NIMCET 2022 — Mathematics PYQ

For a quadratic equation of the form $x^2+px+q=0$, if the roots are $\alpha$ and $\beta$, then:

• Sum of roots: $\alpha +\beta =-p$

• Product of roots: $\alpha \beta =q$

Here, the roots are $\alpha =\tan 30^{\circ }=\frac{1}{\sqrt{3}}$ and $\beta =\tan 15^{\circ }=2-\sqrt{3}$.

1. Finding the values of $p$ and $q$

Sum of roots ($-p$):

$$-p=\tan 30^{\circ }+\tan 15^{\circ }$$

$$-p=\frac{1}{\sqrt{3}}+(2-\sqrt{3})$$

$$-p=\frac{1+2\sqrt{3}-3}{\sqrt{3}}=\frac{2\sqrt{3}-2}{\sqrt{3}}$$

$$p=\frac{2-2\sqrt{3}}{\sqrt{3}}$$

Product of roots ($q$):

$$q=\tan 30^{\circ }\cdot \tan 15^{\circ }$$

$$q=\frac{1}{\sqrt{3}}\cdot (2-\sqrt{3})$$

$$q=\frac{2}{\sqrt{3}}-1$$

2. Calculating $2+q-p$

Substitute the expressions for $p$ and $q$ into the required expression:

$$2+q-p=2+\left(\frac{2}{\sqrt{3}}-1\right)-\left(\frac{2-2\sqrt{3}}{\sqrt{3}}\right)$$

$$=1+\frac{2}{\sqrt{3}}-\frac{2}{\sqrt{3}}+\frac{2\sqrt{3}}{\sqrt{3}}$$

$$=1+2$$

$$=3$$

Correct Option: A) 3