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 respectively the areas of these parts numbered from top to bottom; then S_1: S_2

Correct answer: D. 1: 1: 1 y^2=4 x and x^2=4 y are symmetric about line y=x area bounded between y^2=4 x and y=x is ∫_0^4(2 x -x) d x=(8)/(3) A_ s_2 =(16)/(3) and A_ s_1 =A_ s_3 =(16)/(3) A_ s_1 : A_ s_2 : A_ s_3 :: 1: 1: 1 .

JEE Main 2005

Exam: JEE Main
Year: 2005
Subject: Mathematics
Chapter: Area Under Curves

JEE Main 2005 · Mathematics · Area Under Curves

The parabolas y^2=4 x and x^2=4 y divide the square region bounded...

The parabolas

\mathrm{y}^2=4 \mathrm{x}

and

\mathrm{x}^2=4 \mathrm{y}

divide the square region bounded by the lines

\mathrm{x}=

4, y=4

and the coordinate axes. If

S_1, S_2, S_3

are respectively the areas of these parts numbered from top to bottom; then

\mathrm{S}_1: \mathrm{S}_2: \mathrm{S}_3

Options

A $1: 2: 1$
B $1: 2: 3$
C $2: 1: 2$
D $1: 1: 1$

Correct answer

D. $1: 1: 1$

Step-by-step solution

\mathrm{y}^2=4 \mathrm{x}

and

\mathrm{x}^2=4 \mathrm{y}

are symmetric about line

\mathrm{y}=\mathrm{x}

\Rightarrow

area bounded between

y^2=4 x

and

y=x

\int_0^4(2 \sqrt{x}-x) d x=\frac{8}{3}

\Rightarrow \mathrm{A}_{\mathrm{s}_2}=\frac{16}{3}

and

\mathrm{A}_{\mathrm{s}_1}=\mathrm{A}_{\mathrm{s}_3}=\frac{16}{3}

\Rightarrow A_{s_1}: A_{s_2}: A_{s_3}:: 1: 1: 1

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