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JEE Main · 2024 · 04 Apr 2024 · Shift 1 · Definite Integrals

Let

f (x) = {- 2, x - 2, - 2 \leq x \leq 0 0 < x \leq 2

and

h (x) = f (∣ x ∣) + ∣ f (x) ∣

. Then

\int_{- 2}^{2} h (x) d x

is equal to

(A)4
(B)6
(C)1
(D)2✓ Correct

Answer: (D)

Solution

For

- 2 \leq x \leq 0

:

∣ x ∣ \in [0, 2]

so

f (∣ x ∣) = ∣ x ∣ - 2 = - x - 2

, and

f (x) = - 2 \Rightarrow ∣ f (x) ∣ = 2

; thus

h (x) = (- x - 2) + 2 = - x

.
For

0 < x \leq 2

:

f (∣ x ∣) = x - 2

and

∣ f (x) ∣ = ∣ x - 2∣ = 2 - x

; thus

h (x) = (x - 2) + (2 - x) = 0

.

\int_{- 2}^{2} h (x) d x = \int_{- 2}^{0} (- x) d x + \int_{0}^{2} 0 d x = [- \frac{x ^{2}}{2}]_{- 2}^{0} = 2.

Option D.

JEE Main · 2024 · 04 Apr 2024 · Shift 1 · Single correct option

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