Question

The line $y=m x$ bisects the area enclosed by the curve $y=1+4 x-x^2 \&$ 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)

$y=1+4 x-x^2 \Rightarrow(x-2)^2=-(y-5)$
vertex $(2,5)$
$A=\int\limits_0^{3 / 2}\left(1+4 x-x^2\right) d x=\left(x+\frac{4 x^2}{2}-\frac{x^3}{3}\right)_0^{3 / 2}=\frac{39}{8} .....$(1)
$\triangle O A B=\frac{1}{2} \times \frac{3}{2} \times \frac{3}{2} \times m=\frac{9}{8} m .....$(2)

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