Easy -1.3 This is a straightforward rationalizing the denominator question requiring only multiplication by the conjugate and simplification. It's a standard C1 exercise with a single technique and minimal steps, making it easier than average but not trivial since students must execute the algebraic manipulation correctly.
Although 'denominator = 1' may be implied, the \(13 - 7\sqrt{3}\) must obviously be the final answer (not an intermediate step), to score full marks. (Also M0 M1 A1 A1 is not an option.)
\(a = 13\)
A1
The A marks cannot be scored unless the 1st M mark has been scored, but this 1st M mark could be implied by correct expansions of both numerator and denominator.
\(b = -7\)
A1
Special case: If numerator is multiplied by \((2+\sqrt{3})\) instead of \((2-\sqrt{3})\), the 2nd M can still be scored for at least 3 of these terms correct: \(10 - 2\sqrt{3} + 5\sqrt{3} - (\sqrt{3})^2\). The maximum score in the special case is 1 mark: M0 M1 A0 A0.
$\frac{(5-\sqrt{3})}{(2+\sqrt{3})} \times \frac{(2-\sqrt{3})}{(2-\sqrt{3})}$ | M1 | Multiplying top and bottom by $(2-\sqrt{3})$. (As shown above is sufficient.)
$= \frac{10 - 2\sqrt{3} - 5\sqrt{3} + (\sqrt{3})^2}{...}$ (or $= \frac{10 - 7\sqrt{3} + 3}{...}$) | M1 | Attempt to multiply out numerator $(5-\sqrt{3})(2-\sqrt{3})$. Must have at least 3 terms correct.
$= 13 - 7\sqrt{3}$ (Allow $\frac{13-7\sqrt{3}}{1}$) | A1 | Although 'denominator = 1' may be implied, the $13 - 7\sqrt{3}$ must obviously be the final answer (not an intermediate step), to score full marks. (Also M0 M1 A1 A1 is not an option.)
$a = 13$ | A1 | The A marks cannot be scored unless the 1st M mark has been scored, but this 1st M mark could be implied by correct expansions of both numerator and denominator.
$b = -7$ | A1 |
**Special case:** If numerator is multiplied by $(2+\sqrt{3})$ instead of $(2-\sqrt{3})$, the 2nd M can still be scored for at least 3 of these terms correct: $10 - 2\sqrt{3} + 5\sqrt{3} - (\sqrt{3})^2$. The maximum score in the special case is 1 mark: M0 M1 A0 A0.
**Answer only:** Scores no marks.
**Alternative method:**
$5 - \sqrt{3} = (a + b\sqrt{3})(2 + \sqrt{3})$ | M1 | At least 3 terms correct.
$(a + b\sqrt{3})(2 + \sqrt{3}) = 2a + a\sqrt{3} + 2b\sqrt{3} + 3$ | M1 | Form and attempt to solve simultaneous equations.
$5 = 2a + 3b$
$-1 = a + 2b$
$a = 13, b = -7$ | A1, A1 |
**Total: 4 marks**
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