If the roots of the quadratic equation x^2 + px + q = 0 are tan 30° and tan 15°, respectively then the value of 2 + q - p is-  

(A) 3  

(B) 0  

(C) 1  

(D) 2

$$\begin{gathered} {\mathrm{x}}^2+\mathrm{px}+\mathrm{q}=0 \\ \mathrm{given} ,\mathrm{\alpha } =\tan {30}^{\mathrm{o}}=\frac{1}{\sqrt{3}} \& \mathrm{\beta } =\tan {15}^{\mathrm{o}} \\ =\tan ({45}^{\mathrm{o}}-{30}^{\mathrm{o}}) \\ =2-\sqrt{3} \\ \mathrm{so}, \mathrm{\alpha } +\mathrm{\beta } =\frac{-\mathrm{p}}{1} \\ \frac{1}{\sqrt{3}}+2-\sqrt{3}=-\mathrm{p}\Rightarrow \mathrm{p}=\sqrt{3}-\frac{1}{\sqrt{3}}-2 \\ \therefore \mathrm{p}=\frac{3-1-2\sqrt{3}}{\sqrt{3}}=\frac{2-2\sqrt{3}}{\sqrt{3}}=\frac{2}{\sqrt{3}}-2 \\ \mathrm{\alpha \beta }=\frac{\mathrm{c}}{\mathrm{a}}=\frac{\mathrm{q}}{1} \\ \therefore \mathrm{x}+\mathrm{q}-\mathrm{p} =2+\frac{2}{\sqrt{3}}-1-\left(\frac{2}{\sqrt{3}}-2\right) \\ =2+\cancel{\frac{2}{\sqrt{3}}}-1-\cancel{\frac{2}{\sqrt{3}}}+2 \\ \mathbf{=}\mathbf{ }\mathbf{4}\mathbf{-}\mathbf{1} \\ \mathbf{=}\mathbf{3}\mathbf{ }\mathbf{ans} . \\ \mathbf{opition}\mathbf{ }\mathbf{\left(}\mathbf{a}\mathbf{\right)}\mathbf{ }\mathbf{is}\mathbf{ }\mathbf{correct}\mathbf{ } \end{gathered}$$