Can a differentiable function have positive slope at the end point of its co-domain?

Question

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-

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.

Answer 1

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.

Answer 2

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 $|f(0)-f(-4)|\le4$.