A2 June 2025 Paper 1 Q13
13 The gradient of a curve \(y = \mathrm{f}(x)\) satisfies the differential equation \(x\dfrac{\mathrm{d}y}{\mathrm{d}x} - 2y = 2 + x^2\).
By solving this differential equation, determine the exact \(x\)-coordinate of the stationary point on the curve \(y = \mathrm{f}(x)\). [8]
| Scheme | Marks | AO |
|---|---|---|
| \(\dfrac{\mathrm{d}y}{\mathrm{d}x} - \dfrac{2}{x}y = \dfrac{2 + x^2}{x}\) | M1 | 3.1a |
| \(\mathrm{IF} = \mathrm{e}^{\int -\frac{2}{x}\,\mathrm{d}x}\) | M1 | 1.1 |
| \(= \mathrm{e}^{-2\ln x} = x^{-2}\) | A1 | 1.1 |
| [3] |
Notes
M1: division by \(x\) seen. Condone errors on RHS.
M1: integrating factor
A1: AG An intermediate step must be seen before the final answer. Condone errors on RHS. Requires M1M1.
| Scheme | Marks | AO |
|---|---|---|
| \(\frac{\mathrm{d}}{\mathrm{d}x}\left(yx^{-2}\right) = \frac{2 + x^2}{x^3}\) or \(yx^{-2} = \int \frac{2 + x^2}{x^3}\,\mathrm{d}x\) | M1 | 2.1 |
| \(yx^{-2} = -\frac{1}{x^2} + \ln x \; [+c]\) \(y = -1 + x^2\ln x + cx^2\) | A1 | 1.1 |
| \(0 = -1 + c \Rightarrow c = 1\) | M1 | 1.1 |
| \(y = x^2\ln x + x^2 - 1\) | A1cao | 2.1 |
| \(\frac{\mathrm{d}y}{\mathrm{d}x} = 2x\ln x + x + 2x\) | M1* | |
| \(\frac{\mathrm{d}y}{\mathrm{d}x} = 2x\ln x + 3x\) | A1 | |
| \(2x\ln x + 3x = 0 \Rightarrow x(2\ln x + 3) = 0\) | M1dep | |
| \(x = \mathrm{e}^{-\frac{3}{2}}\) only | A1 | |
| [8] |
Notes
M1: multiplying through by \(x^{-2}\) (allow FT from an incorrect rearrangement in (a) if clear) and writing as an exact derivative or setting up integral
A1: Ignore modulus signs if used
M1: substituting \(x = 1, y = 0\) in leading to a value for \(c\)
A1cao: oe
M1*: differentiating their \(y\), allow a slip
A1: correctly (do not FT), need not be simplified
M1dep: setting \(\frac{\mathrm{d}y}{\mathrm{d}x} = 0\) and meaningfully attempting to solve; soi by correct final answer
A1: www, do not condone \(\frac{\mathrm{d}y}{\mathrm{d}x}\) used for \(\frac{\mathrm{d}}{\mathrm{d}x}\). Must dismiss \(x = 0\) if previously given
Ignore \(x = -\mathrm{e}^{-\frac{3}{2}}\) if stated
Alternative method for last 4 marks
| Scheme | Marks |
|---|---|
| \(\left[\frac{\mathrm{d}y}{\mathrm{d}x} = 0 \Rightarrow\right] -2y = 2 + x^2\) | B1 |
| \(-2(x^2\ln x + x^2 - 1) = 2 + x^2\) | M1 |
| \(x^2(3 + 2\ln x) = 0 \Rightarrow x = \cdots\) | M1 |
| \(x = \mathrm{e}^{-\frac{3}{2}}\) only | A1 |
B1: could be stated anywhere or implied by next line
M1: setting up an equation in \(x\) only, cannot be implied
M1: re-arranging and attempting to solve; soi by correct final answer
A1: www, do not condone \(\frac{\mathrm{d}y}{\mathrm{d}x}\) used for \(\frac{\mathrm{d}}{\mathrm{d}x}\). Must dismiss \(x = 0\) if previously given
Ignore \(x = -\mathrm{e}^{-\frac{3}{2}}\) if stated