Moderate -0.8 This is a straightforward application of the binomial theorem requiring identification of the correct term (r=5) and calculation using the formula. It's simpler than average as it involves direct substitution into a standard formula with no algebraic manipulation or problem-solving required, though the arithmetic with negative coefficients requires care.
e.g. sight of 21 implies this mark. May be seen in a full binomial expansion but must extract this coefficient.
\(3^2\) or \((\pm 2)^5\) seen
B1
Could be implied by their working - may need to check
\(^7C_5 \times 3^2 \times (-2)^5\)
A1
Correct coefficient unsimplified - may be embedded in term involving \(x^5\); condone omission of negative sign. Must deal with the \((-2x)^5\) correctly for this mark so \((-2)^5\) must be seen or implied e.g. \(\pm 32\) or \(\pm 32x^5\) soi.
\(-6048\)
A1
\(-6048x^5\) is A0 but accept if they underline/circle etc the \(-6048\). NOTE: May see full binomial expansion here - look for the term in \(x^5\) and mark as above.
## Question 7:
| Answer | Mark | Guidance |
|--------|------|----------|
| $^7C_5$ or $^7C_2$ soi | M1 | e.g. sight of 21 implies this mark. May be seen in a full binomial expansion but must extract this coefficient. |
| $3^2$ or $(\pm 2)^5$ seen | B1 | Could be implied by their working - may need to check |
| $^7C_5 \times 3^2 \times (-2)^5$ | A1 | Correct coefficient unsimplified - may be embedded in term involving $x^5$; condone omission of negative sign. Must deal with the $(-2x)^5$ correctly for this mark so $(-2)^5$ must be seen or implied e.g. $\pm 32$ or $\pm 32x^5$ soi. |
| $-6048$ | A1 | $-6048x^5$ is A0 but accept if they underline/circle etc the $-6048$. NOTE: May see full binomial expansion here - look for the term in $x^5$ and mark as above. |
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