JEE 2023 — Mathematics PYQ

Q: How do I solve JEE 2023 Mathematics questions?

Practice JEE 2023 Mathematics previous year questions with step-by-step solutions on Mathem Solvex. Each question includes a detailed explanation and video solution to help you master the concept quickly.

Q: What is the difficulty level of JEE Mathematics questions?

JEE Mathematics questions range from moderate to advanced difficulty. Consistent practice with official previous year papers and topic-wise drills on Mathem Solvex is the most effective preparation strategy.

Q: Are JEE previous year papers available for free?

Yes. Mathem Solvex provides free access to JEE previous year question papers and topic-wise PDFs. You can download them or attempt them as timed live mock tests directly on the platform.

Question

JEE 2023 — Mathematics PYQ

JEE | Mathematics | 2023

Let A be the area of the region ${(x, y) : y \geq x 2, y \geq (1 - x) 2, y \leq 2 x (1 - x)}$ . Then 540A is equal to ________.

Choose the correct answer:

Explanation

Solution

1. Curves ko pehchanen:

$C_1: y = x^2$ (Upward parabola)
$C_2: y = (1-x)^2$ (Upward parabola, vertex at $x=1$ )
$C_3: y = 2x(1-x) = 2x - 2x^2$ (Downward parabola)

2. Intersection points nikaalein:

$C_1$ aur $C_2$ : $x^2 = (1-x)^2 \implies x^2 = 1 + x^2 - 2x \implies 2x = 1 \implies x = \frac{1}{2}$
$C_1$ aur $C_3$ : $x^2 = 2x - 2x^2 \implies 3x^2 - 2x = 0 \implies x(3x - 2) = 0 \implies x = 0, \frac{2}{3}$
$C_2$ aur $C_3$ : $(1-x)^2 = 2x - 2x^2 \implies 1 + x^2 - 2x = 2x - 2x^2 \implies 3x^2 - 4x + 1 = 0$

Solving this: $(3x-1)(x-1) = 0 \implies x = \frac{1}{3}, 1$

3. Area ( $A$ ) ki limit set karein:

Symmetry ke hisaab se, region $x = \frac{1}{3}$ se $x = \frac{1}{2}$ tak aur $x = \frac{1}{2}$ se $x = \frac{2}{3}$ tak bata hua hai. Dono parts barabar hain, isliye:

A = 2 \int_{1/3}^{1/2} [y_{upper} - y_{lower}] \, dx

A = 2 \int_{1/3}^{1/2} [2x(1-x) - (1-x)^2] \, dx

A = 2 \int_{1/3}^{1/2} [2x - 2x^2 - (1 + x^2 - 2x)] \, dx

A = 2 \int_{1/3}^{1/2} [-3x^2 + 4x - 1] \, dx

4. Integration karein:

A = 2 \left[ -x^3 + 2x^2 - x \right]_{1/3}^{1/2}

Values put karne par:

A = 2 \left[ \left( -\frac{1}{8} + \frac{2}{4} - \frac{1}{2} \right) - \left( -\frac{1}{27} + \frac{2}{9} - \frac{1}{3} \right) \right]

A = 2 \left[ \left( -\frac{1}{8} \right) - \left( \frac{-1 + 6 - 9}{27} \right) \right]

A = 2 \left[ -\frac{1}{8} + \frac{4}{27} \right] = 2 \left[ \frac{-27 + 32}{216} \right] = 2 \left[ \frac{5}{216} \right] = \frac{5}{108}

5. Final Value:

Hame $540 A$ nikaalna hai:

540 A = 540 \times \frac{5}{108}

540 A = 5 \times 5 = 25

Jawab: 25

Choose the correct answer:

Explore More