JEE Exam > JEE Notes > Mathematics (Maths) for JEE Main & Advanced > JEE Advanced Previous Year Questions (2018 - 2023): Application of Derivatives
JEE Advanced Previous Year Questions (2018 - 2023): Application of Derivatives | Mathematics (Maths) for JEE Main & Advanced PDF Download
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)
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 
Equation of AB 
Normal to both the line's
= 
2022
Q1: Let
Let g : [0, 1] → R be the function defined by 
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)
Ans: (a), (b) & (c)
Given,
This is a infinite G.P.
Given x ∈ [0, 1]
We know,
We know, AM = GM when terms are equal.
∴ Option (A) is correct
And option (D) is wrong as at only at a single point x = 1/2, g(x) is minimum.
Now, 
We already found that at x = 1/2 g(x) is minimum.
Similarly, g′(x) < 0 when x < 1/2
If we put it in number line we get this
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) is
From graph you can see value of g(x) is maximum either at x = 0 or x = 1 in the range x ∈ [0, 1]. 
∴ 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
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) 
(d) 
[JEE Advanced 2020 Paper 1]
Ans: (c)
Given region is
and 
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 
and PQ = RS = 2sin 2x.
∴ Perimeter of the rectangle
For maximum, dy / dx = 0
∴ At x = π/6, the rectangle PQRS have maximum perimeter.
So length of sides
∴ Required area = π/6 x √3 = 
2019
Q1: Let f : R → R be given by f(x) = (x − 1)(x − 2)(x − 5). Define
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, 
So, 
According to wavy curve method
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.
∵ 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, 
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)
for every n [JEE Advanced 2019 Paper 2]
Ans: (a), (b) & (d)
Given,
for every n [JEE Advanced 2019 Paper 2]Ans: (a), (b) & (d)
Given,
Since, for maxima and minima of f(x), f'(x) = 0
is point of local minimum.
is point of local maximum.
From the graph, for points of maxima x1, x2, x3 .... it is clear that
From the graph for points of minima y1, y2, y3 ....., it is clear that
Hence, options (a), (b) and (d) are correct.
2018
Q1: For each positive integer n, let
For x ∈ R, let [x] be the greatest integer less than or equal to x. If
, then the value of [L] is _______. [JEE Advanced 2018 Paper 1]Ans: 1
We have,
[by using integration by parts]
The document JEE Advanced Previous Year Questions (2018 - 2023): Application of Derivatives | Mathematics (Maths) for JEE Main & Advanced 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 - 2023): Application of Derivatives - Mathematics (Maths) for JEE Main & Advanced
| 1. What is the application of derivatives in JEE Advanced? |  |
Ans. The application of derivatives in JEE Advanced involves using the concept of derivatives to solve various types of problems related to rates of change, optimization, and graph analysis. It helps in finding the maximum and minimum values of functions, determining the rate at which a quantity is changing, and analyzing the behavior of functions using their derivatives.
| 2. How are derivatives applied to find the maximum and minimum values of functions in JEE Advanced? | |
Ans. To find the maximum and minimum values of functions in JEE Advanced, derivatives are used. The critical points of a function, where the derivative is either zero or undefined, are first determined. Then, the first and second derivative tests are applied to these critical points to determine whether they correspond to maximum or minimum values. By analyzing the sign changes of the derivative and the concavity of the function, the maximum and minimum values can be identified.
| 3. Can derivatives be used to determine the rate at which a quantity is changing in JEE Advanced? | |
Ans. Yes, derivatives can be used to determine the rate at which a quantity is changing in JEE Advanced. By finding the derivative of a function representing the quantity with respect to time or another variable, the derivative represents the instantaneous rate of change of the quantity at any given point. This concept is particularly useful in problems involving motion, growth, or decay, where the rate of change is of interest.
| 4. How can derivatives be applied to analyze the behavior of functions in JEE Advanced? | |
Ans. Derivatives can be applied to analyze the behavior of functions in JEE Advanced by examining their increasing and decreasing intervals, as well as their concavity. The sign of the derivative indicates whether the function is increasing or decreasing, while the second derivative determines the concavity. By analyzing the sign changes of the derivative and the concavity, critical points, inflection points, and the overall behavior of the function can be determined.
| 5. What are some common types of problems involving the application of derivatives in JEE Advanced? | |
Ans. Some common types of problems involving the application of derivatives in JEE Advanced include optimization problems, related rates problems, curve sketching problems, and problems involving motion or growth/decay. These problems often require finding maximum or minimum values, determining rates of change, analyzing the behavior of functions, or solving real-life application problems using the concepts of derivatives.
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