VITEEE 2012 Paper

Let f(x) = -3 + x - $x^2$ Then, f(x) < 0 for all x because coefficient of $x^2$ < 0 and disc < 0. Thus, LHS of the given equation is always positive whereas the RHS is always less than zero. Hence, the given equation has no solution. Alternate Solution: Given, equation is $\frac{9}{10}$ = - 3 + x - $x^2$ Let y = $\frac{9}{10}$ , therefore y = - 3 + x - $x^2$ y = $\left[ x^2 - x + \frac{1}{4} \right]$ - 3 + $\frac{1}{4}$ ⇒ y = $\frac{11}{4}$ = - $\left(x - \frac{1}{2}\right)^2$ It is clear from the graph that two curves do not intersect. Hence, no solution exists