C4 June 2006 Q2
2.
\[\mathrm{f}(x) = \frac{3x - 1}{(1 - 2x)^2}, \qquad |x| \lt \tfrac{1}{2}.\]
Given that, for \(x \neq \frac{1}{2}\), \(\dfrac{3x - 1}{(1 - 2x)^2} = \dfrac{A}{(1 - 2x)} + \dfrac{B}{(1 - 2x)^2}\), where \(A\) and \(B\) are constants,
| Scheme | Marks |
|---|---|
| \(3x - 1 \equiv A(1 - 2x) + B\) | M1 |
| Let \(x = \frac{1}{2}\); \(\frac{3}{2} - 1 = B \Rightarrow B = \frac{1}{2}\) | |
| Equate \(x\) terms; \(3 = -2A \Rightarrow A = -\frac{3}{2}\) | A1;A1 |
| (3) |
Notes
M1 Considers this complete identity and either substitutes \(x = \frac{1}{2}\), equates coefficients or solves simultaneous equations
A1;A1 \(A = -\frac{3}{2}\); \(B = \frac{1}{2}\)
(No working seen, but \(A\) and \(B\) correctly stated \(\Rightarrow\) award all three marks. If one of \(A\) or \(B\) correctly stated give two out of the three marks available for this part.)
Beware: In part (a) take care to spot that \(A = -\frac{3}{2}\) and \(B = \frac{1}{2}\) are the right way around.
Beware: In ePEN, make sure you aware the marks correctly in part (a). The first A1 is for \(A = -\frac{3}{2}\) and the second A1 is for \(B = \frac{1}{2}\).
| Scheme | Marks |
|---|---|
| \(\mathrm{f}(x) = -\frac{3}{2}(1 - 2x)^{-1} + \frac{1}{2}(1 - 2x)^{-2}\) | M1 |
| \(= -\frac{3}{2}\left\{\underline{1 + (-1)(-2x); + \dfrac{(-1)(-2)}{2!}(-2x)^2 + \dfrac{(-1)(-2)(-3)}{3!}(-2x)^3 + \ldots}\right\}\) | dM1; |
| \(+ \frac{1}{2}\left\{\underline{1 + (-2)(-2x); + \dfrac{(-2)(-3)}{2!}(-2x)^2 + \dfrac{(-2)(-3)(-4)}{3!}(-2x)^3 + \ldots}\right\}\) | A1 A1 |
| \(= -\frac{3}{2}\left\{1 + 2x + 4x^2 + 8x^3 + \ldots\right\} + \frac{1}{2}\left\{1 + 4x + 12x^2 + 32x^3 + \ldots\right\}\) | |
| \(= -1 - x; + 0x^2 + 4x^3\) | A1; A1 |
| (6) | |
| (9 marks) |
Notes
M1 Moving powers to top on any one of the two expressions
dM1 Either \(1 \pm 2x\) or \(1 \pm 4x\) from either first or second expansions respectively
A1 Ignoring \(-\frac{3}{2}\) and \(\frac{1}{2}\), any one correct \(\{\ldots\ldots\}\) expansion.
A1 Both \(\{\ldots\ldots\}\) correct.
A1; A1 \(-1 - x\); \((0x^2) + 4x^3\)
Aliter (b) Way 2
| \(\mathrm{f}(x) = (3x - 1)(1 - 2x)^{-2}\) | M1 |
| \(= (3x - 1) \times \left(1 + (-2)(-2x); + \dfrac{(-2)(-3)}{2!}(-2x)^2 + \dfrac{(-2)(-3)(-4)}{3!}(-2x)^3 + \ldots\right)\) | dM1; A1 |
| \(= (3x - 1)(1 + 4x + 12x^2 + 32x^3 + \ldots)\) | |
| \(= \underline{3x + 12x^2 + 36x^3 - 1 - 4x - 12x^2 - 32x^3} + \ldots\) | A1 |
| \(= -1 - x; + 0x^2 + 4x^3\) | A1; A1 |
| [6] |
M1 Moving power to top
dM1; \(1 \pm 4x\);
A1 Ignoring \((3x - 1)\), correct \((\ldots\ldots\ldots)\) expansion
A1 Correct expansion
A1; A1 \(-1 - x\); \((0x^2) + 4x^3\)
Aliter (b) Way 3: Maclaurin expansion
| \(\mathrm{f}(x) = -\frac{3}{2}(1 - 2x)^{-1} + \frac{1}{2}(1 - 2x)^{-2}\) | M1 |
| \(\mathrm{f}'(x) = -3(1 - 2x)^{-2} + 2(1 - 2x)^{-3}\) | M1; A1 oe |
| \(\mathrm{f}''(x) = -12(1 - 2x)^{-3} + 12(1 - 2x)^{-4}\) \(\mathrm{f}'''(x) = -72(1 - 2x)^{-4} + 96(1 - 2x)^{-5}\) | A1 |
| \(\therefore \mathrm{f}(0) = -1,\ \mathrm{f}'(0) = -1,\ \mathrm{f}''(0) = 0\) and \(\mathrm{f}'''(0) = 24\) | |
| gives \(\mathrm{f}(x) = -1 - x; + 0x^2 + 4x^3 + \ldots\) | A1; A1 |
| [6] |
M1 Bringing both powers to top
M1; A1 oe Differentiates to give \(a(1 - 2x)^{-2} \pm b(1 - 2x)^{-3}\); \(-3(1 - 2x)^{-2} + 2(1 - 2x)^{-3}\)
A1 Correct \(\mathrm{f}''(x)\) and \(\mathrm{f}'''(x)\)
A1; A1 \(-1 - x\); \((0x^2) + 4x^3\)
Aliter (b) Way 4
| \(\mathrm{f}(x) = -3(2 - 4x)^{-1} + \frac{1}{2}(1 - 2x)^{-2}\) | M1 |
| \(= -3\left\{\underline{(2)^{-1} + (-1)(2)^{-2}(-4x); + \dfrac{(-1)(-2)}{2!}(2)^{-3}(-4x)^2 + \dfrac{(-1)(-2)(-3)}{3!}(2)^{-4}(-4x)^3 + \ldots}\right\}\) | dM1; |
| \(+ \frac{1}{2}\left\{\underline{1 + (-2)(-2x); + \dfrac{(-2)(-3)}{2!}(-2x)^2 + \dfrac{(-2)(-3)(-4)}{3!}(-2x)^3 + \ldots}\right\}\) | A1 A1 |
| \(= -3\left\{\frac{1}{2} + x + 2x^2 + 4x^3 + \ldots\right\} + \frac{1}{2}\left\{1 + 4x + 12x^2 + 32x^3 + \ldots\right\}\) | |
| \(= -1 - x; + 0x^2 + 4x^3\) | A1; A1 |
| [6] |
M1 Moving powers to top on any one of the two expressions
dM1; Either \(\frac{1}{2} \pm x\) or \(1 \pm 4x\) from either first or second expansions respectively
A1 Ignoring \(-3\) and \(\frac{1}{2}\), any one correct \(\{\ldots\ldots\}\) expansion.
A1 Both \(\{\ldots\ldots\}\) correct.
A1; A1 \(-1 - x\); \((0x^2) + 4x^3\)