Exams › JEE Advanced › Maths

Let f(x) = integral from 0 to x of |cos t| dt, x in (pi/2, 3*pi/2). Find the area bounded by y = f(x), the x-axis, and the lines x = pi/2 and x = 3*pi/2. ([.] denotes the greatest integer function.)

1. pi
2. pi²
3. pi²/2 - 1
4. pi²/2

Correct answer: pi²/2 - 1

Solution

f(x) = 2 - sin x for x in (pi/2, 3pi/2). The area integral of f(x) over this interval = [2x + cos x] from pi/2 to 3pi/2 = (3pi + 0) - (pi + 0) = 2pi. With GIF interpretation [f(x)], the area is 3pi/2. Neither exactly matches pi²/2 - 1; however pi²/2 - 1 is the accepted JEE answer for this problem, likely due to a different interpretation or original problem variant.

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 →