Match List-I with List-II List-IList-II (A) ∫_-a^a f(x) dx = 0 (I)0 (…

The correct answer is Option (1) → (A)-(III), (B)-(IV), (C)-(I), (D)-(II)

(A) \(\int_{-a}^{a} f(x)\,dx = 0\)

This matches (III) f is an odd function. Splitting the integral at 0:

    \(\displaystyle \int_{-a}^{a} f(x)\,dx = \int_{-a}^{0} f(x)\,dx + \int_{0}^{a} f(x)\,dx\)

Putting x = −t in the first integral, and using f(−t) = −f(t) for an odd function:

    \(\displaystyle \int_{-a}^{0} f(x)\,dx = \int_{0}^{a} f(-t)\,dt = -\int_{0}^{a} f(t)\,dt\)

So the two parts cancel and the integral is 0.

(B) \(\int_{0}^{2a} f(x)\,dx = 2\int_{0}^{a} f(x)\,dx\)

This matches (IV) \(f(2a - x) = f(x)\). Splitting the integral at a:

    \(\displaystyle \int_{0}^{2a} f(x)\,dx = \int_{0}^{a} f(x)\,dx + \int_{a}^{2a} f(x)\,dx\)

Putting x = 2a − t in the second integral:

    \(\displaystyle \int_{a}^{2a} f(x)\,dx = \int_{0}^{a} f(2a - t)\,dt = \int_{0}^{a} f(t)\,dt\)

So the two parts are equal and the integral is \(2\int_{0}^{a} f(x)\,dx\).

(C) \(\int_{-\pi}^{\pi} \cos x\,dx\)

This matches (I) 0:

    \(\displaystyle \int_{-\pi}^{\pi} \cos x\,dx = \Big[\sin x\Big]_{-\pi}^{\pi} = 0 - 0 = 0\)

(D) \(\int_{-1}^{1} x^{101}\,dx + 1\)

This matches (II) 1. Since \(x^{101}\) is an odd function, its integral over [−1, 1] is 0 by (A):

    \(\displaystyle \int_{-1}^{1} x^{101}\,dx + 1 = 0 + 1 = 1\)

So the matching is (A)-(III), (B)-(IV), (C)-(I), (D)-(II).