COLUMN-I COLUMN-II
A)Area bounded by f (x) = max
(sin x, cos x) between $x = 0$ and $x = \frac{π}{2}$ and the x-axis is
P)7/12
B)Area bounded by y = x + sin x and its inverse function between the ordinates x = 0 and $x = 2 π$ is Q)4/3
C)Area bounded by the curves $y^{2} = x^{2} (1 - x^{2})$ is R)8
D)Area bounded by the curves $\frac{y}{x} = \log x$ and $\frac{y}{2} = - x^{2} + x$ is S) $\sqrt{2}$

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$A - S, B - Q, C - R, D - P$

$A - S, B - R, C - Q, D - P$

$A - P, B - R, C - Q, D - S$

answer is A.

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Detailed Solution

a) $4 y^{2} = x^{2} (4 - x^{2})$
$A = 4 \int_{0}^{1} \frac{x}{2} \sqrt{4 - x^{2}} d x = \frac{16}{3}$
b) Question Image
c) A= 8
d) $x = y^{2} - 1; x = | y | \sqrt{1 - y^{2}}$

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COLUMN-I	COLUMN-II
A)Area bounded by f (x) = max (sin x, cos x) between $x = 0$ and $x = \frac{π}{2}$ and the x-axis is	P)7/12
B)Area bounded by y = x + sin x and its inverse function between the ordinates x = 0 and $x = 2 π$ is	Q)4/3
C)Area bounded by the curves $y^{2} = x^{2} (1 - x^{2})$ is	R)8
D)Area bounded by the curves $\frac{y}{x} = \log x$ and $\frac{y}{2} = - x^{2} + x$ is	S) $\sqrt{2}$