AP DSC School Assistant Math Model Test 2

Concept:The given expression is equated to a rationalized form of a surd. By rationalizing the denominator twice, we can express the right-hand side in the form $P + \sqrt{3}Q + \sqrt{5}R + \sqrt{15}S$. Then comparing coefficients gives the value of $P$.Explanation:We start with the equation:$$P + \sqrt{3}Q + \sqrt{5}R + \sqrt{15}S = \frac{1}{(1+\sqrt{3})+\sqrt{5}} \quad \text{... (1)}$$First rationalize the RHS by multiplying numerator and denominator by the conjugate of the denominator, $(1+\sqrt{3}) - \sqrt{5}$:$$\frac{1}{(1+\sqrt{3})+\sqrt{5}} \times \frac{(1+\sqrt{3})-\sqrt{5}}{(1+\sqrt{3})-\sqrt{5}} = \frac{1+\sqrt{3}-\sqrt{5}}{(1+\sqrt{3})^2 - (\sqrt{5})^2}$$Using $(a+b)(a-b) = a^2 - b^2$:$$= \frac{1+\sqrt{3}-\sqrt{5}}{(1+3+2\sqrt{3}) - 5} = \frac{1+\sqrt{3}-\sqrt{5}}{2\sqrt{3} - 1}$$Now rationalize again by multiplying numerator and denominator by the conjugate $2\sqrt{3} + 1$:$$\frac{1+\sqrt{3}-\sqrt{5}}{2\sqrt{3} - 1} \times \frac{2\sqrt{3} + 1}{2\sqrt{3} + 1} = \frac{(1+\sqrt{3}-\sqrt{5})(2\sqrt{3}+1)}{(2\sqrt{3})^2 - 1^2}$$Denominator becomes $12 - 1 = 11$. Expand the numerator:$$(1+\sqrt{3}-\sqrt{5})(2\sqrt{3}+1) = 1\cdot2\sqrt{3} + 1\cdot1 + \sqrt{3}\cdot2\sqrt{3} + \sqrt{3}\cdot1 - \sqrt{5}\cdot2\sqrt{3} - \sqrt{5}\cdot1$$Simplify term by term:$2\sqrt{3} + 1 + 2\cdot3 + \sqrt{3} - 2\sqrt{15} - \sqrt{5}$$= 2\sqrt{3} + 1 + 6 + \sqrt{3} - 2\sqrt{15} - \sqrt{5}$$= 7 + 3\sqrt{3} - \sqrt{5} - 2\sqrt{15}$Thus the RHS becomes:$$\frac{7 + 3\sqrt{3} - \sqrt{5} - 2\sqrt{15}}{11} = \frac{7}{11} + \frac{3\sqrt{3}}{11} - \frac{\sqrt{5}}{11} - \frac{2\sqrt{15}}{11}$$Comparing with the left side $P + \sqrt{3}Q + \sqrt{5}R + \sqrt{15}S$, we identify:$P = \frac{7}{11}$, $Q = \frac{3}{11}$, $R = -\frac{1}{11}$, $S = -\frac{2}{11}$.Answer:$P = \frac{7}{11}$, which corresponds to option A.