Moderate -0.3 This is a straightforward application of the binomial expansion with n = -1/3 and bx = 6x. It requires substituting into the standard formula and simplifying coefficients through arithmetic, but involves no problem-solving or conceptual challenges beyond direct recall of the generalized binomial theorem.
2 Expand \(\frac { 1 } { \sqrt [ 3 ] { } ( 1 + 6 x ) }\) in ascending powers of \(x\), up to and including the term in \(x ^ { 3 }\), simplifying the coefficients.
State a correct unsimplified version of the \(x\) or \(x^2\) or \(x^3\) term in the expansion of \((1+6x)^{-\frac{1}{3}}\)
M1
EITHER method
State correct first two terms \(1 - 2x\)
A1
Obtain term \(8x^2\)
A1
Obtain term \(-\frac{112}{3}x^3\) \(\left(37\frac{1}{3}x^3\right)\) in final answer
A1
Differentiate expression and evaluate \(f(0)\) and \(f'(0)\), where \(f'(x) = k(1+6x)^{-\frac{4}{3}}\)
M1
OR method
Obtain correct first two terms \(1 - 2x\)
A1
Obtain term \(8x^2\)
A1
Obtain term \(-\frac{112}{3}x^3\) in final answer
A1
Total: 4
## Question 2:
| Answer | Mark | Guidance |
|--------|------|----------|
| State a correct unsimplified version of the $x$ or $x^2$ or $x^3$ term in the expansion of $(1+6x)^{-\frac{1}{3}}$ | M1 | EITHER method |
| State correct first two terms $1 - 2x$ | A1 | |
| Obtain term $8x^2$ | A1 | |
| Obtain term $-\frac{112}{3}x^3$ $\left(37\frac{1}{3}x^3\right)$ in final answer | A1 | |
| Differentiate expression and evaluate $f(0)$ and $f'(0)$, where $f'(x) = k(1+6x)^{-\frac{4}{3}}$ | M1 | OR method |
| Obtain correct first two terms $1 - 2x$ | A1 | |
| Obtain term $8x^2$ | A1 | |
| Obtain term $-\frac{112}{3}x^3$ in final answer | A1 | |
**Total: 4**
---
2 Expand $\frac { 1 } { \sqrt [ 3 ] { } ( 1 + 6 x ) }$ in ascending powers of $x$, up to and including the term in $x ^ { 3 }$, simplifying the coefficients.\\
\hfill \mbox{\textit{CAIE P3 2017 Q2 [4]}}