Match List-I with List-IIList-IList-II(A) ∫_-2^2cos^4xsin^5x dx (I) π…

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

For an odd function, \( \int_{-a}^{a}f(x)\,dx=0 \); for an even function, \( \int_{-a}^{a}f(x)\,dx=2\int_{0}^{a}f(x)\,dx \).

(A) \( \int_{-2}^{2}\cos^4x\sin^5x\,dx \)

This matches (III) 0.

\( \cos^4(-x)\sin^5(-x)=-\cos^4x\sin^5x \), so the integrand is odd and the integral is 0.

(B) \( \int_{-1}^{1}|x|\,dx \)

This matches (IV) 1.

|x| is even:

    \( \displaystyle \int_{-1}^{1}|x|\,dx=2\int_{0}^{1}x\,dx \)

    \( \displaystyle =2\cdot\frac{1}{2}=1 \)

(C) \( \int_{0}^{\pi/2}\cos^2x\,dx \)

This matches (II) \( \frac{\pi}{4} \).

    \( \displaystyle \int_{0}^{\pi/2}\frac{1+\cos 2x}{2}\,dx=\frac{1}{2}\left[x+\frac{\sin 2x}{2}\right]_{0}^{\pi/2} \)

    \( \displaystyle =\frac{1}{2}\cdot\frac{\pi}{2} \)

    \( \displaystyle =\frac{\pi}{4} \)

(D) \( \int_{-\pi/2}^{\pi/2}(x^3+x\cos x+\tan^5x+1)\,dx \)

This matches (I) \( \pi \).

\( x^3 \), \( x\cos x \) and \( \tan^5x \) are odd functions, so their parts give 0 by the odd-function property. Only the constant remains:

    \( \displaystyle \int_{-\pi/2}^{\pi/2}1\,dx=\frac{\pi}{2}-\left(-\frac{\pi}{2}\right) \)

    \( =\pi \)

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

Note: It seems that there is a mistake in the original NTA paper, which prints the limits of (D) as \( -\frac{\pi}{2} \) and \( \frac{\pi}{2} \), where \( \tan^5x \) is not defined, so, strictly, the integral in (D) does not converge. The intended value \( \pi \) comes from the odd-function property, as used in the working. The question can still be solved because (A), (B) and (C) have only one possible match each, and Option (1) is the only option containing (A)-(III), (B)-(IV) and (C)-(II).