October 2020 Paper 1 Q7
7. Given that \(k\) is a positive constant and \(\displaystyle \int_1^k \left(\frac{5}{2\sqrt{x}} + 3\right) \mathrm{d}x = 4\)
| Scheme | Marks | AO |
|---|---|---|
| \(x^n \to x^{n+1}\) | M1 | 1.1b |
| \(\displaystyle\int \left(\frac{5}{2\sqrt{x}} + 3\right) \mathrm{d}x = 5\sqrt{x} + 3x\) | A1 | 1.1b |
| \(\left[5\sqrt{x} + 3x\right]_1^k = 4 \Rightarrow 5\sqrt{k} + 3k - 8 = 4\) | dM1 | 1.1b |
| \(3k + 5\sqrt{k} - 12 = 0\) * | A1* | 2.1 |
| (4) |
Notes
M1: For \(x^n \to x^{n+1}\) on correct indices. This can be implied by the sight of either \(x^{\frac{1}{2}}\) or \(x\)
A1: \(5\sqrt{x} + 3x\) or \(5x^{\frac{1}{2}} + 3x\) but may be unsimplified. Also allow with + \(c\) and condone any spurious notation.
dM1: Uses both limits, subtracts, and sets equal to 4. They cannot proceed to the given answer without a line of working showing this.
A1*: Fully correct proof with no errors (bracketing or otherwise) leading to given answer.
| Scheme | Marks | AO |
|---|---|---|
| \(3k + 5\sqrt{k} - 12 = 0 \Rightarrow \left(3\sqrt{k} - 4\right)\left(\sqrt{k} + 3\right) = 0\) | M1 | 3.1a |
| \(\sqrt{k} = \dfrac{4}{3}, (-3)\) | A1 | 1.1b |
| \(\sqrt{k} = \ldots \Rightarrow k = \ldots\) oe | dM1 | 1.1b |
| \(k = \dfrac{16}{9}, \cancel{9}\) | A1 | 2.3 |
| (4) | ||
| (8 marks) |
Notes
M1: For a correct method of solving. This could be as the scheme, treating as a quadratic in \(\sqrt{k}\) and using allowable method to solve including factorisation, formula etc.
Allow values for \(\sqrt{k}\) to be just written down, e.g. allow \(\sqrt{k} = \pm\dfrac{4}{3}, (\pm 3)\)
Alternatively score for rearranging to \(5\sqrt{k} = 12 - 3k\) and then squaring to get \(\ldots k = (12 - 3k)^2\)
A1: \(\sqrt{k} = \dfrac{4}{3}, (-3)\)
Or in the alt method it is for reaching a correct 3TQ equation \(9k^2 - 97k + 144 = 0\)
dM1: For solving to find at least one value for \(k\). It is dependent upon the first M mark.
In the main method it is scored for squaring their value(s) of \(\sqrt{k}\)
In the alternative scored for solving their 3TQ by an appropriate method
A1: Full and rigorous method leading to \(k = \dfrac{16}{9}\) only. The 9 must be rejected.