question_answer

The line \[\operatorname{y} = mx\] bisects the area enclosed by lines \[\operatorname{x} =0\], \[\operatorname{y} = 0 and x = 3/2\] and the curve \[\operatorname{y} = 1 + 4x - {{x}^{2}}\]. 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 :

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