question_answer

The line y = mx bisects the area enclosed by lines \[x=0,\text{ }y=0\] and \[x=3/2\] and the curve\[y=1+4x-{{x}^{2}}\]. Then the value of m is

A) \[\frac{13}{6}\]

B) \[\frac{13}{2}\]

C) \[\frac{13}{5}\]

D) \[\frac{13}{7}\]

Correct Answer: A

Solution :

[a] \[y=1+4x-{{x}^{2}}=5-{{(x-2)}^{2}}\] We have \[\int\limits_{0}^{3/2}{(1+4x-{{x}^{2}})dx=2\int\limits_{0}^{3/2}{mx\,\,dx}}\] \[=\frac{3}{2}+2\left( \frac{9}{4} \right)-\frac{1}{3}\left( \frac{27}{8} \right)=m.\frac{9}{4}\] On solving we get \[m=\frac{13}{6}\]