Moderate -0.5 This is a straightforward binomial expansion requiring identification of the term containing x from (2/x - 3x)^5. Students must recognize that the general term is C(5,r)(2/x)^(5-r)(-3x)^r and solve for r when the power of x equals 1, then calculate the coefficient. While it requires careful algebraic manipulation of powers, it's a standard textbook exercise with no conceptual difficulty beyond routine application of the binomial theorem.
For \(\left(\frac{2}{x}-3x\right)^5\) term in \(x\) is \(10\) or \(5C_3\) or \(5C2 \times \left(\frac{2}{x}\right)^2 \times (-3x)^3\) or \(\left(\frac{2}{x}\right)^5\frac{5.4.3}{3!}\left(-\frac{3}{2}x^2\right)^3\) or \((-3x)^5\frac{5.4}{2!}\left(\frac{2}{3x^2}\right)^2\)
B2,1
3 elements required. \(-1\) for each error with or without \(x\)'s. Can be seen in an expansion.
\(-1080\) identified
B1
Allow \(-1080x\). Allow if expansion stops at this term. Allow from expanding brackets.
Total: 3
**Question 1:**
| Answer | Marks | Guidance |
|--------|-------|----------|
| For $\left(\frac{2}{x}-3x\right)^5$ term in $x$ is $10$ or $5C_3$ or $5C2 \times \left(\frac{2}{x}\right)^2 \times (-3x)^3$ or $\left(\frac{2}{x}\right)^5\frac{5.4.3}{3!}\left(-\frac{3}{2}x^2\right)^3$ or $(-3x)^5\frac{5.4}{2!}\left(\frac{2}{3x^2}\right)^2$ | **B2,1** | 3 elements required. $-1$ for each error with or without $x$'s. Can be seen in an expansion. |
| $-1080$ identified | **B1** | Allow $-1080x$. Allow if expansion stops at this term. Allow from expanding brackets. |
| **Total: 3** | | |
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