Question bank/JEE Main/Mathematics

If the line y = mx, bisects the area of the region
$\left\{ (x,y):0 \leq x \leq \frac{3}{2},0 \leq y \leq 1 + 4x - \ \ \ \ \ \ \ \ \ \ \ \ \ \ x^{2} \right\}$, then m equals.
nks cjkcj Hkkxksa esa foHkkftr djrh gS] rks m

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

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[IMAGE] As per the question given,
Area of OAB = Area of region OBCD = Area of OACD - Area of OAB
or 2(Area of region OAB) = (Area of OACD)
$\therefore$
$2\left\lbrack \frac{1}{2} \times \frac{3}{2} \times \frac{3}{2}m \right\rbrack = \int_{0}^{3/2}\mspace{2mu}\left( 1 + 4x - x^{2} \right)dx$
$\therefore$
$\frac{9}{4}m = \frac{3}{2} + 2\left( \frac{3}{2} \right)^{2} - \frac{1}{3}\left( \frac{3}{2} \right)^{3}$
$= \frac{3}{2} + \frac{9}{2} - \frac{9}{8} = 6 - \frac{9}{8}\$ or
 $\frac{9m}{4} = \frac{39}{8}$
$\Rightarrow$ $m = \frac{39}{18} = \frac{13}{6}$
$\therefore$ $m = \frac{13}{6}$

Source: Grass-G / M-21 Mathematics Paper 19 Dec Hindi Eng.docx