Exams › JEE Advanced › Maths

Find the area of the region satisfying both 12x - y² >= 0 and y >= 2x.

1. (a) 3
2. (b) 4
3. (c) 6
4. (d) 9

Correct answer: (a) 3

Solution

The region 12x - y² >= 0 is the interior of the rightward-opening parabola y² = 12x. The region y >= 2x is above the line through the origin. Intersections: substitute y = 2x into y² = 12x: 4x² = 12x => x = 0 or x = 3. Points: (0,0) and (3,6). For x in [0,3], the bounded region lies between y = 2x (lower) and y = sqrt(12x) (upper). Area = integral from 0 to 3 of [sqrt(12x) - 2x] dx = [2*sqrt(12)/3 * x^(3/2) - x²] from 0 to 3 = (4*sqrt(3)/3 * 3*sqrt(3)) - 9 = 12 - 9 = 3.

Related JEE Advanced Maths questions

• What is the area enclosed by the curves y = 2 − |2 − x| and y = 3/|x|?
• If f(x) and g(x) are continuous functions over the interval a ≤ x ≤ b, and p(x) represents the greater of f(x) and g(x), while q(x) represents the lesser of f(x) and g(x), what is the expression for the area enclosed by the curves y = p(x), y = q(x), and the vertical lines x = a and x = b?
• The area enclosed by the curves y = sin x + cos x and y = |cos x − sin x| over the interval [0, π/2] is -
• Let f: [1/2, 1] → R (the set of all real numbers) be a positive, non-constant and differentiable function such that f'(x) < 2 f(x) and f(1/2) = 1. Then the value of ∫[1/2 to 1] f(x) dx lies in the interval -
• Let S = {(x, y) ∈ R × R: x ≥ 0, y ≥ 0, y² ≤ 4x, y² ≤ 12 - 2x and 3y + √8x ≤ 5√8}. If the area of the region S is α√2, then α is equal to -
• The area of the region enclosed by the curves y = |(x + 2)/(x - 2)| and y = |(x - 2)/(x + 2)| together with the x-axis equals 4 ln(a/e). Find the value of a.

⚔️ Practice JEE Advanced Maths free + battle 1v1 →