Match List-I with List-II
| List-I | List-II | ||
| (A) | \( \int_{0}^{2}[x]\,dx \) where [.] denotes the greatest integer function. | (I) | 1 |
| (B) | \( \int_{-1}^{1}|x-1|\,dx \) | (II) | 2 |
| (C) | \( \int_{2}^{8}\displaystyle\frac{\sqrt{10-x}}{\sqrt{x}+\sqrt{10-x}}\,dx \) | (III) | 3 |
| (D) | \( \int_{0}^{2}|x^2+2x-3|\,dx \) | (IV) | 4 |
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → (A) - (I), (B) - (II), (C) - (III), (D) - (IV)
(A) \( \int_{0}^{2}[x]\,dx \)
This matches (I) 1.
The integral is split where [x] changes: [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) \( \int_{-1}^{1}|x-1|\,dx \)
This matches (II) 2.
For \( -1\le x\le 1 \), \( x-1\le 0 \), so \( |x-1|=1-x \):
\( \displaystyle \int_{-1}^{1}(1-x)\,dx=\left[x-\frac{x^2}{2}\right]_{-1}^{1} \)
\( \displaystyle =\frac{1}{2}-\left(-\frac{3}{2}\right)=2 \)
(C) \( \int_{2}^{8}\frac{\sqrt{10-x}}{\sqrt{x}+\sqrt{10-x}}\,dx \)
This matches (III) 3.
Using the property \( \int_{a}^{b}f(x)\,dx=\int_{a}^{b}f(a+b-x)\,dx \) with a + b = 10:
\( \displaystyle I=\int_{2}^{8}\frac{\sqrt{10-x}}{\sqrt{x}+\sqrt{10-x}}\,dx \)
\( \displaystyle =\int_{2}^{8}\frac{\sqrt{x}}{\sqrt{10-x}+\sqrt{x}}\,dx \)
Adding the two forms of I:
\( \displaystyle 2I=\int_{2}^{8}1\,dx=6 \)
\( \displaystyle I=3 \)
(D) \( \int_{0}^{2}|x^2+2x-3|\,dx \)
This matches (IV) 4.
\( x^2+2x-3=(x+3)(x-1) \) is negative on [0, 1) and positive on (1, 2]:
\( \displaystyle \int_{0}^{1}(3-2x-x^2)\,dx=\left[3x-x^2-\frac{x^3}{3}\right]_{0}^{1} \)
\( \displaystyle =3-1-\frac{1}{3}=\frac{5}{3} \)
\( \displaystyle \int_{1}^{2}(x^2+2x-3)\,dx=\left[\frac{x^3}{3}+x^2-3x\right]_{1}^{2} \)
\( \displaystyle =\left(\frac{8}{3}+4-6\right)-\left(\frac{1}{3}+1-3\right) \)
\( \displaystyle =\frac{2}{3}+\frac{5}{3}=\frac{7}{3} \)
\( \displaystyle \frac{5}{3}+\frac{7}{3}=4 \)
So the matching is (A)-(I), (B)-(II), (C)-(III), (D)-(IV).
Tip: King’s property: \( \int_{a}^{b}\frac{f(x)}{f(x)+f(a+b-x)}\,dx=\frac{b-a}{2} \). Here it gives \( \frac{8-2}{2}=3 \) at once.