Exams › JEE Main › Maths

The area enclosed between the curves y = sec⁻¹(x), y = cot⁻¹(x) and the line x = 1 can be expressed as which of the following integral forms?

1. integral[1 to (1+sqrt(5))/2] (cot⁻¹ x - sec⁻¹ x) dx
2. integral[0 to a] sec(x) dx + integral[a to pi/4] cot(x) dx - pi/4, where sin(a) = cos²(a)
3. integral[0 to a] sec(x) dx + integral[a to pi/4] cot(x) dx - pi/4 + 1, where sin(a) = cos²(a)
4. integral[1 to (1+sqrt(5))/2] (sec⁻¹ x - cot⁻¹ x) dx

Correct answer: integral[1 to (1+sqrt(5))/2] (cot⁻¹ x - sec⁻¹ x) dx

Solution

For x >= 1, cot⁻¹(x) is above sec⁻¹(x) until they meet at x = (1+sqrt5)/2; the area is the integral of (cot⁻¹ x - sec⁻¹ x) over [1, (1+sqrt5)/2].

Related JEE Main Maths questions

• For every b > 1, the area enclosed by the x-axis, the graph of y = f(x), and the vertical lines x = 1 and x = b is given by √(b² + 1) − √2. Then f(x) must be:
• The graphs of y = sin x and y = cos x meet at infinitely many points, forming repeated bounded regions of equal area. The area of one such region is
• Find the area bounded by the parametric curve x = a cos³ t, y = b sin³ t together with the positive x-axis and positive y-axis.
• Find the area enclosed by the curves y = e^x, y = e^(-x), and the vertical line x = 1, measured in square units.
• A straight line of the form y = mx divides into two equal parts the region bounded by the y-axis, the horizontal line y = 3/2, and the parabola y = 1 + 4x - x². What is the value of m?
• Find the area enclosed by the curve y = cos² x, the x-axis, and the vertical lines x = 0 and x = π over the interval (0, π).

⚔️ Practice JEE Main Maths free + battle 1v1 →