Limit, Continuity and Differentiability JEE 2020 Solution

What did the JEE Advanced 2020 question ask about the functional equation f(x+y)=f(x)+f(y)+f(x)f(y)?

f and g map reals to reals. They satisfy f(x+y) = f(x) + f(y) + f(x)f(y) for all real x, y. Also f(x) = x g(x) for all x and the limit as x approaches 0 of g(x) equals 1. The question requires deciding which statements hold on the differentiability of f, the differentiability of g if g(0)=1, f'(0)=1, and f'(1)=1.

The limit forces f(x) = e^x - 1 everywhere. This explicit exponential confirms A, B and D while ruling out C.

What is the official step-by-step solution to this JEE Advanced 2020 functional equation question?

Given f(x) = x g(x) and the limit of g(x) as x approaches 0 equals 1, first compute the limit of f(x).

{lim}_{x \to 0} f (x) = {lim}_{x \to 0} x \cdot g (x) = 0

because x tends to 0 while g(x) tends to 1.

Fix any y and let x tend to 0 in the functional equation. Then

{lim}_{x \to 0} f (x + y) = {lim}_{x \to 0} [f (x) + f (y) + f (x) f (y)] = f (y)

using the fact that the limit of f(x) is 0. This proves f is continuous at every real number. From f(x) = x g(x) it follows that f(0) = 0.

To check differentiability, start with the definition

f^{'} (x) = {lim}_{h \to 0} \frac{f (x + h) - f (x)}{h} .

Substitute the functional equation to obtain f(x+h) = f(x) + f(h) + f(x)f(h), so

f (x + h) - f (x) = f (h) (1 + f (x)) .

The difference quotient simplifies to

\frac{f (h)}{h} \cdot (1 + f (x)) .

Since f(h) = h g(h), this becomes g(h) times (1 + f(x)). Taking the limit as h tends to 0 and using the given limit of g(h) equals 1 yields $f^{'} (x) = 1 + f (x) .$

Thus f is differentiable at every real x, so option A holds. At x = 0,

f^{'} (0) = 1 + f (0) = 1 + 0 = 1,

so option D holds.

Solve the differential equation f'(x) = 1 + f(x) with f(0) = 0. Separate variables:

\frac{f^{'} (x)}{1 + f (x)} = 1 .

Integrate both sides:

\int \frac{d f}{1 + f} = \int d x

which gives $\ln (1 + f (x)) = x + C .$

Apply f(0) = 0 to find C = 0, so

1 + f (x) = e^{x}, f (x) = e^{x} - 1 .

Differentiate to obtain f'(x) = e^x, hence f'(1) = e which is not equal to 1, so option C is false.

For g, when x ≠ 0,

g (x) = \frac{e^{x} - 1}{x} .

With g(0) = 1 the function is g(x) = (e^x - 1)/x for x ≠ 0 and g(0) = 1. It is clearly differentiable for x ≠ 0. At x = 0,

g^{'} (0) = {lim}_{h \to 0} \frac{g (h) - 1}{h} = {lim}_{h \to 0} \frac{\frac{e^{h} - 1}{h} - 1}{h} = {lim}_{h \to 0} \frac{e^{h} - 1 - h}{h^{2}} .

The series e^h = 1 + h + h^2/2 + higher terms. Thus e^h - 1 - h = h^2/2 + higher terms, so the limit equals 1/2. Therefore g is differentiable at 0 and option B holds. The correct choices are A, B, and D.

What common mistake leads students to pick option C in this JEE Advanced 2020 question?

Students compute f'(x) = 1 + f(x) but then substitute x=1 directly into the DE without solving it or using f(0)=0 to obtain the actual f(1). They incorrectly treat the derivative as constantly 1 because lim g(h)=1 dominates the difference quotient and forget the multiplicative (1 + f(x)) factor grows with x.

This produces the false conclusion that f'(1) must equal 1. Solving the DE first reveals f(1) = e - 1 so f'(1) = e.

Why does the limit as x approaches 0 of g(x) equal to 1 control continuity and differentiability in this question?

The limit of g(x) as x approaches 0 equals 1 is the only extra condition. It is used to prove lim f(x)=0 which gives continuity at arbitrary y by fixing y and taking x to 0.

It supplies the exact factor 1 in the expression f'(x) = lim g(h) times (1 + f(x)). Once the DE is solved for f(x) = e^x - 1 it enables the explicit g(x)=(e^x-1)/x whose Taylor remainder gives g'(0)=1/2 from the series.

Remove that one limit and the entire chain collapses.

What other JEE Advanced questions use similar functional equations with limits and differentiability?

Here are three practice questions that test the identical techniques.

Question 1 (JEE Advanced style): Let f satisfy f(x+y)+f(x-y)=2f(x)cosh y and f(x)=x g(x), lim x→0 g(x)=0; prove f is odd and differentiable with f'(x)=f(x).

Question 2 (similar difficulty): Given f(x+y)=f(x)+f(y)+2xy and lim x→0 [f(x)/x^2]=1, find f''(0) and check differentiability of f(x)/x at x=0.

Question 3 (prior year): Functional equation f(x+y)=f(x)f(y) with f(1)=e and lim h→0 [f(h)-1]/h =1; compute f'(2).

You can search every JEE Main paper from 2002 and every Advanced paper from 2007, by year, subject or chapter, each with a worked solution (free) to locate these and their worked solutions.

How do you solve this type of JEE Advanced functional equation and limit question quickly?

First obtain f(0) and lim f(x) from the product definition. Convert the functional equation into a difference quotient before taking any derivative limit. Solve the resulting first-order DE using the initial condition at zero.

For the auxiliary function g always check the derivative at zero separately using the explicit form or series. The explicit exponential form decides all four options in one stroke. Expect to solve it in around 240 seconds.

If a similar question blocks you during practice, 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).

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).

For a worked example of the same idea, see P-Block Elements JEE 2010: Matrix Match on Halide Hydrolysis.

Frequently asked questions

What is the solution to f(x+y)=f(x)+f(y)+f(x)f(y) with lim x->0 g(x)=1 in JEE Advanced 2020?

The condition gives lim x->0 f(x)=0 so f is continuous everywhere and f(0)=0. The functional equation yields the difference quotient that simplifies to f'(x)=1+f(x). Solving the DE with initial condition produces f(x)=e^x-1.

Is f differentiable everywhere in the JEE Advanced 2020 limit continuity question?

Yes. Substituting the functional equation into the derivative definition and using lim g(h)=1 as h->0 gives f'(x)=1+f(x) at every real x. Thus f is differentiable everywhere and f'(0)=1.

Why is f'(1) not equal to 1 in the limit continuity and differentiability JEE 2020 problem?

From f'(x)=1+f(x) students sometimes assume f'(1)=1 without solving. The actual solution with f(0)=0 is f(x)=e^x-1, so f(1)=e-1 and f'(1)=e which is not 1. This makes option C false.

Is g differentiable at 0 when g(0)=1 in the 2020 JEE Advanced question?

Yes. After solving, g(x)=(e^x-1)/x for x≠0 and g(0)=1. The derivative limit at 0 uses the Taylor series of e^h to yield lim (e^h-1-h)/h^2=1/2. Therefore g is differentiable at 0.

How does lim x->0 g(x)=1 help prove continuity and differentiability in JEE 2020?

It forces lim x->0 f(x)=0 which, when plugged into the functional equation with fixed y, shows f is continuous at every point. The same limit turns the difference quotient into g(h)(1+f(x)) whose limit is exactly 1+f(x), proving differentiability everywhere.

Limit, Continuity and Differentiability JEE 2020 Question Solution