Question

If the line, $y=mx$, bisects the area of the region
$\{(x,y):0\le x\le \frac{3}{2},0\le y\le 1+4x-x^2\}$, then $m$ equals:

• A. $\frac{9}{8}$
• B. $\frac{13}{3}$
• C. $\frac{13}{6}$
• D. $\frac{39}{16}$

Answer 1

The correct option is C $\frac{13}{6}$

We have $y=-(x^2-4x-1)$

As per the question given,

Area of $OAB=$ Area of region $OBCD=$ Area of $OACD$ $-$ Area of $OAB$

$\Rightarrow 2\left(\text{Area of region}OAB\right)=\left(\text{Area of}OACD\right)$

Therefore, $2[\frac{1}{2}\times \frac{3}{2}\times \frac{3}{2}m]={\int }_0^{3/2}(1+4x-x^2)dx$

$\Rightarrow \frac{9}{4}m=\frac{3}{2}+2{\left(\frac{3}{2}\right)}^2-\frac{1}{3}{\left(\frac{3}{2}\right)}^3$

$=\frac{3}{2}+\frac{9}{2}-\frac{9}{8}$

$=6-\frac{9}{8}$

Thus $\frac{9m}{4}=\frac{39}{8}$

$\Rightarrow m=\frac{39}{18}=\frac{13}{6}$

$\Rightarrow m=\frac{13}{6}$