Mathematics - Quadratic Equation Question with Solution | TestHub

Let $\mathrm{f}(\mathrm{x})={\mathrm{ax}}^2+(\mathrm{a}-2)\mathrm{x}-2$

$\mathrm{f}(\mathrm{x})$ is negative for two integral values of x , so graph should be vertically upward parabola i.e., $\mathrm{a}>0$

Let two roots of $\mathrm{f}(\mathrm{x})=0$ are $\alpha$ and $\beta$ then $\alpha ,\beta =\frac{-(\mathrm{a}-2)\pm (\mathrm{a}+2)}{2\mathrm{a}}$ $\Rightarrow \alpha =-1,\beta =\frac{2}{\mathrm{a}}\Rightarrow 1<\beta \le 2\Rightarrow 1<\frac{2}{\mathrm{a}}\le 2\Rightarrow \mathrm{a}\in [1,2]$