June 2025 Paper 1 Q12
12
The gradient of a curve is given by \(\dfrac{\mathrm{d}y}{\mathrm{d}x} = \dfrac{2 + 4x}{x(1 + x)(1 - x)\tan y}\) and the curve passes through the point \(\left(\frac{1}{2}, \frac{1}{4}\pi\right)\).
| Scheme | Marks | AO |
|---|---|---|
| \(\dfrac{2 + 4x}{x(1 + x)(1 - x)} = \dfrac{A}{x} + \dfrac{B}{1 + x} + \dfrac{C}{1 - x}\) | B1 | 3.1a |
| \(A(1 + x)(1 - x) + Bx(1 - x) + Cx(1 + x) = 2 + 4x\) | M1 | 1.1 |
| \(A = 2\), \(B = 1\), \(C = 3\) | A1 | 1.1 |
| \(\dfrac{2 + 4x}{x(1 - x^2)} = \dfrac{2}{x} + \dfrac{1}{1 + x} + \dfrac{3}{1 - x}\) | A1 | 2.1 |
| [4] |
Notes
B1: Identify correct partial fractions
M1: Clear fractions and attempt (at least) one coefficient. From using their 3 distinct linear factors
A1: Obtain one correct value www
A1: Obtain fully correct partial fractions. Possibly implied by \(A = 2\) etc
| Scheme | Marks | AO |
|---|---|---|
| \(\displaystyle\int \tan y\,\mathrm{d}y = \int \frac{2 + 4x}{x(1 + x)(1 - x)}\,\mathrm{d}x\) | M1* | 3.1a |
| \(\displaystyle\int \left(\frac{2}{x} + \frac{1}{1 + x} + \frac{3}{1 - x}\right)\mathrm{d}x =\) \(2\ln|x| + \ln|1 + x| - 3\ln|1 - x| \quad (+\,c)\) | B1FT | 2.1 |
| \(\displaystyle\int \tan y\,\mathrm{d}y = \ln|\sec y|\) | B1 | 3.1a |
| \(\ln\sec y = \ln\dfrac{kx^2(1 + x)}{(1 - x)^3}\) \(\sec y = \dfrac{kx^2(1 + x)}{(1 - x)^3}\) or \(\cos y = \dfrac{(1 - x)^3}{kx^2(1 + x)}\) | M1dep* | 3.1a |
| A1FT | 2.1 | |
| \(\sqrt{2} = \dfrac{\frac{1}{4} \times \frac{3}{2}k}{\frac{1}{8}}\), \(\frac{1}{2}\sqrt{2} = \dfrac{\frac{1}{8}}{\frac{1}{4} \times \frac{3}{2} \times k}\) so \(k = \frac{1}{3}\sqrt{2}\) | M1dep* | 1.1 |
| \(\cos y = \dfrac{3\sqrt{2}(1 - x)^3}{2x^2(1 + x)}\) | A1 | 2.1 |
| [7] |
Notes
M1*: Attempt to separate variables to obtain \(\int \mathrm{f}(y)\,\mathrm{d}y = \int \mathrm{g}(x)\,\mathrm{d}x\)
Condone no integral signs
BOD if no d\(x\) and/or d\(y\) or still present as a single operator
B1FT: Correct integration of their 3 linear partial fractions from part (a). B0 if any \(y\) term still on RHS
Condone no constant of integration
Condone brackets not modulus signs
Brackets / modulus could be implied by further working, otherwise B0
B1: Correct integration of \(\tan y\). B0 if any \(x\) term on LHS, or if \(\tan y\) is still in term on RHS
\(\ln|\sec y|\) or \(-\ln|\cos y|\)
M1dep*: Correct process to combine all terms and remove logs. Must deal correctly with indices and correctly use rules for combining logs
Must include constant of integration (could be \(\mathrm{e}^c\) rather than \(k\)). but could be numerical already (possibly incorrect)
A1FT: Obtain a correct equation no longer involving logs. FT on an incorrect numerical constant of integration only
NB \((x - 1)^3\) not \((1 - x)^3\) is A0
All algebraic terms must be fully correct (no FT on these)
M1dep*: Use \(\left(\frac{1}{2}, \frac{1}{4}\pi\right)\) in a correct equation to attempt to find \(k\). Could be awarded before combining terms
A1: Obtain fully correct equation, in required form. Surd may not be rationalised, but no fractions within fractions