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Considering the following Columns.
COLUMN-I COLUMN-II
A)Area enclosed by y = [x] and y = {x}, where [.] and {.} represent greatest integer and fractional part functions. P)32/5 sq. units
B)The area bounded by the curves $y^{2} = x^{3}$ and $| y | = 2 x$ . Q)1 sq. units
C)The smaller area included between the curves $\sqrt{x} + \sqrt{| y |} = 1$ and $| x | + | y | = 1$ . R)4 sq. units
D)Area bounded by the curves $y = [\frac{x^{2}}{64} + 2]$ (where [.] denotes the greatest integer function), y = x – 1 and x = 0 above the x–axis. S)2/3 sq. units

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$A \to Q, B \to P, C \to S, D \to R$

$A \to R, B \to P, C \to S, D \to R$

$A \to Q, B \to P, C \to R, D \to R$

$A \to Q, B \to Q, C \to S, D \to R$

answer is A.

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Detailed Solution

A : y = [x] and y = {x}
Question Image
Area = $\frac{1}{2} \times 2 \times 1 = \frac{1}{2} \times 2 = 1$
B : $y^{2} = x^{3}, | y | = 2 x$

Point A : ${(2 x)}^{2} = x^{3}$
4 = x, y = 8
Area =
= $2 [16 - \frac{64}{5}] = \frac{32}{5}$
C : $\sqrt{x} + \sqrt{| y |} = 1 a n d | x | + | y | = 1$

$A = 2 [\frac{1}{2} \times 1 \times 1 - \int_{0}^{1} {(1 - \sqrt{x})}^{2}] d x$
= $2 [\frac{1}{2} - {(x + \frac{x^{2}}{2} - \frac{4}{3} x^{3 / 2}) |}_{0}^{1}]$
= $2 [\frac{1}{2} - \frac{1}{6}] = \frac{2}{3}$
D : $y = [\frac{x^{2}}{64} + 2]$
Question Image
$A = \frac{1}{2} \times 4 \times 2 = 4$
$A \to Q, B \to P, C \to S, D \to R$

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