Question

If $\tan \frac{\pi }{12}$ is one of the roots of the equation $2x^2-bx+c=0,(b,c\in Q)$ then $(b+c)$ equals

• A. 4
• B. 6
• C. 10
• D. 20

Answer 1

Answer: (C)

$\tan \frac{\pi }{12}=2-\sqrt{3}$, Conjugate $=2+\sqrt{3}$
$\Rightarrow$ Sum of roots $=\frac{b}{2}=2-\sqrt{3}+2+\sqrt{3}=4$
$\Rightarrow$ Product of roots $=\frac{c}{2}=(2-\sqrt{3})(2+\sqrt{3})=1$
$\Rightarrow b=8,c=2$