
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
= 

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

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 = 

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
for every n [JEE Advanced 2019 Paper 2]

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.

, then the value of [L] is _______. [JEE Advanced 2018 Paper 1]
[by using integration by parts]

| 1. What is the importance of the application of derivatives in JEE Advanced? | ![]() |
| 2. How can I effectively prepare for the application of derivatives section in JEE Advanced? | ![]() |
| 3. Are there any specific types of problems related to derivatives that frequently appear in JEE Advanced? | ![]() |
| 4. What are some common mistakes to avoid while solving application of derivatives problems in JEE Advanced? | ![]() |
| 5. How can I apply derivatives to solve real-world problems in JEE Advanced? | ![]() |