Question bank/JEE Main/Mathematics

The line y = mx bisects the area enclosed by the curve y = 1 +
4x - x^2 and the lines x = 0, x = $\frac{3}{2}$ and y = 0.Then the
value of m is
x = 0, x = $\frac{3}{2}$ y =
0,oa y = 0 ls ifjc) Hkkftr djrh gS] rc
m

A$\frac{13}{6}$B$\frac{6}{13}$C$\frac{3}{2}$D4

Show answer and solution▾

[IMAGE] y = 1 + 4x - x^2 $\Rightarrow$ (x -2)^2 = - (y - 5)
vertex (2, 5)
Area OACD =
$\int_{0}^{3/2}\mspace{2mu}\left( 1 + 4x - x^{2} \right)dx = \left( x + \frac{4x^{2}}{2} - \frac{x^{3}}{3} \right)_{0}^{3/2} = \frac{39}{8}$........(A)
$\Delta OAB = \frac{1}{2} \times \frac{3}{2} \times \frac{3}{2} \times m = \frac{9}{8}m$........(B)
From (A) and (B)
$\frac{9}{8}m = \frac{1}{2}\left( \frac{39}{8} \right)$ $\Rightarrow$
$m = \frac{39}{2 \times 9} = \frac{13}{6}$
Hindi
y = 1 + 4x - x^2 $\Rightarrow$ (x -2)^2 = - (y - 5)
'kh"kZ (2, 5)
OACD =Qy =
$\int_{0}^{3/2}\mspace{2mu}\left( 1 + 4x - x^{2} \right)dx = \left( x + \frac{4x^{2}}{2} - \frac{x^{3}}{3} \right)_{0}^{3/2} = \frac{39}{8}$........(A)
$\Delta OAB = \frac{1}{2} \times \frac{3}{2} \times \frac{3}{2} \times m = \frac{9}{8}m$........(B)
(A) rFkk (B)
$\frac{9}{8}m = \frac{1}{2}\left( \frac{39}{8} \right)$ $\Rightarrow$
$m = \frac{39}{2 \times 9} = \frac{13}{6}$

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