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Q.

The parabolas $y^{2} = 4 x and x^{2} = 4 y$ divide the square region bounded by the lines .x = 4, y = 4 and the coordinate axes. If $S_{1}, S_{2}, S_{3}$ are the areas of these parts numbered from top to bottom, respectively, then

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a

$S_{2} : S_{3} \equiv 1 : 2$

b

$S_{1} : S_{2} \equiv 1 : 1$

c

$S_{1} : S_{3} \equiv 1 : 1$

d

$S_{1} : (S_{1} + S_{2}) = 1 : 2$

answer is A, C, D.

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Detailed Solution

$y^{2} = 4 x and x^{2} = 4 y$ O(0, 0) and A (4,4).

Question Image

$\begin{matrix} Now S_{3} = \int_{0}^{4} \frac{x^{2}}{4} dx = \frac{1}{4} {[\frac{x^{3}}{3}]}_{0}^{4} = \frac{1}{12} [64 - 0] = \frac{16}{3} . \\ S_{2} = \int_{0}^{4} 2 \sqrt{x} dx - S_{3} = 2 {[\frac{x^{3 / 2}}{3 / 2}]}_{0}^{4} - \frac{16}{3} \\ = \frac{4}{3} [8 - 0] - \frac{16}{3} = \frac{16}{3} . \end{matrix}$

And $S_{1} = 4 \times 4 - (S_{2} + S_{3}) = 16 - (\frac{16}{3} + \frac{16}{3}) = \frac{16}{3}$

Hence, $S_{1} : S_{2} : S_{3} = 1 : 1 : 1$

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