JEE Advanced Previous Year Questions (2018 - 2024): Definite Integrals and Applications of Integrals | Mathematics (Maths) for JEE Main & Advanced PDF Download

2024

Q1: Let f: [0, π2] → [0,1] be the function defined by f(x) = sin²x and let g: [0, π2] → [0, ∞) be the function defined by

The Value of  __________. [JEE Advanced 2024 Paper 2]
Ans: 0

apply kings

add both

Now,

Q2: Let f: [0, π2] → [0,1] be the function defined by f(x) = sin²x and let  g: [0, π2] → [0, ∞) be the function defined by
The value of  is ________. [JEE Advanced 2024 Paper 2]
Ans: 0.25
Now,

Using

Now
= 0.25

Q3: Let the function f : [1, ∞) → ℝ be defined by

Define . Let α denote the number of solutions of the equation g(x) = 0 in the interval (1, 8) and

Then the value of α + β is ______.      [JEE Advanced 2024 Paper 2]
Ans: 5


Apply L'pital
= g'(1⁺)1 = f(1⁺)
β = 2
∴ α + β = 5

Q4: Set S = {(x, y) ∈ ℝ × ℝ : x ≥ 0, y ≥ 0, y² ≤ 4x, y² ≤ 12 - 2x and 3y +√8x ≤ 5 √8}. If the area of the region S is α √2 , then α is equal to:

(a) 172
(b) 173
(c) 174
(d) 175 [JEE Advanced 2024 Paper 2 ]
Ans: (b)
Point of intersection of all curves is (2, 2 √2).
Area = A₁ + A₂

α √2 = 17√23
α = 173

2023

Q1: For x ∈ R, let . Then the minimum value of the function f : R →R defined by  is : [JEE Advanced 2023 Paper 2]
Ans:

Q2: Let n ≥ 2 be a natural number and f : [0, 1] → R be the function defined by

If n is such that the area of the region bounded by the curves x = 0, x = 1, y = 0 and y = f(x) is 4 , then the maximum value of the function f is :                      [JEE Advanced 2023 Paper 1]
Ans: 8
f(x) is decreasing in
increasing in
decreasing in
increasing in

Area =

n = 8

Q2: Let f :(0, 1)→ R be the function defined as  where n ∈ N. Let g:(0, 1) → R be a function such that  for all x ∈ (0, 1). Then
(a) does NOT exist
(b) is equal to 1
(c) is equal to 2
(d) is equal to 3                    [JEE Advanced 2023 Paper 1]
Ans: (c)

Now (According to the question)

 (Using Sandwich Theorem)

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2022

Q1: The greatest integer less than or equal to
is ___________. [JEE Advanced 2022 Paper 2]
Ans: 5

When,

When,

Now,

= 4 x 1.58 - 1
= 6.32 - 1
= 5.32
Greatest integer value fo
I = [5.32] = 5

Q2: Consider the functions f, g : R → R defined by

If  α is the area of the region , then the value of 9α is:                      [JEE Advanced 2022 Paper 2]
Ans: 6

This represent upward parabola.

∴  Graph is

Intersection point of f(x) and g(x) at first quadrant,

In first quadrant x = 1/2
When, x = 1/2

2021

Q1: Let f₁ : (0, ∞) → R and f₂ : (0, ∞) → R be defined by
 , x > 0 and , where, for any positive integer n and real numbers a₁, a₂, ....., a_n,  denotes the product of a₁, a₂, ....., a_n. Let m_i and n_i, respectively, denote the number of points of local minima and the number of points of local maxima of function f_i, i = 1, 2 in the interval (0, ∞).
The value of 2m₁ + 3n₁+ m₁n₁ is ___________. [JEE Advanced 2021 Paper 2]
Ans: 57.00

Q2: Let f₁ : (0, ∞) → R and f₂ : (0, ∞) → R be defined by , x > 0 and , where, for any positive integer n and real numbers a₁, a₂, ....., a_n, denotes the product of a₁, a₂, ....., a_n. Let m_i and n_i, respectively, denote the number of points of local minima and the number of points of local maxima of function f_i, i = 1, 2 in the interval (0, ∞).
The value of 6m₂ + 4n₂+ 8m₂n₂ is ___________. [JEE Advanced 2021 Paper 2]
Ans: 6.00

Sign Scheme for f₁'(x)
From sign scheme of f₁'(x), we observe that f(x) has local minima at x = 4k + 1, k∈W i.e. f₁'(x) changes sign from −ve to + ve which are x = 1, 5, 9, 13, 17, 21 and f(x) has local maxima at x = 4k + 3, k ∈ W i.e. f₁'(x) changes sign from + ve to − ve, which are x = 3, 7, 11, 15, 19.
So, m₁ = number of local minima points = 6
and n₁ = number of local maxima points = 5
Hence, 2m₁ + 3n₁ + m₁n₁
= 2 × 6 + 3 × 5 + 6 × 5
= 57

Clearly, m₂ = 1 and n₂ = 0
So, 6m₂ + 4n₂+ 8m₂n₂ 
= 6 + 0 + 0
= 6

Q3: Let  be functions such that f(0) = g(0) = 0,

Which of the following statements is TRUE?
(a)
(b) For every x > 1, there exists an α ∈ (1, x) such that
(c) For every x > 0, there exists a β ∈ (0, x) such that
(d) f is an increasing function on the interval (0, 3/2)    [JEE Advanced 2021 Paper 2]
Ans: (c)

∴ f is increasing for x∈(0, 1) and f is decreasing for x ∈ (1, ∞). Hence, option (d) is incorrect.
Now,

Hence, option (a) is incorrect.

Now,

By LMVT,

Hence, option (c) is correct.

Q4: Let  and  be functions such that and f(x)=sin²x, for all . Define
The value of  is _____________. [JEE Advanced 2021 Paper 2]
Ans: 2.00

Q5: Let  and  be functions such that and f(x)=sin²x, for all . Define
The value of  is _____________.                            [JEE Advanced 2021 Paper 2]
Ans: 1.50

Adding Eqs. (i) and (ii), we get

From figure,

Q6: The area of the region 

is
(a) 11/32
(b) 35/96
(c) 37/96
(d) 13/32      [JEE Advanced 2021 Paper 1]
Ans: (a)

Required area = Shaded region
On solving x + y = 2 and x = 3y, we get

On solving y = 0 and x + y = 2, we get
B ≡ (2, 0)
On solving x = 9/4 and x = 3y, we get

Required area = Area of ∆OCD − Area of ∆OBA

2020

Q1: Let f : R → R be a differentiable function such that its derivative f' is continuous and f(�) = −6. If F : [0, π] → R is defined by , and if , then the value of f(0) is ______ [JEE Advanced 2020 Paper 2]
Ans: 4
It is given that, for functions

Now,

{by integration by parts}

=

Q2: Let the functions f : R → R and g : R → R be defined by
f(x) = e^x − 1 − e^{−|x − 1|}
and g(x) = 1/2(e^x − 1 + e^1 − x).
The the area of the region in the first quadrant bounded by the curves y = f(x), y = g(x) and x = 0 is
(a)
(b)
(c)
(d)  [JEE Advanced 2020 Paper 1]
Ans: (a)
The given functions f : R → R and g : R → R be defined by 

For point of intersection of curves f(x) and g(x) put f(x) = g(x)

So, required area is

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2019

equals ______. [JEE Advanced 2019 Paper 2]
Ans: 

Now, on adding integrals (i) and (ii), we get

Now,

So,

Q2: If , then 27I² equals .______ [JEE Advanced 2019 Paper 1]
Ans: 4
Given,

On applying property

On adding integrals (i) and (ii), we get

Put,

So,

=

Q3: The area of the region {(x, y) : xy ≤ 8, 1 ≤ y  ≤  x²} is
(a)
(b)
(c)
(d)  [JEE Advanced 2019 Paper 1]
Ans: (c)
The given region
{(x, y) : xy ≤ 8, 1 ≤ y ≤ x²}.
From the figure, region A and B satisfy the given region, but only A is bounded region, so area of bounded region

[∴ Points P(1, 1), Q(2, 4) and R(8, 1)]

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2018

Q1: The value of the integral

is ______. [JEE Advanced 2018 Paper 2]
Ans: 2
Let,

Put,

When,