Column – I Column – IIAThe area between the curves y = 2x4 − x2, the x-axis and the ordinates of two minimum of the curve isP2 sin1BThe area bounded by the curve x = at2, y = 2at the x-axis 1 ≤ t ≤ 3Q7/120CThe area of the circle centerd at (1, 2) and passing through (4, 6) isR 104 a23DThe area of the curve bounded by y=cosx, y = 0, |x| = 1 isS25π

A) y=2x4−x2⇒dydx=8x3−2x=2x4x2−1=0 Thus required area =−2∫01/2 2x4−x2dx=7120B) equation is y2=4axArea=∫a9a4axdx=104a23

Column – I Column – IIAThe area between the curves y = 2x4 − x2, the x-axis and the ordinates of two minimum of the curve isP2 sin1BThe area bounded by the curve x = at2, y = 2at the x-axis 1 ≤ t ≤ 3Q7/120CThe area of the circle centerd at (1, 2) and passing through (4, 6) isR 104 a23DThe area of the curve bounded by y=cosx, y = 0, |x| = 1 isS25π

A) y=2x4−x2⇒dydx=8x3−2x=2x4x2−1=0 Thus required area =−2∫01/2 2x4−x2dx=7120B) equation is y2=4axArea=∫a9a4axdx=104a23

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Column – I Column – II
A The area between the curves $y = 2 x^{4} - x^{2},$ the x-axis and the ordinates of two minimum of the curve is P $2 \sin 1$
B The area bounded by the curve $x = a t^{2}, y = 2 a t$ the x-axis $(1 \leq t \leq 3)$ Q $7 / 120$
C The area of the circle centerd at (1, 2) and passing through (4, 6) is R $\frac{104 a^{2}}{3}$
D The area of the curve bounded by $y = \cos x, y = 0, | x | = 1$ is S $25 π$

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

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

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

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

answer is B.

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Detailed Solution

A)
$y = 2 x^{4} - x^{2} \Rightarrow \frac{d y}{d x} = 8 x^{3} - 2 x = 2 x (4 x^{2} - 1) = 0 Thus required area = - 2 \int_{0}^{1 / 2} (2 x^{4} - x^{2}) d x = \frac{7}{120}$
B) equation is $y^{2} = 4 a x$
$A r e a = \int_{a}^{9 a} \sqrt{4 a x} d x = \frac{104 a^{2}}{3}$

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