Limit, Continuity and Differentiability JEE 2016 Solution

Limit, Continuity and Differentiability JEE 2016: Official Solution

JEE Advanced 2016 Mathematics Limit, Continuity and Differentiability Limit as a sum and monotonicity using derivatives

By Founder, JEEnius - IIT Kanpur Alumni · Aug 29, 2026 · 3 min read

Hard 5 min target

Let $f (x) = {lim}_{n \to \infty} {\frac{\prod_{r = 1}^{n} (1 + \frac{r x}{n})}{\prod_{r = 1}^{n} (1 + {(\frac{r x}{n})}^{2})}}^{\frac{x}{n}}$ , for all $x > 0$ . Then

Show answerAnswer

B) $f (\frac{1}{3}) \leq f (\frac{2}{3})$

C) $f^{'} (2) \leq 0$

Explanation

Given

$f (x) = {lim}_{n \to \infty} {\frac{\prod_{r = 1}^{n} (1 + \frac{r x}{n})}{\prod_{r = 1}^{n} (1 + {(\frac{r x}{n})}^{2})}}^{\frac{x}{n}}$

Take logarithm.

$\ln f (x) = {lim}_{n \to \infty} \frac{x}{n} [\sum_{r = 1}^{n} \ln (1 + \frac{r x}{n}) - \sum_{r = 1}^{n} \ln (1 + {(\frac{r x}{n})}^{2})]$

This is a Riemann sum. Let

$t = \frac{r}{n}$

Then as $n \to \infty$ ,

$\frac{1}{n} \sum_{r = 1}^{n} g (\frac{r}{n}) \to \int_{0}^{1} g (t) d t$

So,

$\ln f (x) = x \int_{0}^{1} [\ln (1 + x t) - \ln (1 + x^{2} t^{2})] d t$

Now put

$u = x t$

Then

$d u = x d t$

As $t$ goes from $0$ to $1$ , $u$ goes from $0$ to $x$ .

Therefore,

$\ln f (x) = \int_{0}^{x} [\ln (1 + u) - \ln (1 + u^{2})] d u$

Hence,

$f (x) = e^{\int_{0}^{x} [\ln (1 + u) - \ln (1 + u^{2})] d u}$

Differentiate using the fundamental theorem of calculus.

$\frac{f^{'} (x)}{f (x)} = \ln (1 + x) - \ln (1 + x^{2})$

So,

$f^{'} (x) = f (x) \ln (\frac{1 + x}{1 + x^{2}})$

Since $f (x) > 0$ , the sign of $f^{'} (x)$ depends on

$\ln (\frac{1 + x}{1 + x^{2}})$

For $0 < x < 1$ ,

$1 + x > 1 + x^{2}$

So,

$\frac{1 + x}{1 + x^{2}} > 1$

Thus,

$f^{'} (x) > 0$

Therefore $f$ is increasing on $(0, 1)$ . Hence,

$f (\frac{1}{3}) \leq f (\frac{2}{3})$

So option B is correct.

Also, since $\frac{1}{2} < 1$ and $f$ is increasing on $(0, 1)$ ,

$f (\frac{1}{2}) < f (1)$

So option A is false.

For option C,

$f^{'} (2) = f (2) \ln (\frac{1 + 2}{1 + 2^{2}})$

$f^{'} (2) = f (2) \ln (\frac{3}{5})$

Since $f (2) > 0$ and $\ln (\frac{3}{5}) < 0$ ,

$f^{'} (2) < 0$

Therefore,

$f^{'} (2) \leq 0$

So option C is correct.

For option D,

$\frac{f^{'} (x)}{f (x)} = \ln (\frac{1 + x}{1 + x^{2}})$

At $x = 3$ ,

$\frac{f^{'} (3)}{f (3)} = \ln (\frac{4}{10})$

$\frac{f^{'} (3)}{f (3)} = \ln (\frac{2}{5})$

At $x = 2$ ,

$\frac{f^{'} (2)}{f (2)} = \ln (\frac{3}{5})$

Since

$\frac{2}{5} < \frac{3}{5}$

we get

$\ln (\frac{2}{5}) < \ln (\frac{3}{5})$

Thus,

$\frac{f^{'} (3)}{f (3)} < \frac{f^{'} (2)}{f (2)}$

So option D is false.

Hence, the correct options are B and C.

Watch the full solution, worked step by step.

What was the JEE Advanced 2016 Paper 2 multi-correct question from limit continuity and differentiability?

The question defines

f (x) = {lim}_{n \to \infty} {\frac{\prod_{r = 1}^{n} (1 + \frac{r x}{n})}{\prod_{r = 1}^{n} (1 + {(\frac{r x}{n})}^{2})}}^{\frac{x}{n}}

for all x > 0. It asks which of these four statements are true: A) f(1/2) ≥ f(1), B) f(1/3) ≤ f(2/3), C) f'(2) ≤ 0, D) f'(3)/f(3) ≥ f'(2)/f(2). This is a multiple correct type from Paper 2. Only B and C are correct.

How do you solve the 2016 JEE Advanced limit as a sum question step by step using the official method?

Take natural log first. This pulls the exponent x/n down and turns the product into a difference of two sums:

\ln f (x) = {lim}_{n \to \infty} \frac{x}{n} [\sum_{r = 1}^{n} \ln (1 + \frac{r x}{n}) - \sum_{r = 1}^{n} \ln (1 + {(\frac{r x}{n})}^{2})] .

Recognise each sum as a Riemann sum with t = r/n. This converts to

\ln f (x) = x \int_{0}^{1} [\ln (1 + x t) - \ln (1 + x^{2} t^{2})] d t .

Substitute u = x t so du = x dt. Limits change from t = 0 to 1 into u = 0 to x. This yields

\ln f (x) = \int_{0}^{x} [\ln (1 + u) - \ln (1 + u^{2})] d u .

Thus f(x) equals the exponential of that integral. Differentiate by FTC to obtain

\frac{f^{'} (x)}{f (x)} = \ln (\frac{1 + x}{1 + x^{2}}) .

For 0 < x < 1, 1+x > 1+x^2 so the argument exceeds 1, the logarithm is positive, f'(x) > 0 and f is strictly increasing on (0,1). Hence f(1/3) < f(2/3) so B is true and f(1/2) < f(1) so A is false.

At x = 2, ln(3/5) < 0. Since f(2) > 0, f'(2) < 0 and option C is true.

At x = 3, f'(3)/f(3) = ln(2/5) while f'(2)/f(2) = ln(3/5). Because 2/5 < 3/5 < 1, both logarithms are negative and ln(2/5) < ln(3/5) < 0. Thus f'(3)/f(3) < f'(2)/f(2) so D is false.

The past-paper archive holds every JEE Advanced paper from 2007 by chapter with worked solutions.

What mistake with inequalities or substitution makes students pick the wrong options in this limit continuity and differentiability jee 2016 question?

Reversing the quadratic inequality while checking when 1+x > 1+x^2 by solving x^2 - x < 0 as x > 1 instead of 0 < x < 1 leads to the false conclusion that f is decreasing on (0,1) and therefore selecting A as true.

Failing to compare the actual numerical arguments 2/5 versus 3/5 before applying the increasing property of ln, instead only noting that g(x) = ln((1+x)/(1+x^2)) is decreasing and wrongly inferring the ratio at larger x is larger, leads to selecting D.

Why does the logarithmic derivative simplify everything in this question?

Once f'(x)/f(x) = ln((1+x)/(1+x^2)) is obtained, every option reduces to a sign test or direct numerical comparison without ever evaluating the integral for f(x) itself.

This mirrors standard logarithmic differentiation of products or quotients. On (0,1), (1+x) - (1+x^2) > 0 so the logarithm is positive and f increases. For x > 1, f' is negative at both 2 and 3 but ln(2/5) < ln(3/5) < 0, so the ratio at x = 3 is smaller than at x = 2.

What similar practice questions test Riemann-sum limits and monotonicity from an integral-defined function?

Evaluate

{lim}_{n \to \infty} {(\prod_{k = 1}^{n} (1 + \frac{2 k}{n}))}^{1 / n} .

Convert to exponential of integral of ln(1+2t) from 0 to 1.

Let g(x) = integral from 0 to x of [ln(1+t^2) - ln(1+t)] dt and f(x) = e^{g(x)}. Find intervals where f is increasing by testing sign of g'(x) = ln((1+x^2)/(1+x)).

Suppose h(x) is defined by a limit expression identical in form to the 2016 question but with an extra factor of x^3 inside the products. Check whether h'(4) ≤ 0 without computing h(4).

What should you remember when you next meet a product-limit monotonicity question from JEE Advanced 2016 Paper 2 Mathematics?

The technique of substituting after writing as integral from 0 to 1 then scaling to 0 to x is reusable for any similar product limit.

Read the official solution to the 2018 LCD question here: official solution to the 2018 question.

Use the photograph a doubt feature for any stuck PYQ. Apply the f'(x)/f(x) check on the next monotonicity problem you see.

Frequently asked questions

How to solve limit continuity and differentiability jee 2016 paper 2 question?

Take natural log of f(x) so the exponent comes down and each product becomes a sum. Recognise these as Riemann sums with t=r/n converting to integral from 0 to 1, then substitute u=xt to get integral from 0 to x. By FTC, f'(x)/f(x) = ln((1+x)/(1+x^2)). Use the sign of this expression to test each option.

Which options are correct in limit continuity and differentiability jee 2016?

Only B and C are correct. For 0<x<1, ln((1+x)/(1+x^2))>0 so f is increasing, making f(1/3)<f(2/3) (B true) and f(1/2)<f(1) (A false). At x=2, ln(3/5)<0 so f'(2)<0 (C true). At x=3, ln(2/5)<ln(3/5)<0 so f'(3)/f(3)<f'(2)/f(2) (D false).

What is f'(x)/f(x) for the function in JEE Advanced 2016 LCD question?

After expressing ln f(x) as integral from 0 to x of [ln(1+u)-ln(1+u^2)] du, the fundamental theorem of calculus gives f'(x)/f(x) = ln((1+x)/(1+x^2)). This ratio is enough to answer every option by checking signs and numerical comparisons. You never need the explicit antiderivative or value of f(x).

Why is option A wrong in limit continuity and differentiability jee 2016?

Option A claims f(1/2) ≥ f(1). On (0,1), 1+x > 1+x^2 so ln((1+x)/(1+x^2)) > 0, f'(x) > 0 and f is strictly increasing. Therefore f(1/2) < f(1) and A is false. The common error is solving x^2 - x < 0 as x > 1 instead of 0 < x < 1.

Why use f'(x)/f(x) in the 2016 JEE Advanced limit continuity question?

Once you have f'(x)/f(x) = ln((1+x)/(1+x^2)), every statement reduces to a sign check or direct comparison of two numbers. On (0,1) the log is positive so f increases; at x=2 and x=3 both values are negative but ln(2/5) < ln(3/5) < 0. This avoids computing the actual integral for f(x).

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