October 2021 Paper 3 Q10
10
(a) Express \(\dfrac{1}{(4x + 1)(x + 1)}\) in partial fractions. [3]
(b) A curve passes through the point \((0, 2)\) and satisfies the differential equation
\(\dfrac{\mathrm{d}y}{\mathrm{d}x} = \dfrac{y}{(4x + 1)(x + 1)}\), for \(x > -\dfrac{1}{4}\).
Show by integration that \(y = A\left(\dfrac{4x + 1}{x + 1}\right)^B\) where \(A\) and \(B\) are constants to be determined. [6]
\(\dfrac{\mathrm{d}y}{\mathrm{d}x} = \dfrac{y}{(4x + 1)(x + 1)}\), for \(x > -\dfrac{1}{4}\).
Show by integration that \(y = A\left(\dfrac{4x + 1}{x + 1}\right)^B\) where \(A\) and \(B\) are constants to be determined. [6]
| Scheme | Marks | AO |
|---|---|---|
| \(\dfrac{1}{(4x+1)(x+1)} = \dfrac{A}{4x+1} + \dfrac{B}{x+1}\) \(1 = A(x + 1) + B(4x + 1)\) | M1 | 1.1a |
| \(x = -1 \Rightarrow 1 = -3B \Rightarrow B = -\frac{1}{3}\) | A1 | 1.1 |
| \(x = -\frac{1}{4} \Rightarrow 1 = \frac{3}{4}A \Rightarrow A = \frac{4}{3}\) So \(\dfrac{1}{(4x+1)(x+1)} = \dfrac{4}{3(4x+1)} - \dfrac{1}{3(x+1)}\) | A1 | 1.1 |
| [3] |
Notes
M1: Method mark is implied by correct answer
A1: Final solution needed for A1.
Can be recovered in 10(b)
| Scheme | Marks | AO |
|---|---|---|
| \(\displaystyle\int\frac{1}{y}\,\mathrm{d}y = \int\frac{1}{(4x+1)(x+1)}\,\mathrm{d}x\) | M1 | 3.1a |
| \(\displaystyle\int\frac{1}{y}\,\mathrm{d}y = \int\left\{\frac{4}{3(4x+1)} - \frac{1}{3(x+1)}\right\}\mathrm{d}x\) | M1 | 2.2a |
| \(\ln|y| = \frac{1}{3}\ln|4x + 1| - \frac{1}{3}\ln|x + 1| + c\) | M1 | 1.1 |
| When \(x = 0\), \(y = 2 \Rightarrow c = \ln 2\) | B1 | 2.2a |
| \(\ln|y| = \ln 2\left(\dfrac{|4x+1|}{|x+1|}\right)^{\frac{1}{3}}\) | M1 | 1.1 |
| \(y = 2\left(\dfrac{4x+1}{x+1}\right)^{\frac{1}{3}}\) | A1 | 2.1 |
| [6] |
Notes
M1: Separation of variables – both sides seen
Integral signs, \(\mathrm{d}y\) and \(\mathrm{d}x\) needed
M1: Use of their (a). RHS only needed
Condone no \(\mathrm{d}x\)
M1: Integration. One correct \(x\) term (ft their partial fractions)
Condone missing modulus signs
B1: Finding constant (FT their (a))
M1: For use of \(a\ln m = \ln m^a\) and \(\ln m - \ln n = \ln\dfrac{m}{n}\)
Not dep on \(c\)
Condone missing modulus signs throughout
A1: Answer with correct values of \(A\) and \(B\)