Exams › JEE Main › Maths

Define f, g: R -> R by f(x) = e^(x-1) - e^(-|x-1|) and g(x) = (1/2)(e^(x-1) + e^(1-x)). Find the area of the region in the first quadrant bounded by y = f(x), y = g(x) and x = 0.

1. (2 - sqrt(3)) + (1/2)(e - e⁻¹)
2. (2 + sqrt(3)) + (1/2)(e - e⁻¹)
3. (2 - sqrt(3)) + (1/2)(e + e⁻¹)
4. (2 + sqrt(3)) + (1/2)(e + e⁻¹)

Correct answer: (2 - sqrt(3)) + (1/2)(e - e⁻¹)

Solution

This standard JEE Advanced 2022 problem evaluates to (2 - sqrt3) + (1/2)(e - e⁻¹) after finding the intersection and integrating the difference of the curves over the bounded region.

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 →