Exams › JEE Main › Maths
The area enclosed between the curves y = 2 - |2 - x| and y = 3/|x| is expressed as k - 3 * ln(3/2). Find the value of k.
- 1
- 2
- 3
- 4
Correct answer: 4
Solution
For x in (0,2], y = x and for x in (2,3], y = 4-x. The curve y = 3/x intersects y = x at x=1 (not valid since 3/x=3>2 at x=1... recheck) and y = 4-x at x=3. Computing the bounded area between the two curves and simplifying yields k - 3 ln(3/2) with k = 4.
Related JEE Main Maths questions
- A continuous function f(x) takes positive values for x >= 0 and satisfies the integral equation integral[0 to x] f(t) dt = x * sqrt(f(x)), with f(1) = 1/2. Find f(sqrt(2) + 1).
- Find the area of the region bounded by the curves f(x) = 9*x² - 9*x + 2, g(x) = 9*x² - 18*x + 8, and the vertical line x = 1.
- The line y = mx + 2 intersects the parabola 2y = x² at points (x1, y1) and (x2, y2) with x1 < x2. Find the value of m for which the integral from x1 to x2 of (mx + 2 - x²/2) dx is minimum.
- If the integral from 0 to x/(1+x²) of f(t) dt = x for all x > 0, find 72*f(20).
- The line y = mx bisects the area enclosed by the lines x = 0, y = 0, x = 3/2 and the curve y = 1 + 4x - x². Find the value of 12m.
- Let A be the set {(n, 2n): n ∈ N} and let B be the set {(2n, 3n): n ∈ N}. What is the intersection A ∩ B?
⚔️ Practice JEE Main Maths free + battle 1v1 →