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

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

List-I	List-II
(A) $\int\limits_{-a}^{a} f(x) \, dx = 0$	(III) $f$ is an odd function
(B) $\int\limits _{0}^{2a} f(x) \, dx = 2 \int\limits _{0}^{a} f(x) \, dx$	(IV) $f(2a - x) = f(x)$
(C) $\int\limits _{-\pi}^{\pi} \cos x \, dx$	(I) 0
(D) $\int\limits _{-1}^{1} x^{101} \, dx + 1$	(II) 1

Given List-I and List-II:

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

If $f$ is an odd function, i.e., $f(-x) = -f(x)$, then:

$\int_{-a}^a f(x) \, dx = 0$ because the areas from $-a$ to $0$ and $0$ to $a$ cancel each other out. Hence, (A) matches with (III) $f$ is an odd function.

(B) $\displaystyle \int_0^{2a} f(x) \, dx = 2 \int_0^a f(x) \, dx$

If $f$ satisfies $f(2a - x) = f(x)$ (symmetry about $x = a$), then:

$\int_0^{2a} f(x) \, dx = \int_0^a f(x) \, dx + \int_a^{2a} f(x) \, dx$ Substitute $t = 2a - x$ in the second integral:

$\int_a^{2a} f(x) \, dx = \int_{a}^{0} f(2a - t) (-dt) = \int_0^a f(t) \, dt$ Thus, $\int_0^{2a} f(x) \, dx = \int_0^a f(x) \, dx + \int_0^a f(x) \, dx = 2 \int_0^a f(x) \, dx$ Hence, (B) matches with (IV) $f(2a - x) = f(x)$.

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

Since $\cos x$ is an even function:

$\int_{-\pi}^\pi \cos x \, dx = 2 \int_0^\pi \cos x \, dx = 2[\sin x]_0^\pi = 2(0 - 0) = 0$ Hence, (C) matches with (I) $0$.

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

Since $x^{101}$ is an odd function (because 101 is odd):

$\int_{-1}^1 x^{101} \, dx = 0$ Therefore, $\int_{-1}^1 x^{101} \, dx + 1 = 0 + 1 = 1$ Hence, (D) matches with (II) $1$.