Moderate -0.3 This is a straightforward rationalizing denominators question requiring students to multiply by conjugates and simplify. While it involves multiple terms and some algebraic manipulation, it's a standard technique with no conceptual difficulty—slightly easier than average due to being purely procedural with clear steps.
For dealing appropriately with fractions; clearing fractions or making \(k\) the subject with RHS as single fraction: \(k=\frac{(\sqrt{5}+\sqrt{7}+2\sqrt{6})(\sqrt{5}+\sqrt{7})}{(\sqrt{5}+\sqrt{6})(\sqrt{6}+\sqrt{7})}\)
Expanding brackets and collecting like surds: \(k=\frac{12+2\sqrt{5}\sqrt{7}+2\sqrt{5}\sqrt{6}+2\sqrt{6}\sqrt{7}}{6+\sqrt{5}\sqrt{7}+\sqrt{5}\sqrt{6}+\sqrt{6}\sqrt{7}}\)
\(2(\sqrt{5}\sqrt{6}+\sqrt{5}\sqrt{7}+\sqrt{6}\sqrt{7}+6)=k(\sqrt{5}\sqrt{6}+\sqrt{5}\sqrt{7}+\sqrt{6}\sqrt{7}+6)\), so \(k=2\)
A1
Finding \(k\) convincingly after M1
Total: [3]
## Question 3:
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\frac{\sqrt{6}-\sqrt{5}}{6-5}+\frac{\sqrt{7}-\sqrt{6}}{7-6}$ oe | M1 | Rationalising denominators. Minimum working needed for M1. Accept 1 for "$6-5$" and $-1$ for "$5-6$" etc |
| $\sqrt{7}-\sqrt{5}$ or $\frac{\sqrt{5}-\sqrt{7}}{-1}$ | A1 | |
| $\frac{k}{\sqrt{5}+\sqrt{7}}=\frac{k(\sqrt{7}-\sqrt{5})}{7-5}=\sqrt{7}-\sqrt{5}$ | | AO 2.2a |
| so $k=2$ | A1 | Finding $k$ convincingly after M1 |
**Alternative:**
| Answer | Marks | Guidance |
|--------|-------|----------|
| $(\sqrt{6}+\sqrt{7})(\sqrt{5}+\sqrt{7})+(\sqrt{5}+\sqrt{6})(\sqrt{5}+\sqrt{7})=k(\sqrt{5}+\sqrt{6})(\sqrt{6}+\sqrt{7})$ | M1 | For dealing appropriately with fractions; clearing fractions or making $k$ the subject with RHS as single fraction: $k=\frac{(\sqrt{5}+\sqrt{7}+2\sqrt{6})(\sqrt{5}+\sqrt{7})}{(\sqrt{5}+\sqrt{6})(\sqrt{6}+\sqrt{7})}$ |
| $(2\sqrt{5}\sqrt{6}+2\sqrt{5}\sqrt{7}+2\sqrt{6}\sqrt{7}+12)=k(\sqrt{5}\sqrt{6}+\sqrt{5}\sqrt{7}+\sqrt{6}\sqrt{7}+6)$ | A1 | Expanding brackets and collecting like surds: $k=\frac{12+2\sqrt{5}\sqrt{7}+2\sqrt{5}\sqrt{6}+2\sqrt{6}\sqrt{7}}{6+\sqrt{5}\sqrt{7}+\sqrt{5}\sqrt{6}+\sqrt{6}\sqrt{7}}$ |
| $2(\sqrt{5}\sqrt{6}+\sqrt{5}\sqrt{7}+\sqrt{6}\sqrt{7}+6)=k(\sqrt{5}\sqrt{6}+\sqrt{5}\sqrt{7}+\sqrt{6}\sqrt{7}+6)$, so $k=2$ | A1 | Finding $k$ convincingly after M1 |
**Total: [3]**
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