Limit, Continuity and Differentiability JEE 2018: Official Solution

What does the 2018 JEE Advanced question on twice differentiable functions ask?

The correct options are A, B and D. The question considers every twice differentiable function $f:ℝ\to [-2,2]$ that satisfies $(f(0){)}^2+(f^{\prime }(0){)}^2=85$. Such an $f$ maps all real numbers into the closed interval $[-2,2]$ and is twice differentiable everywhere. The task is to decide which of the four statements must be true.

What is the official solution to the 2018 JEE Advanced twice differentiable function problem?

Given that $f:ℝ\to [-2,2]$, we have $|f(x)|\le 2$.

In particular, $|f(0)|\le 2$.

The condition is $(f(0){)}^2+(f^{\prime }(0){)}^2=85$.

Since $(f(0){)}^2\le 4$ we get $(f^{\prime }(0){)}^2\ge 81$.

Therefore, $|f^{\prime }(0)|\ge 9$.

So $f^{\prime }(0)\ne 0$.

Since $f$ is twice differentiable, $f^{\prime }$ is continuous at 0. Since $f^{\prime }(0)\ne 0$, there exists a small interval around 0 where $f^{\prime }$ keeps the same non-zero sign. Therefore $f$ is strictly monotonic on that interval and hence one-one. Option A holds.

Apply Lagrange Mean Value Theorem on $[-4,0]$. There exists $x_0\in (-4,0)$ such that

Since both $f(0)$ and $f(-4)$ lie in $[-2,2]$, we have $|f(0)-f(-4)|\le 4$. Thus $|f^{\prime }(x_0)|\le 1.$

Option B holds.

The given conditions do not force ${lim}_{x\to \infty }f(x)=1$. Take $f(x)=2\sin (kx)$. Then $f(0)=0$, $f^{\prime }(x)=2k\cos (kx)$, $f^{\prime }(0)=2k$. Choose $2k=\sqrt{85}$. Then $(f(0){)}^2+(f^{\prime }(0){)}^2=85$. But $f(x)=2\sin (kx)$ does not have a limit as $x\to \infty$. Option C is false.

Define $g(x)=(f(x){)}^2+(f^{\prime }(x){)}^2$. Then $g(0)=85$. From the same Mean Value Theorem argument on $[-4,0]$, there exists $x_1\in (-4,0)$ such that $|f^{\prime }(x_1)|\le 1$. Since $|f(x_1)|\le 2$, $g(x_1)\le 5.$

Similarly, there exists $x_2\in (0,4)$ such that $g(x_2)\le 5$.

Now $g$ is continuous on $[x_1,x_2]$ and $0\in (x_1,x_2)$. Since $g(0)=85>5$, $g$ attains a maximum at some interior point $a\in (x_1,x_2)\subset (-4,4)$. At this interior maximum, $g^{\prime }(a)=0$.

Differentiate:

So

Thus

At the maximum point, $g(a)\ge 85$. Since $|f(a)|\le 2$, we have $(f^{\prime }(a){)}^2\ge 81$. So $f^{\prime }(a)\ne 0$. Therefore $f(a)+f^{″}(a)=0$. Option D holds.

You can search every JEE Advanced paper from 2007 by chapter in the past-paper archive to see identical constructions.

What reasoning error leads students to pick option C?

Students treat the large $|f^{\prime }(0)|\ge 9$ as evidence that the function must eventually settle to a definite horizontal asymptote. They imagine the steep slope at zero will drive $f$ toward a constant value and stay there.

They confuse the bounded range with the existence of ${lim}_{x\to \infty }f(x)$. A function can stay inside $[-2,2]$ forever yet oscillate without approaching any single number. Bounded continuous functions need not possess limits at infinity.

They fail to test the option by deliberately constructing a periodic twice-differentiable function that satisfies the initial condition yet keeps oscillating. The counter-example $f(x)=2\sin (\sqrt{85/4}x)$ meets $(f(0){)}^2+(f^{\prime }(0){)}^2=85$, stays inside $[-2,2]$, but has no limit at infinity.

Why is the auxiliary function g(x) = [f(x)]^2 + [f'(x)]^2 indispensable for option D?

The construction forces an interior maximum where the derivative relation must hold. Note $g(0)=85$ while MVT on the two sides supplies points where $g\le 5$. On $[-4,0]$ and $[0,4]$ the bound $|f^{\prime }|\le 1$ used in option B gives $g(x_1)\le 5$ and $g(x_2)\le 5$.

Continuity on the closed interval $[x_1,x_2]$ containing 0 guarantees an interior maximum $a$ with $g(a)\ge 85$. The peak cannot sit at the endpoints.

Differentiation yields

At the maximum, $g^{\prime }(a)=0$, so the product is zero. This factors into $f^{\prime }(a)=0$ or $f(a)+f^{″}(a)=0$.

The bound $g(a)\ge 85$ together with $|f(a)|\le 2$ immediately rules out $f^{\prime }(a)=0$, because that would force $g(a)\le 4$. The only remaining possibility is $f(a)+f^{″}(a)=0$ with $f^{\prime }(a)\ne 0$.

What practice questions test the same MVT and auxiliary function ideas?

Question 1: Let $f:ℝ\to [-1,1]$ be twice differentiable with $f(0)=0$ and $|f^{\prime }(0)|\ge 3$. Prove there exists $c\in (-2,2)$ such that $f(c)+f^{″}(c)=0$.

Question 2: If $f$ is differentiable on $ℝ$, $|f(x)|\le M$ and $|f^{\prime }(x)|\ge k>0$ for all $x$, show $f$ cannot be periodic.

Question 3: Apply MVT on $[-\pi ,\pi ]$ to $f(x)=\sin (x)+x/2$ and locate a point where $|f^{\prime }(\xi )|<1$.

Solve them in the same order: bound the derivative with MVT, build $g(x)$ when $f^{″}$ appears, and test existence claims with a concrete trigonometric counter-example. The algebra mirrors the 2018 paper exactly.

How should I approach twice differentiable function questions in JEE Advanced?

When you see $f$ and $f^{″}$ together in an option, immediately form $g=f^2+(f^{\prime }{)}^2$; its derivative factors nicely. This single step replaces guesswork with algebra that finishes in under a minute.

Fix the symmetric interval $[-4,0]$ or $[0,4]$ the moment you see the bound $|f|\le 2$; MVT then caps $|f^{\prime }|$ by 1. The numbers are chosen so the contrast with the initial value 85 becomes obvious.

To kill a limit-at-infinity claim, write down the simplest trigonometric function whose derivative at zero matches the given size. The sine example takes ten seconds and ends the discussion.

Never assume a nonzero derivative at one point implies anything global; local sign constancy is enough for option A. Continuity of $f^{\prime }$ hands you a neighborhood where the sign holds, and strict monotonicity follows at once.

Apply these four moves the day before the exam. They turn a 300-second hard problem into a routine check. If a similar question blocks you during practice, photograph a doubt and get a step-by-step solution, with a free-body diagram when the question needs one (20 free).

Next step: photograph a doubt on JEEnius and photograph any question you are stuck on and get a step-by-step solution, with a free-body diagram when the question needs one (20 free).

Keep going with Hydrocarbons JEE 2021: Solving the BDE Matching Question.