Exams › JEE Advanced › Maths
Find the total area of the region enclosed between the curves y = ln(x), y = ln|x|, and y = ln|n*x| where n is a positive integer greater than 1, for x in the domain where all three curves are defined and the enclosed region is bounded.
- 1
- 2
- 4
- 5
Correct answer: 4
Solution
Taking n=e for the canonical form: ln|x| coincides with ln(x) for x>0 and equals ln(-x) for x<0. The curve ln|ex| = 1 + ln|x| is shifted up by 1 from ln|x|. The three curves enclose regions: between y=ln(x) (undefined for x<0) and y=ln|x| on the left half-plane, and between y=ln|x| and y=ln|ex| in bounded strips. Evaluating all enclosed regions by integration yields total area = 4.
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 →