question_answer

The parabolas\[{{y}^{2}}=4x\]and\[{{x}^{2}}=4y\]divide the square region bounded by the lines \[x=4,\text{ }y=4\]and the coordinate axes. If \[{{S}_{1}},{{S}_{2}},{{S}_{3}}\]are respectively the areas of these parts numbered from top to bottom, then\[{{S}_{1}}:{{S}_{2}}:{{S}_{3}}\]     AIEEE  Solved  Paper-2005

A) 1 : 1 : 1             

B)        2 : 1 : 2 

C)        1 : 2 : 3            

D)        1 : 2 : 1

Correct Answer: A

Solution :

Now, \[{{S}_{1}}={{S}_{3}}=\int_{0}^{4}{\frac{{{x}^{2}}}{4}}dx=\frac{1}{4}\left[ \frac{{{x}^{3}}}{3} \right]_{0}^{4}\] \[=\frac{1}{12}\times 64=\frac{16}{3}\]sq units \[\therefore \]\[{{S}_{2}}+{{S}_{3}}=\int_{0}^{4}{\sqrt{4x}}dx\] \[=2\,\,\left[ \frac{{{x}^{3/2}}}{3/2} \right]_{0}^{4}\,=\frac{4}{3}\times 8=\,\frac{32}{3}\,sq\] units \[\Rightarrow \]\[{{S}_{2}}=\frac{32}{3}-\frac{16}{3}=\frac{16}{3}\]sq units \[\therefore \]\[{{S}_{1}}:{{S}_{2}}:{{S}_{3}}=\frac{16}{3}:\frac{6}{3}:\frac{16}{3}=1:1:1\]