Standard +0.3 This is a straightforward integration problem requiring evaluation of a definite integral with standard functions (polynomial and power of x), followed by algebraic manipulation to reach the given cubic equation. The integration is routine C2 material, and while the algebra involves expanding and collecting terms, it follows a clear path with no conceptual obstacles or novel problem-solving required.
Substitutes \(k\) and \(3\), subtracts correct way around, sets equal to \(10k\). Condone lack of bracketing
\(\times k \Rightarrow k^3 - 6 - 7k = 10k^2\)
dM1
Multiplies by \(k\) to achieve a cubic in \(k\). Dependent on both M's and achieving *two* correct powers
\(k^3 - 10k^2 - 7k - 6 = 0\)
A1*
AG. Achieves result with no errors or omissions. Bracketing error withholds this mark
## Question 8:
| Working/Answer | Mark | Guidance |
|---|---|---|
| $\int\!\left(2x + \frac{6}{x^2}\right)dx = \int(2x + 6x^{-2})\,dx = x^2 - \frac{6}{x}$ | M1A1 | M1: Attempts to integrate, obtains at least one correct power. A1: Result which may be left unsimplified |
| $\left(k^2 - \frac{6}{k}\right) - \left(3^2 - \frac{6}{3}\right) = 10k$ | M1 | Substitutes $k$ and $3$, subtracts correct way around, sets equal to $10k$. Condone lack of bracketing |
| $\times k \Rightarrow k^3 - 6 - 7k = 10k^2$ | dM1 | Multiplies by $k$ to achieve a cubic in $k$. Dependent on both M's and achieving *two* correct powers |
| $k^3 - 10k^2 - 7k - 6 = 0$ | A1* | AG. Achieves result with no errors or omissions. Bracketing error withholds this mark |
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