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JEE Advanced Previous Year Questions (2018 - 2025): Application of Derivatives

2023


Q1: Let Q be the cube with the set of vertices {(x1, x2, x3) ∈ R3: x1, x2, x3 ∈ {0, 1}}. Let F be the set of all twelve lines containing the diagonals of the six faces of the cube Q. Let S be the set of all four lines containing the main diagonals of the cube Q; for instance, the line passing through the vertices (0, 0, 0) and (1, 1, 1) is in S. For lines ℓ1 and ℓ2, let d(ℓ1, ℓ2) denote the shortest distance between them. Then the maximum value of d(ℓ1, ℓ2), as ℓ1 varies over F and ℓ2 varies over S, is : [JEE Advanced 2023 Paper 1]
(a) 1√6
(b) 1√8
(c) 1√3
(d) 1√12
Ans: 
(a)
2023

DR'S of OG = 1, 1, 1
DR'S of AF =-1, 1, 1
DR'S of CE = 1, 1, -1
DR'S of BD = 1, -1, 1
Equation OG 2023

Equation of AB 2023

Normal to both the line's
= 2023

2023

2022


Q1: Let 2022 Let g : [0, 1] → R be the function defined by 

2022

Then, which of the following statements is/are TRUE ?
(a) The minimum value of g(x) is 27/6
(b) The maximum value of g(x) is 1 + 21/3
(c) The function g(x) attains its maximum at more than one point
(d) The function g(x) attains its minimum at more than one point             [JEE Advanced 2022 Paper 2]
Ans:
(a), (b) & (c)

Given,

2022

This is a infinite G.P.

2022
Given x ∈ [0, 1]
We know,

2022

We know, AM = GM when terms are equal. 

2022
∴ Option (A) is correct
And option (D) is wrong as at only at a single point x = 1/2, g(x) is minimum.
Now, 2022

2022

We already found that at x = 1/2 g(x) is minimum. 

2022
Similarly, g′(x) < 0 when x < 1/2
If we put it in number line we get this

2022

We know g′(x) represent slope of curve g(x) and it is negative when x < 1/2 and positive when x > 1/2 and zero when x = 1/2
∴ Graph of g(x) is2022

From graph you can see value of g(x) is maximum either at x = 0 or x = 1 in the range x ∈ [0, 1].
2022
∴ We get maximum value at x = 0 and x = 1 both.
∴ B and C options are correct.

2020


Q1: Consider the rectangles lying the region

2020
and having one side on the X-axis. The area of the rectangle which has the maximum perimeter among all such rectangles, is
(a) 3π / 2
(b) π
(c) 2020
(d) 2020 [JEE Advanced 2020 Paper 1]
Ans:
(c)
Given region is

2020 and 2020

On drawing the diagram,
Let the side PS on the X-axis, such that P(x, 0), and Q(x, 2sin(2x)), so length of the sides 2020 and PQ = RS = 2sin 2x.

2020

∴ Perimeter of the rectangle

2020

For maximum, dy / dx = 0
2020
 At x = π/6, the rectangle PQRS have maximum perimeter.
So length of sides 

2020
 Required area = π/6 x √3 = 2020

2019


Q1: Let f : R  R be given by f(x) = (x - 1)(x - 2)(x - 5). Define

2019
Then which of the following options is/are correct?
(a) F(x) ≠ 0 for all x ∈ (0, 5)
(b) F has a local maximum at x = 2
(c) F has two local maxima and one local minimum in (0, ∞)
(d) F has a local minimum at x = 1                         [JEE Advanced 2019 Paper 2]
Ans:
(a), (b) & (d)
Given, f : R  and
f(x) = (x - 1)(x - 2)(x - 5)
Since, 2019
So, 2019

According to wavy curve method

2019

F'(x) changes, it's sign from negative to positive at x = 1 and 5, so F(x) has minima at x = 1 and 5 and as F'(x) changes, it's sign from positive to negative at x = 2, so F(x) has maxima at x = 2.

2019

 AT the point of maxima x = 2, the functional value F(2), = -10/3, is negative for the interval, x (0, 5), so F(x)  0 for any value of x (0, 5),
Hence, options (a), (b) and (d) are correct. 

Q2: Let, 2019
Let x1 < x2 < x3 < ... < xn < ... be all the points of local maximum of f and y1 < y2 < y3 < ... < yn < ... be all the points of local minimum of f.
Then which of the following options is/are correct?
(a) |xn - yn|>1

(b) xn+1 - xn > 2 for every n
(c) x1 < y1
(d)2019 for every n [JEE Advanced 2019 Paper 2]
Ans:
(a), (b) & (d)
Given,
2019

2019

Since, for maxima and minima of f(x), f'(x) = 0

2019

2019

2019

is point of local minimum.

2019

is point of local maximum.
From the graph, for points of maxima x1, x2, x3 .... it is clear that 

2019

From the graph for points of minima y1, y2, y3 ....., it is clear that 

2019

Hence, options (a), (b) and (d) are correct.

2018


Q1: For each positive integer n, let 2018
For x R, let [x] be the greatest integer less than or equal to x. If 2018, then the value of [L] is _______.             [JEE Advanced 2018 Paper 1]
Ans: 1
We have,

2018

[by using integration by parts]

2018

The document JEE Advanced Previous Year Questions (2018 - 2025): Application of Derivatives is a part of the JEE Course Mathematics (Maths) for JEE Main & Advanced.
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FAQs on JEE Advanced Previous Year Questions (2018 - 2025): Application of Derivatives

1. What is the importance of the application of derivatives in JEE Advanced?
Ans. The application of derivatives is crucial in JEE Advanced as it helps in analyzing the behavior of functions, finding maxima and minima, and understanding the concepts of motion and rates of change. This knowledge is essential for solving complex problems in calculus and related topics.
2. How can I effectively prepare for the application of derivatives section in JEE Advanced?
Ans. To effectively prepare for this section, students should practice a variety of problems from previous years, focus on understanding the core concepts, and utilize resources like textbooks and online tutorials. Regular mock tests and timed practice sessions can also enhance problem-solving speed and accuracy.
3. Are there any specific types of problems related to derivatives that frequently appear in JEE Advanced?
Ans. Yes, JEE Advanced often includes problems on finding local maxima and minima, determining points of inflection, and solving real-world application problems involving rates of change. Students should familiarize themselves with these types of questions to perform well.
4. What are some common mistakes to avoid while solving application of derivatives problems in JEE Advanced?
Ans. Common mistakes include miscalculating derivatives, overlooking critical points, and failing to check the second derivative for concavity. Students should also be cautious about interpreting the problem correctly and ensuring they apply the correct mathematical principles.
5. How can I apply derivatives to solve real-world problems in JEE Advanced?
Ans. To apply derivatives to real-world problems, students should focus on formulating the problem mathematically, using derivatives to determine rates of change, and applying optimization techniques to find maximum or minimum values in practical scenarios.
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