JEE Main 2020 · 4 September · morning · Single correct · +4 / −1

Let f(x) = | x - 2 | and g(x) = f(f(x)), x ∈ [ 0,4 ]. Then
∫₀³ ( g(x) - f(x) ) dx is equal to:

  1. (a) 1
  2. (b) 0
  3. (c) 1/2
  4. (d) 3/2
Answer and solution

Answer (a) 1

Solution

Maths sir · 2 steps

Break at x = 2, write g and f as plain lines on each piece, then integrate g − f piece by piece.

  1. Open the inner modulus. For 0 ≤ x ≤ 2, f(x) = 2 - x, so g(x) = |(2-x) - 2| = |x| = x. For 2 ≤ x ≤ 3, f(x) = x - 2, so g(x) = |(x-2) - 2| = |x - 4| = 4 - x.
    Intervalf(x)g(x)g(x) - f(x)Integral
    [0, 2]2 - xx2x - 2[x² - 2x]₀² = 0
    [2, 3]x - 24 - x6 - 2x[6x - x²]₂³ = 9 - 8 = 1

    ∫₀³ (g(x) - f(x))dx = 0 + 1 = 1

  2. So. The integral equals 1, which is option (a).

The idea to keep. For a modulus inside a modulus, open the inner one first and split at its breakpoint. On each piece the function becomes a straight line, and the integral becomes routine.

Answer from the NTA final answer key. Solution by PrepWiser.

Study Definite Integration

Asked about 230 times in JEE Main since 2020. This question is from 2020.