JEE Advanced Previous Year Questions (2018 - 2025): Limits, Continuity and Differentiability

Table of Contents
1. 2024
2. 2023
3. 2022
4. 2021
5. 2020
6. 2019
7. 2018
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2024

Q1: Let k ∈ ℝ. If  then the value of k is:
(a) 1
(b) 2
(c) 3
(d) 4     [JEE Advanced 2024 Paper 2]
Ans: 

Q2: Let f : ℝ → ℝ be a function defined by

Then which of the following statements is TRUE?
(a) f(x) = 0 has infinitely many solutions in the interval [1/10¹⁰, ∞).
(b) f(x) = 0 has no solutions in the interval [1/π, ∞).
(c) The set of solutions of f(x) = 0 in the interval (0, 1/10¹⁰) is finite.
(d) f(x) = 0 has more than 25 solutions in the interval (1/π², 1/π).   [JEE Advanced 2024 Paper 2]
Ans: (d)
Option-A: f(x) = x² sin(π/x²)
f(x) = 0 ⇒ sin(π/x²) = 0 ⇒ π/x² = nπ, n ∈ ℕ
x² = 1/n ⇒ x = 1/√n
1/√n ≥ 1/10¹⁰ ⇒ √n ≥ 10¹⁰ ⇒ n ≤ 10²⁰, finite number of solutions.

Option-B: x = 1/√n
1/√n > 1/π ⇒ x > √n ⇒ n < π², Number of solutions is 9.

Option-C: x = 1/√n
1/√n < 1/10¹⁰ ⇒ √n > 10¹⁰ ⇒ n > 10²⁰, infinite number of solutions.

Option-D: 1/π² < 1/√n < 1/π ⇒ √n ∈ (π, π²) ⇒ n ∈ (π², π⁴), Definitely more than 25 solutions.

Q3: Let f : ℝ → ℝ and g : ℝ → ℝ be functions defined by

Let a, b, c, d ∈ ℝ. Define the function h : ℝ → ℝ by
h(x) = a f(x) + b (g(x) + g(1/2 - x)) + c(x - g(x)) + d g(x), x ∈ ℝ.
Match each entry in List-I to the correct entry in List-II.

The correct option is:
(a) (P) → (4), (Q) → (3), (R) → (1), (S) → (2)
(b) (P) → (5), (Q) → (2), (R) → (4), (S) → (3)
(c) (P) → (5), (Q) → (3), (R) → (2), (S) → (4)
(d) (P) → (4), (Q) → (2), (R) → (1), (S) → (3)    [JEE Advanced 2024 Paper 1 ]
Ans: (c)

(P) Now a = 0, b = 1, c = 0, d = 0.

Hence Range of h(x) is {0, 1}

Hence h(x) $ℎ\left(�\right)$ is differentiable on R

Range of h(x) is [0, 1]

Q4: Let S be the set of all (α, β) ∈ ℝ × ℝ such that

Then which of the following is (are) correct?
(a) (-1, 3) ∈ S
(b) (-1, 1) ∈ S
(c) (1, -1) ∈ S
(d) (1, -2) ∈ S    [JEE Advanced 2024 Paper 2]
Ans: (b), (c)

2023

Q1: Let f : (0, 1)→ R be the function defined as , where [x] denotes the greatest integer less than or equal to x. Then which of the following statements is(are) true?(a) The function f is discontinuous exactly at one point in (0, 1)
(b) There is exactly one point in (0, 1) at which the function f is continuous but NOT differentiable
(c) The function f is NOT differentiable at more than three points in (0, 1)
(d) The minimum value of the function f is -1/512                  [JEE Advanced 2023 Paper 2]
Ans:

f (x) is discontinuous at x = 3/4 only

f (x) is non-differentiable at x = 1/2 and 3/4
Minimum values of f(x) occur at x = 5/12 whose value is  -1/432

2022

Then, the value of  is equal to :
(a)
(b)
(c)
(d)                  [JEE Advanced 2022 Paper 2]
Ans: (b)

Q2: If  , then the value of 6β is ___________. [JEE Advanced 2022 Paper 2]
Ans: 5
Given,

[Neglecting higher power of x] 

Q3: Let α be a positive real number. Let  be the functions defined by

Then the value of  is_______.      [JEE Advanced 2022 Paper 1]
Ans:  0.49 to 0.51
 [As f(x) is continuous function so we can write this]
Now,

From graph you can see

= 2

2021

Q1: Let f : R → R be defined by Then which of the following statements is (are) TRUE?

Sign scheme for f'(x)
Here, f is decreasing in the interval (-2, -1) and f is increasing in the interval (1, 2).
Now, and

∴ Range =

Hence, f(x) is into.f(x) has local maxima at x = -4
and local minima at x = 0. 

2020

Q1: Let f : R → R and g : R → R be functions satisfying f(x + y) = f(x) + f(y) + f(x)f(y)
and f(x) = xg(x) for all x, y ∈ R.
If , then which of the following statements is/are TRUE?
(a) f is differentiable at every x∈R
(b) If g(0) = 1, then g is differentiable at every x ∈ R
(c) The derivative f'(1) is equal to 1
(d) The derivative f'(0) is equal to 1          [JEE Advanced 2020 Paper 2]
Ans: (a), (b) & (d)
The given function f : R → R is satisfying as 

Therefore, f(x) = e^x - 1 is differentiable at every x ∈ R.
and Now,

LHD (at x = 0) of

and, RHD (at x = 0) of

So, if g(0) = 1, then g is differentiable at every x ∈ R.

Q2: Let the function f : R → R be defined by f(x) = x³ - x² + (x - 1)sin x and let g : R → R be an arbitrary function. Let fg : R → R be the product function defined by (fg)(x) = f(x)g(x). Then which of the following statements is/are TRUE?
(a) If g is continuous at x = 1, then fg is differentiable at x = 1
(b) If f g is differentiable at x = 1, then g is continuous at x = 1
(c) If g is differentiable at x = 1, then fg is differentiable at x = 1
(d) If f g is differentiable at x = 1, then g is differentiable at x = 1     [JEE Advanced 2020 Paper 1]
Ans: (a) & (c)
Given functions f : R → R be defined
by f(x) = x³ - x² + (x + 1) sin x and g : R → R be an arbitrary function.
Now, let g is continuous at x = 1, then 

{∵ f(1) = 0 and g is continuous at x = 1, so g(1 - h) = g(1)}

Similarly,  

∵ RHD and LHD of function fg at x = 1 is finitely exists and equal, so fg is differentiable at x = 1
Now, let (fg)(x) is differentiable at x = 1, so 

It does not mean that g(x) is continuous or differentiable at x = 1.
But if g is differentiable at x = 1, then it must be continuous at x = 1 and so fg is differentiable at x = 1. 

Q3: The value of the limit  is ________.                    [JEE Advanced 2020 Paper 2]Ans: 8
The Limit

= 8

Q4: let e denote the base of the natural logarithm. The value of the real number a for which the right hand limit is equal to a non-zero real number, is ____ .    [JEE Advanced 2020 Paper 1]
Ans: 1
The right hand limit

The above limit will be non-zero, if a = 1. And at a = 1, the value of the limit is

2019

                 
(a) -6
(b) -7
(c) 8
(d) -9                      [JEE Advanced 2019 Paper 2]
Ans: (c) & (d)

Hence, options (c) and (d) are correct.

Q2: Let f : R → R be given by 

So,

At x = 1, f"(1^-) = 2 > 0 and f"(1⁺) = 4-8 = -4 < 0
∴ f'(x) is not differentiable at x = 1 and f'(x) has a local maximum at x = 1.
For x ∈ (-∞, 0)
f'(x) = 5x⁴ + 20x³ + 30x² + 20x + 3
and since
f'(-1) = 5-20 + 20 + 30 - 20 + 3 = -2 < 0
So, f(x) is not increasing on x ∈(-∞, 0).
Now, as the range of function f(x) is R, so f is onto function.
Hence, options (b), (c) and (d) are correct. 

2018

Q1: For every twice differentiable function f : R → [-2, 2] with (f(0))² + (f′(0))² = 85, which of the following statement(s) is(are) TRUE?
(a) There exist r, s ∈ R, where r < s, such that f is one-one on the open interval (r, s)
(b) There exists x₀ ∈ (-4, 0) such that |f'(x₀)| ≤ 1
(c) (d) There exists α ∈ (-4, 4) such that f(α) + f"(α) = 0 and f'(α) ≠ 0      [JEE Advanced 2018 Paper 1]Ans:

Range of f is [-2, 2]

Hence, |f'(x₀)| = 1.
Hence, statement is true.
(c) As no function is given, then we assume 

Now, 
and does not exists.Hence, statement is false.
(d) From option b, 

hence,

Now, let p ∈ (-4, 0) for which g(p) = 5
Similarly, let q be smallest positive number q ∈ (0, 4)
such that g(q) = 5
Hence, by Rolle's theorem is (p, q)
g'(c) = 0 for α ∈ (-4, 4) and since g(x) is greater than 5 as we move from x = p to x = q
and f(x))² ≤ 4
⇒ (f'(x))² ≥ 1 in (p, q)
Thus, g'(c) = 0
⇒ f'f + f'f" = 0
So, f(α) + f"(α) = 0 and f'(α) ≠ 0
Hence, statement is true. 

Q2: Let f : R → R and g : R → R be two non-constant differentiable functions. If f'(x) = (e^{(f(x) - g(x))}) g'(x) for all x ∈ R and f(1) = g(2) = 1, then which of the following statement(s) is (are) TRUE?
(a) f(2) < 1 - loge²
(b) f(2) > 1 - loge²
(c) g(1) > 1 - loge²
(d) g(1) < 1 - loge² [JEE Advanced 2018 Paper 1]
Ans: (b) & (c)
We have,

On integrating both side, we get

At x = 1

At x = 2

From Eqs. (i) and (ii)

We know that, e^-x is decreasing
∴ -f(2) < log_e 2-1
f(2) > 1 - log_e 2

⇒ g(1) > -log_e 2 

Q3: Let  and  be functions defined by
(i)
(ii) the inverse trigonometric function tan^-1x assumes values in (-π/2, π/2)
(iii) , where for t ∈ R, [t] denotes the greatest integer less than or equal to t,
(iv)

(i) Given,
f₁ : R → R and f₁(x) =
∴ f₁(x) is continuous at x = 0
Now,

At x = 0
f₁'(x) does not exists.
∴ f₁(x) is not differential at x = 0
Hence, option (2) for P. 

(ii) Given,

Clearly, f₂(x) is not continuous at x = 0.
∴ Option (1) for Q.

(iii) Given, f₃(x) = [sin(log_e(x + 2))], where [ ] is G.I.F.
and f₃ : (-1, e^π/2 - 2) → R
It is given, 

It is differentiable and continuous at x = 0.
∴ Option (4) for R
(iv) Given, 

Now,

thus

Again,

does not exists.
Since, does not exists.
Hence, f'(x) is not continuous at x = 0.
∴ Option (3) for S. 

Q4: The value of  is ________. [JEE Advanced 2018 Paper 1]
Ans: 8

= 2² x 2
= 8