JEE Main 2020 · 3 September · morning · Single correct · +4 / −1
∫_(- π)^π | π - | x | |dx is equal to :
- (a) π ²
- (b) 2π ²
- (c) √2 π ²
- (d) (π ²)/2
Answer and solution
Answer (a) π ²
Solution
Maths sir · 4 steps
Inside [-π, π] the outer modulus does nothing, and the graph is just a triangle.
- Drop outer modulus. For every x in [-π, π], |x| ≤ π, so π - |x| ≥ 0. That means |π - |x|| = π - |x| on the whole interval.
- Use evenness. The function π - |x| is even, so the integral from -π to π is twice the integral from 0 to π. On [0, π], |x| = x.
2∫₀^π (π - x) dx = 2[π x - x²/2]₀^π = 2(π² - π²/2) = π²
- Check by area. The graph of y = π - |x| is a triangle with base 2π and height π. Its area is 1/2 × 2π × π = π², which matches.
- So. The integral equals π², which is option (a).
The idea to keep. With nested modulus, first check the sign of the inside on the given interval. Often a modulus vanishes, and an even integrand lets you double the integral from 0.
Answer from the NTA final answer key. Solution by PrepWiser.