Question

The line $y=mx$ bisects the area enclosed by the curve $y=1+4x-x^2\&$ the lines $x=\frac{3}{2},x=0\&y=0$. Then the value of $m$ is:

• A. $\frac{13}{6}$
• B. $\frac{6}{13}$
• C. $\frac{3}{2}$
• D. 4

Answer 1

Answer: (A)

$y=1+4x-x^2\Rightarrow (x-2{)}^2=-(y-5)$
vertex $(2,5)$
$A=\underset{0}{\overset{3/2}{\int }}\left(1+4x-x^2\right)dx={\left(x+\frac{4x^2}{2}-\frac{x^3}{3}\right)}_0^{3/2}=\frac{39}{8}.....$(1)
$△OAB=\frac{1}{2}\times \frac{3}{2}\times \frac{3}{2}\times m=\frac{9}{8}m.....$(2)

From (1) & (2) $\frac{9}{8}m=\frac{1}{2}\left(\frac{39}{8}\right)$
$\Rightarrow m=\frac{39}{2\times 9}=\frac{13}{6}$