VITEEE 2013 Paper

The equation of curve is y = x - $x^2$ ⇒ $x^2$ - x = y ⇒ $\left(x-\frac{1}{2}\right)^2$ = - $\left(y-\frac{1}{4}\right)$ which is a parabola whose vertex is $\left(\frac{1}{2},\frac{1}{4}\right)$ Hence, finding the point of intersection of the curve and the line, x - $x^2$ = mx ⇒ x (1 - x - m) = 0 i.e, x = 0 or x = 1 - m ∴ $\frac{9}{2}$ = $\int\limits_{0}^{1-m}$ (x - $x^2$ - mx) dx = $\left\{\frac{x^2}{2} - \frac{x^3}{3} - \frac{m x^2}{2}\right\}_{0}^{1-m}$ (1 - m) $\frac{(1-m)^2}{2}$ - $\frac{(1-m)^3}{3}$ = $\frac{(1-m)^3}{6}$ ∴ $(1-m)^3$ = $\frac{6 \times 9}{2}$ = 27 ⇒ 1 - m = $(27)^{\frac{1}{3}}$ = 3 ⇒ m = - 2 Also $(1-m^{3})$ - $(3)^{3}$ = 0 ∴ $(1-m^{3})$ = $3^{3}$ ⇒ 1 - m = 3 or m = - 2