For every twice differentiable function $f : \bf R \rightarrow [–2, 2]$ with $(f(0))^2 + (f'(0))^2 = 85$, which of the following statements are TRUE?

Question

For every twice differentiable function $f : \mathbf R \rightarrow [–2, 2]$ with $(f(0))^2 + (f'(0))^2 = 85$, which of the following statement(s) is (are) TRUE ?

(A) There exist $r, s\in \bf R$, 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 $|f'(x_0)| < 1$

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

(D) There exists $\alpha\in(–4, 4)$ such that $f(\alpha) + f''(\alpha) = 0$ and $f '(\alpha)\ne0$

My attempt is as follows:-

$A)$ If function is continuous, then definitely in some part of the interval it will be increasing or decreasing hence will be one-one on the open interval $(r,s)$

$f(x)$ cannot be a constant function as in that case $f'(x)=0$ which means $f(0)=\sqrt{85}$, but it doesn't belong to co-domain of $f$

B) By mean value theorem we can say that there exists some $c\in(-4,0)$ for which $4f'(c)=f(0)-f(-4)$

Suppose if $f(0)=2$ and $f(-4)=-2$, then we can only say that there exists $c$ such that $f'(c)=1$. It will not be necessary that there exists a $c$ for which $|f'(c)|<1$

C) $\lim\limits_{x\to\infty}f(x) = 1$, this is not at all necessary.

D) If function is not constant in $(-4,4)$, then we can safely say that at some point $\alpha$, $f(\alpha)\ne0$

But how to know $f(\alpha)+f''(\alpha)=0$

Actual answer is $A,B,D$ but I am able to ascertain only option $(A)$. Please help me in this.

Question source JEE Advanced 2018.

Answer 1

A) While you correctly note that $f$ cannot be constant, continuous and non-constant alone does not lead to the result. However, from $|f(0)|\le 2$, we get $|f'(0)|\ge 9$. As $f$ is twice differentiable, $f'$ must be continuous, hence non-zero and of constant sign in a neighbourhood of $0$ - in other words monotonic and ultimately injective on that neighbourhood

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B) If we had $f(0)=\pm2$, it would need to be a local extremum, but we already know $f'(0)\ne 0$. Therefore $|f(4)-f(0)|<4$ and $f'(x_0)|<1$ for the point obtained from the Mean Value Theorem.

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C) You do not show why it is not at all necessary (something that also brought you trouble in B). If $f$ has the property of the problem statement, then so has $-f$. But if all such functions have limit $1$, then $-f$ has limit$1$ and $f$ has limit $-1$, contradiction. Actually, such $f$ need not converge at all, but showing so with an explicit counterexample seems to be harder than just showing that the limit cannot always equal $1$. (Had the problem statement said $0$ instead of $1$, this simple trick would not work)

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D) I think this one is a bit tricky. As seen above, $|f'(0)|\ge 9$. To simplify the argument, consider only the case that $f'(0)\ge 9$. In B, we found $x_0\in(-4,0)$ with $f'(x_0)<1$. Let $$x_1= \sup\{\,x\le0\mid f'(x)<1\,\}.$$ As witnessed by $x_0$, we have $x_1>4$. From $f'(0)>1$ and continuity of $f'$, we have $x_1<0$ and $f'(x_1)=1$. By the same argument we have $0<x_2<4$ and $f'(x_2)=1$ for $$x_2= \inf\{\,x\ge0\mid f'(x)<1\,\}.$$ The MWT applied to $f'$ give us $\xi_1\in(x_1,0)$ with $f''(\xi_1)=\frac{f'(0)-f'(x_1)}{0-x_1}>2$ and $\xi_2\in(0,x_2)$ with $f''(\xi_2)<-2$. This implies $$f(\xi_0)+f''(\xi_0)>0>f(\xi_1)+f''(\xi_1).$$ As $f+f''$ is the derivative of $\int_0^x f(t)\,\mathrm dt+f'(x)$, Darboux's theorem tells us that $$f(\alpha)+f''(\alpha)=0$$ for some $\alpha\in(\xi_1,\xi_2)\subset(x_1,x_2)\subset (-4,4)$.

As $x_1<\alpha<x_2$, it follows that from their definitions of infimum/supremum that $$f'(\alpha)\ge 1.$$

Answer 2

Option (B) $$ f(x) \in [-2,2]$$ $$ \implies f(0) - f(-4) \in [-4,4] $$ By Lagrange's Mean Value Theorem, there exists an $x_0 \in (-4,0)$ such that $$ f'(x_0) = \dfrac{f(0)-f(-4)}{4} \in [-1,1]$$ Thus, option (B) is correct.

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Option (D)

Consider $g(x)= (f(x))^2 + (f'(x))^2$

From the previous result, there exists an $\alpha \in (0,4)$ such that $|f'(\alpha)| \le 1$ and a $\beta \in (-4,0)$ such that $|f'(\beta)| \le 1$ $$ g(\alpha) = \underbrace{(f(\alpha))^2}_{\le 4} + \underbrace{(f'(\alpha))^2}_{\le 1} $$ Therefore, $g(\alpha) \le 5$ and similarly, $g(\beta) \le 5$ Now, as mentioned in the question, $g(0) = 85 \ge 5$

Since $g(x)$ is a continuous function, $g(x)$ must pass through a maxima for the above to happen, i.e. there must be at least one $\gamma \in (-4,4)$ such that $g'(\gamma)=0$ and $g(\gamma) \ge 85$ $$ g'(\gamma) = 0 \implies (f'(\gamma))(f(\gamma) + f''(\gamma)) = 0$$ If $f'(\gamma)=0$ then $g(\gamma) = (f(\gamma))^2 + (f'(\gamma))^2 = (f(\gamma))^2 \le 4 $

But this is a contradiction as $g(\gamma) \ge 85$

Therefore, $f'(\gamma) \ne 0 \implies f(\gamma) + f''(\gamma) = 0$

Thus, option (D) is correct.