Match the LIST-I with LIST-II (where [x] denotes greatest integer fun…

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

For a greatest integer function, the interval is split at the points where the value jumps; on each piece the integrand is a constant.

(A) \( \int_{0}^{2}[x]\,dx \)

This matches (II) 1.

[x] = 0 on [0, 1) and [x] = 1 on [1, 2):

    \( \displaystyle \int_{0}^{2}[x]\,dx=\int_{0}^{1}0\,dx+\int_{1}^{2}1\,dx \)

    \( \displaystyle =0+1=1 \)

(B) \( \int_{0}^{1.5}\left[x^2\right]dx \)

This matches (IV) \( 2-\sqrt{2} \).

\( [x^2] \) jumps where \( x^2=1 \) and \( x^2=2 \), that is at x = 1 and \( x=\sqrt{2}\approx 1.414 \):

    \( \displaystyle [x^2]=0\text{ on }[0,1) \)

    \( \displaystyle [x^2]=1\text{ on }[1,\sqrt{2}) \)

    \( \displaystyle [x^2]=2\text{ on }[\sqrt{2},1.5] \)

    \( \displaystyle \int_{0}^{1.5}\left[x^2\right]dx \)

    \( \displaystyle =0+1\cdot(\sqrt{2}-1)+2\,(1.5-\sqrt{2}) \)

    \( \displaystyle =\sqrt{2}-1+3-2\sqrt{2} \)

    \( \displaystyle =2-\sqrt{2} \)

(C) \( \int_{-\frac{\pi}{4}}^{\frac{\pi}{4}}\sin x\,dx \)

This matches (I) 0.

\( \sin x \) is an odd function and the limits are symmetric about 0, so the integral is 0.

(D) \( \int_{-\frac{\pi}{4}}^{\frac{\pi}{4}}\cos x\,dx \)

This matches (III) \( \sqrt{2} \).

\( \cos x \) is an even function, so

    \( \displaystyle \int_{-\frac{\pi}{4}}^{\frac{\pi}{4}}\cos x\,dx \)

    \( \displaystyle =2\int_{0}^{\frac{\pi}{4}}\cos x\,dx \)

    \( \displaystyle =2\left[\sin x\right]_{0}^{\frac{\pi}{4}} \)

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

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

Shortcut: (C) is 0 by the odd-function property, which leaves only Options (1) and (2); (A) = 1 then decides Option (1).

Tip: For \( [x^2] \) the break points are at \( x=\sqrt{1},\sqrt{2},\sqrt{3},\ldots \), not at the integers.