Area Under Curves Set: Paragraph 1 | JEE Advanced 2023 P2

JEE Advanced 2023 Paper 2 Mathematics: Paragraph 1

Let

f(x)=x^3-6x^2+9x

. The graph of

y=f(x)

has a local maximum at

x=1

with value

4

, touches the

x

-axis at

x=3

(local minimum) and crosses it at

x=0

. The horizontal line

y=4

touches the graph at

x=1

and meets it again at

x=4

Question 1

Numerical answer

The area of the finite region bounded by the graph of

y=f(x)

and the line

y=4

is (give the answer correct to two decimal places)

Show answer and solution

Answer: 6.75

Since

f(x)-4=(x-1)^2(x-4)

, we have

4-f(x)=(x-1)^2(4-x)\ge0

[1,4]

. Area

=\int_1^4 (4-f)\,dx=\int_1^4(x-1)^2(4-x)dx

. With

u=x-1

\int_0^3u^2(3-u)du=27-\frac{81}{4}=\frac{27}{4}=6.75

Numerical answer

The region bounded by

y=f(x)

, the

x

-axis, and the lines

x=0

and

x=3

is revolved about the

x

-axis. If

V

is the volume of the solid generated, the value of

V/\pi

is (give the answer correct to two decimal places)

Show answer and solution

Answer: 20.83

[0,3]

f(x)=x(x-3)^2\ge0

V/\pi=\int_0^3 f^2dx=\int_0^3 x^2(x-3)^4dx

. Expanding or using the Beta integral with

x=3u

3^7\int_0^1u^2(1-u)^4du=2187\cdot\frac{2!\,4!}{7!}=2187\cdot\frac{48}{5040}=\frac{729}{35}\approx20.83

Original question written in the JEE Advanced 2023 Paper 2 pattern and checked twice for the correct answer. Not an official question paper. How we write and check questions · Corrections policy · Report a mistake