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
=
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 Xaxis. 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 Xaxis, 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 x_{1} < x_{2} < x_{3} < ... < x_{n} < ... be all the points of local maximum of f and y_{1} < y_{2} < y_{3} < ... < y_{n} < ... be all the points of local minimum of f.
Then which of the following options is/are correct?
(a) x_{n} − y_{n}>1
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 x_{1}, x_{2}, x_{3} .... it is clear that
From the graph for points of minima y_{1}, y_{2}, y_{3} ....., it is clear that
Hence, options (a), (b) and (d) are correct.
[by using integration by parts]
209 videos443 docs143 tests

1. What is the application of derivatives in JEE Advanced? 
2. How are derivatives applied to find the maximum and minimum values of functions in JEE Advanced? 
3. Can derivatives be used to determine the rate at which a quantity is changing in JEE Advanced? 
4. How can derivatives be applied to analyze the behavior of functions in JEE Advanced? 
5. What are some common types of problems involving the application of derivatives in JEE Advanced? 
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