question_answer

The area enclosed by the curves \[y={{x}^{2}},y={{x}^{3}},x=0\]and \[x=0\]where \[p>1,\]is 1/6. The p equals     JEE Main Online Paper (Held On 12 May 2012)

A) 8/3                        

B)                        16/3

C)                        2                                             

D)                        4/3

Correct Answer: D

Solution :

                Given curves are \[y={{x}^{2}}\]and\[y={{x}^{3}}\] Also, x = 0 and x = p,p > 1 \[=\int\limits_{0}^{1}{\left( {{x}^{2}}-{{x}^{3}} \right)dx+\int\limits_{1}^{p}{\left( {{x}^{3}}-{{x}^{2}} \right)dx}}\] \[\frac{1}{6}=\frac{{{x}^{3}}}{3}-\frac{{{x}^{4}}}{4}\left| _{0}^{1}+\frac{{{x}^{4}}}{4}-\frac{{{x}^{3}}}{3} \right|_{1}^{p}\] \[\Rightarrow \]\[\frac{1}{6}=\left( \frac{1}{3}-\frac{1}{4} \right)+\left( \frac{{{p}^{4}}}{4}-\frac{{{p}^{3}}}{3}-\frac{1}{4}+\frac{1}{3} \right)\] \[\Rightarrow \]\[\frac{1}{6}-\frac{1}{3}+\frac{1}{4}+\frac{1}{4}-\frac{1}{3}=\frac{3{{p}^{4}}-4{{p}^{3}}}{12}\] \[\Rightarrow \]\[\frac{{{p}^{4}}\left( 3p-4 \right)}{12}=0\Rightarrow {{p}^{3}}(3p-4)=0\]\[\Rightarrow \]or\[\frac{4}{3}\]Since, it is given that p > 1 \[\therefore p\]can not be zero. Hence,\[p=\frac{4}{3}\]