Which function satisfies $: \mathbb R → [−2, 2]$ with $((0))^ 2 + ( ′ (0))^ 2 = 85$,also $f$ is twice differentiable?

Question

Let $f:\mathbb R\to[−2,2]\,$ be a twice differentiable function with $$\big((0)\big)^2+\big(′(0)\big)^2=85.$$ Which of the following statements are necessarily TRUE?

(A) There exist , ∈ ℝ, where < , such that is one-one on the open interval (, )

(B) There exists $_0\in(−4, 0),\,$ such that $|f'(_0)|\le 1$

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

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

Solution:I tried it by using the function $f'(x)=x^3+2x+9$,as it is satisfying $((0))^ 2 + ( ′ (0))^ 2 = 85$. On checking all the 4 options on the chosen function ,i'm getting $A,D$ as correct options.

My query is that the chosen function is not satisfying the definition of function(every element of the domain has a unique image in codomain) as $f(3)=43$ is not in $[-2,2]$.

Please provide a function which staisfies the definition in the problem.

What is the source of this question? this version of it appeared a while ago. — lulu, Jun 22 '18 at 20:57
@lulu:https://storage.googleapis.com/candidate-portal-bucket/QuestionPapers/Paper1_ENGLISH.pdf see question 4 in maths section — Styles, Jun 22 '18 at 21:00
Consider a function like $f(x)=\sin (\sqrt {85}, x)$. At least that is good enough to show that $c$ is not always true. — lulu, Jun 22 '18 at 21:06
In addition to your function not matching the prerequisites, a function satisfying a statement is not proof that the statement is true "for every twice differentiable function". — Henrik supports the community, Jun 22 '18 at 21:13

Calvin Khor · Answer 1 · 2022-01-27T07:30:34.290

Solution for first two. (C) is shown false by the counterexample of lulu $f(x)=\sin (\sqrt{85}x$), so that leaves (D) which has answers here: 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? .

(A) True. Since $f\in C^2$, $f'\in C^1$, and $|f'(0)|^2=85 - |f(0)|^2 \ge 85 - 4 = 81>0$. So $f'(0)>9$ or $f'(0)<-9$. In either case, by continuity of $f'$, $f'(x)$ has a fixed sign on a small neighbourhood of $0$, and therefore $f$ is injective restricted to this neighbourhood.

(B) True. Suppose not. Then $|f'(x)|> 1$ on $(-4,0)$ . Then we have $$f(0) = f(-4)+ \int_{-4}^0 f'(s)ds$$ By continuity, $f'$ has a fixed sign. If $f'<0$, then $$ f(-4) = f(0)-\int_{-4}^0 f'(s)ds = f(0) + \int_{-4}^0 |f'(s)|ds > -2 + 4 =2 $$ contradicting the assumption on the range of $f$. Similarly, if $f'>0$, then we see that $f(-4) < -2$, which cannot happen.

McSuperbX1 · Answer 2 · 2019-11-15T10:55:00.857

1

Try this function: f(x) = 2sin($\frac{\sqrt{85}x}{2}$)

P.S This question appeared in the JEE Advanced 2018 (Paper - 1)

edited Nov 15 '19 at 10:55

answered Jan 12 '19 at 10:18

McSuperbX1

117

score -1 · Answer 3 · answered Jan 12 '19 at 10:30

-1

(A) Not true - Try for example $f(x)=\sqrt{85}$

(B) Not true - Try for example $f(x)=\sqrt{85}x$

(C) Not true - Try for example $f(x)=\sqrt{85}$

(D) Not true - Try for example $f(x)=\sqrt{85}$

answered Jan 12 '19 at 10:30

Yiorgos S. Smyrlis

86,581

4

The range should be in $[-2,2]$ – Shashi Jan 12 '19 at 12:24
3

Note that the $f(x)$ has a co-domain from $-2$ to $+2$. None of the functions you've mentioned satisfy the restriction for all real values of $x$. – McSuperbX1 Apr 16 '20 at 16:32
(Your answer) Not true. – Eric Duminil Jan 27 '22 at 07:48

Which function satisfies $: \mathbb R → [−2, 2]$ with $((0))^ 2 + ( ′ (0))^ 2 = 85$,also $f$ is twice differentiable?

3 Answers3