Moderate -0.8 This is a straightforward application of the binomial theorem requiring only substitution into the formula C(6,4) × 5² × (2x)⁴ and simplification. It's a single-step calculation with no problem-solving required, making it easier than average but not trivial since students must correctly identify which term contains x⁴ and handle the arithmetic carefully.
Condone inclusion of \(x^4\) e.g. \((2x)^4\); condone omission of brackets in \(2x^4\) if 16 used
M3 for \(15 \times 5^2 \times 2^4\)
M3
Allow M3 for correct term seen (often all terms written down) but then wrong term evaluated or all evaluated and correct term not identified; \(15\times5^2\times(2x)^4\) earns M3 even if followed by \(15\times25\times2\) calculated
or M2 for two of these elements correct with multiplication or all three elements correct but without multiplication (e.g. in list or with addition signs)
M2
or M1 for 15 soi or for 1 6 15 … seen in Pascal's triangle
M1
No MR for wrong power evaluated but SC for fourth term evaluated
SC2 for \(20000\ [x^3]\)
SC2
## Question 6:
| Answer | Marks | Guidance |
|--------|-------|----------|
| 6000 | 4 marks total | Condone inclusion of $x^4$ e.g. $(2x)^4$; condone omission of brackets in $2x^4$ if 16 used |
| M3 for $15 \times 5^2 \times 2^4$ | M3 | Allow M3 for correct term seen (often all terms written down) but then wrong term evaluated or all evaluated and correct term not identified; $15\times5^2\times(2x)^4$ earns M3 even if followed by $15\times25\times2$ calculated |
| or M2 for two of these elements correct with multiplication or all three elements correct but without multiplication (e.g. in list or with addition signs) | M2 | |
| or M1 for 15 soi or for 1 6 15 … seen in Pascal's triangle | M1 | No MR for wrong power evaluated but SC for fourth term evaluated |
| SC2 for $20000\ [x^3]$ | SC2 | |
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