Standard +0.3 This is a straightforward application of the binomial expansion for fractional powers combined with a simple product. Students need to expand (1+3x)^(1/3) to the x³ term using the standard formula, then multiply by (3-x) and collect coefficients. While it requires careful algebraic manipulation and knowledge of the generalised binomial theorem, it's a routine textbook exercise with no novel insight required, making it slightly easier than average.
State unsimplified term in \(x^2\), or its coefficient: \((1+3x)^{\frac{1}{3}}\left(\dfrac{\frac{1}{3}\times\frac{-2}{3}}{2}(3x)^2\right)\)
B1
Symbolic binomial coefficients are not sufficient for the B marks
State unsimplified term in \(x^3\), or its coefficient: \((1+3x)^{\frac{1}{3}}\left(\dfrac{\frac{1}{3}\times\frac{-2}{3}\times\frac{-5}{3}}{6}(3x)^3\right)\)
B1
Multiply by \((3-x)\) to give 2 terms in \(x^3\), or their coefficients
M1
\(\left(3\times\frac{10}{6}+1\right)\) Ignore errors in terms other than \(x^3\); \(3\times x^3\text{coeff} - x^2\text{coeff}\) and no other term in \(x^3\)
Obtain answer 6
A1
Not \(6x^3\)
4
## Question 1:
| Answer | Mark | Guidance |
|--------|------|----------|
| State unsimplified term in $x^2$, or its coefficient: $(1+3x)^{\frac{1}{3}}\left(\dfrac{\frac{1}{3}\times\frac{-2}{3}}{2}(3x)^2\right)$ | B1 | Symbolic binomial coefficients are not sufficient for the B marks |
| State unsimplified term in $x^3$, or its coefficient: $(1+3x)^{\frac{1}{3}}\left(\dfrac{\frac{1}{3}\times\frac{-2}{3}\times\frac{-5}{3}}{6}(3x)^3\right)$ | B1 | |
| Multiply by $(3-x)$ to give 2 terms in $x^3$, or their coefficients | M1 | $\left(3\times\frac{10}{6}+1\right)$ Ignore errors in terms other than $x^3$; $3\times x^3\text{coeff} - x^2\text{coeff}$ and no other term in $x^3$ |
| Obtain answer 6 | A1 | Not $6x^3$ |
| | **4** | |
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