Let $\alpha, \beta$ be the roots of the equation $x^2-\mathrm{p} x+\mathrm{r}=0$ and $\frac{\alpha}{2}, 2 \beta$ be the roots of the equation $x^2-q x+r=0$. Then the value of r is

Let $\quad \overline{\mathrm{a}}=\alpha \hat{\mathrm{i}}+3 \hat{\mathrm{j}}-\hat{\mathrm{k}}, \quad \overline{\mathrm{b}}=3 \hat{\mathrm{i}}-\beta \hat{\mathrm{j}}+4 \hat{\mathrm{k}} \quad$ and $\overline{\mathrm{c}}=\hat{\mathrm{i}}+2 \hat{\mathrm{j}}-2 \hat{\mathrm{k}}$, where $\alpha, \beta \in \mathbb{R}$, be three vectors. If the projection at $\overline{\mathrm{a}}$ on $\overline{\mathrm{c}}$ is $\frac{10}{3}$ and $\overline{\mathrm{b}} \times \overline{\mathrm{c}}=-6 \hat{\mathrm{i}}+10 \hat{\mathrm{j}}+7 \hat{\mathrm{k}}$, then the value of $\alpha^2+\beta^2-\alpha \beta$ is equal to

The function $\mathrm{f}(x)=2 x^3-6 x+5$ is an increasing function, if

Let $2 \sin ^2 x+3 \sin x-2>0$ and $x^2-x-2<0$ ($x$ is measured in radians). Then $x$ lies in the interval