WAEC Further Mathematics 2021 Paper

Concept:Rationalise each denominator first, then simplify the sum and take its reciprocal.Explanation:Rationalise $\frac{1}{2-\sqrt{3}}$ by multiplying numerator and denominator by $2+\sqrt{3}$:$\frac{1}{2-\sqrt{3}} \times \frac{2+\sqrt{3}}{2+\sqrt{3}} = \frac{2+\sqrt{3}}{(2)^2-(\sqrt{3})^2} = 2+\sqrt{3}$.Rationalise $\frac{2}{2+\sqrt{3}}$ by multiplying numerator and denominator by $2-\sqrt{3}$:$\frac{2}{2+\sqrt{3}} \times \frac{2-\sqrt{3}}{2-\sqrt{3}} = \frac{2(2-\sqrt{3})}{(2)^2-(\sqrt{3})^2} = 4-2\sqrt{3}$.Add the two simplified terms:$(2+\sqrt{3}) + (4-2\sqrt{3}) = 6-\sqrt{3}$.The given expression has power $-1$, so take the reciprocal:$\left(6-\sqrt{3}\right)^{-1} = \frac{1}{6-\sqrt{3}}$.Rationalise this final denominator:$\frac{1}{6-\sqrt{3}} \times \frac{6+\sqrt{3}}{6+\sqrt{3}} = \frac{6+\sqrt{3}}{(6)^2-(\sqrt{3})^2} = \frac{6+\sqrt{3}}{33}$.Answer:$\frac{1}{33}(6+\sqrt{3})$ (Option C)