Match List-I with List-II

List-I	List-II
(A) $\int_{-a}^a f(x) dx = 0$	(I) 0
(B) $\int_0^{2a} f(x) dx = 2\int_0^a f(x) dx$	(II) 1
(C) $\int_{-\pi}^{\pi} \cos x dx$	(III) $f$ is an odd function
(D) $\int_{-1}^1 x^{101} dx + 1$	(IV) $f(2a-x) = f(x)$

Choose the correct answer from the options given below:

For $\int_{-a}^a f(x) dx = 0$:

This integral equals zero when $f$ is an odd function, where $f(-x) = -f(x)$.

The negative side (from $-a$ to $0$) exactly cancels out the positive side (from $0$ to $a$).

(A) matches with (III)

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For $\int_0^{2a} f(x) dx = 2\int_0^a f(x) dx$:

The integral from $0$ to $2a$ is exactly double the integral from $0$ to $a$.

This occurs when the function is symmetric about $x = a$, satisfying the condition $f(2a - x) = f(x)$.

When the graph is folded at $x = a$, both halves are identical, so the area from $a$ to $2a$ equals the area from $0$ to $a$.

(B) matches with (IV)

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For $\int_{-\pi}^{\pi} \cos x dx$:

$\int_{-\pi}^{\pi} \cos x \, dx = [\sin x]_{-\pi}^{\pi}$

$= \sin(\pi) - \sin(-\pi)$

$= 0 - 0$

$= 0$

(C) matches with (I)

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For $\int_{-1}^1 x^{101} dx + 1$:

The function $x^{101}$ is an odd function (odd power).

Since $x^{101}$ is odd:

$\int_{-1}^1 x^{101} dx = 0$

Therefore:

$\int_{-1}^1 x^{101} dx + 1 = 0 + 1$

$= 1$

(D) matches with (II)

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Final Matching:

(A) → (III) - Odd function property

(B) → (IV) - Symmetric function property

(C) → (I) - Equals 0

(D) → (II) - Equals 1