Question

The parabolas $y^2=4x$ and $x^2=4y$ divide the square region bounded the lines $x=4,y=4$ and the coordinate axes. If $S_1,S_2,S_3$ respectively the areas of these parts numbered from top to bottom, then

• A. $S_1,S_2,S_3$ are in A.P
• B. $S_1,S_2,S_3$ are in G.P
• C. $S_1^3+S_3^3>2S_2^3$
• D. $S_1=S_3$

Answer 1

Answer: (A), (B), (D)

The correct options are
A $S_1,S_2,S_3$ are in A.P
B $S_1,S_2,S_3$ are in G.P
D $S_1=S_3$

$S_1=S_3$ from the figure below.
$S_2={\int }_0^4(2\sqrt{x}-\frac{x^2}{4})dx$
$={[\frac{4}{3}{x}^{\frac{3}{2}}-\frac{x^3}{12}]}_0^4$
$=\frac{32}{3}-\frac{16}{3}=\frac{16}{3}$.sq.unit.
$S_1=S_3=\frac{1}{2}(16-\frac{16}{3})$
$=8-\frac{8}{3}=\frac{16}{3}=S_2$
$\because S_1=S_2=S_3$
$\therefore S_1,S_2,S_3$ are in A.P