level of generality of the question it is really not, because $|f(0)-f(-4)|$ can be as close to $4$ as you want without reaching it. $\endgroup$
    – Ian
    May 21, 2021 at 19:03

2 Answers 2


You’re correct in saying that $f(0) =2$ and $f’(0) =\pm 9$ cannot be true simultaneously, as then $f$ violates its range restriction on a neighborhood of $0$ due to it being differentiable.

But $f(0)$ can still be arbitrarily close to $2$. As long as there is some difference between $f(0)$ and $2$, it gets sufficient ‘space’ to bend inwards back into $[-2,2]$. The smaller the difference, the larger the second derivative at $0$ must be.

So, $|f’(x_0)|$ cannot equal $1$, but can be arbitrarily close to $1$, i.e. there exists some $x_0$ in $(-4,0)$ such that $$|f’(x_0)| \lt 1$$ and (B) is still correct.

  • $\begingroup$ But B says $|f'(x_0)|"\leq" 1$ $\endgroup$
    – Robin
    May 21, 2021 at 18:52
  • $\begingroup$ @Robin That is a looser condition: $a\lt b \implies a\le b$. $\endgroup$
    – Vishu
    May 21, 2021 at 18:53
  • 1
    $\begingroup$ Ah yes, got it, thanks $\endgroup$
    – Robin
    May 21, 2021 at 18:55

The answer never assumes that $f(0)=2$ or that $f(0)=-2$. It only says that $|f(0)-f(-4)|\le4$ because $-2\le f(x)\le2$.

A minor point: the answer only states that $f(0)-f(-4)\le4$, but it really should have stated that $|f(0)-f(-4)|\le4$.


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