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JEE Main · 2020 · 04 Sep 2020 · Shift 1 · Definite Integrals

Let

f (x) = ∣ x - 2∣

and

g (x) = f (f (x)), x \in [0, 4]

. Then

\int_{0}^{3} (g (x) - f (x)) d x

is equal to:

(A) $\frac{1}{2}$
(B) $\frac{3}{2}$
(C)1✓ Correct
(D)0

Answer: (C)

Solution

Here

f (x) = ∣ x - 2∣

and

g (x) = f (f (x)) = ∣ x - 2∣ - 2

. On

[0, 2]

,

∣ x - 2∣ = 2 - x

, so

g (x) = ∣2 - x - 2∣ = x

; on

[2, 3]

,

∣ x - 2∣ = x - 2

, so

g (x) = ∣ x - 4∣ = 4 - x

. Then

\int_{0}^{3} g d x = \int_{0}^{2} x d x + \int_{2}^{3} (4 - x) d x = 2 + 1.5 = 3.5

, while

\int_{0}^{3} f d x = \int_{0}^{2} (2 - x) d x + \int_{2}^{3} (x - 2) d x = 2 + 0.5 = 2.5

. Hence

\int_{0}^{3} (g (x) - f (x)) d x = 3.5 - 2.5 = 1

.

JEE Main · 2020 · 04 Sep 2020 · Shift 1 · Single correct option

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