C4 January 2009 Q3
3.
\[\mathrm{f}(x) = \dfrac{27x^2 + 32x + 16}{(3x + 2)^2(1 - x)},\quad |x| \lt \dfrac{2}{3}\]
Given that \(\mathrm{f}(x)\) can be expressed in the form
\[\mathrm{f}(x) = \dfrac{A}{(3x + 2)} + \dfrac{B}{(3x + 2)^2} + \dfrac{C}{(1 - x)},\]
| Scheme | Marks |
|---|---|
| \(27x^2 + 32x + 16 \equiv A(3x + 2)(1 - x) + B(1 - x) + C(3x + 2)^2\) | M1 |
| \(x = -\tfrac{2}{3},\quad 12 - \tfrac{64}{3} + 16 = \left(\tfrac{5}{3}\right)B \Rightarrow \tfrac{20}{3} = \left(\tfrac{5}{3}\right)B \Rightarrow B = 4\) \(x = 1,\quad 27 + 32 + 16 = 25C \Rightarrow 75 = 25C \Rightarrow C = 3\) | M1 A1 |
| Equate \(x^2\): \(27 = -3A + 9C \Rightarrow 27 = -3A + 27 \Rightarrow 0 = -3A \Rightarrow A = 0\) \(x = 0,\quad 16 = 2A + B + 4C\) \(\qquad \Rightarrow 16 = 2A + 4 + 12 \Rightarrow 0 = 2A \Rightarrow A = 0\) | B1 |
| (4) |
Notes
M1: Forming this identity
M1: Substitutes either \(x = -\tfrac{2}{3}\) or \(x = 1\) into their identity or equates 3 terms or substitutes in values to write down three simultaneous equations.
A1: Both \(B = 4\) and \(C = 3\) (Note the A1 is dependent on both method marks in this part.)
B1: Compares coefficients or substitutes in a third \(x\)-value or uses simultaneous equations to show \(A = 0\).
Aliter 3. (a) Way 2
| Scheme | Marks |
|---|---|
| \(27x^2 + 32x + 16 \equiv A(3x + 2)(1 - x) + B(1 - x) + C(3x + 2)^2\) | M1 |
| \(\begin{aligned} x^2 \text{ terms}&\colon\ 27 = -3A + 9C &&(1)\\ x \text{ terms}&\colon\ 32 = A - B + 12C &&(2)\\ \text{constants}&\colon\ 16 = 2A + B + 4C &&(3)\end{aligned}\) | M1 |
| (2) + (3) gives \(48 = 3A + 16C\) (4) | |
| (1) + (4) gives \(75 = 25C \Rightarrow C = 3\) | |
| (1) gives \(27 = -3A + 27 \Rightarrow 0 = -3A \Rightarrow A = 0\) | |
| (2) gives \(32 = -B + 36 \Rightarrow B = 36 - 32 = 4\) | A1 B1 |
| (4) |
M1: Forming this identity. M1: equates 3 terms. A1: Both \(B = 4\) and \(C = 3\). B1: Decide to award B1 for \(A = 0\)
3. (a) If the candidate assumes \(A = 0\) and writes the identity \(27x^2 + 32x + 16 \equiv B(1 - x) + C(3x + 2)^2\) and goes on to find \(B = 4\) and \(C = 3\) then the candidate is awarded M0M1A0B0.
3. (a) If the candidate has the incorrect identity \(27x^2 + 32x + 16 \equiv A(3x + 2) + B(1 - x) + C(3x + 2)^2\) and goes on to find \(B = 4\), \(C = 3\) and \(A = 0\) then the candidate is awarded M0M1A0B1.
3. (a) If the candidate has the incorrect identity \(27x^2 + 32x + 16 \equiv A(3x + 2)^2(1 - x) + B(1 - x) + C(3x + 2)^2\) and goes on to find \(B = 4\), \(C = 3\) and \(A = 0\) then the candidate is awarded M0M1A0B1.
| Scheme | Marks |
|---|---|
| \(\mathrm{f}(x) = \dfrac{4}{(3x + 2)^2} + \dfrac{3}{(1 - x)}\) | |
| \(= 4(3x + 2)^{-2} + 3(1 - x)^{-1}\) | M1 |
| \(= 4\left[2^{-2}\left(1 + \tfrac{3}{2}x\right)^{-2}\right] + 3(1 - x)^{-1}\) (corrected from the printed mark scheme: the scheme prints \(4\left[2\left(1 + \tfrac{3}{2}x\right)^{-2}\right]\)) | |
| \(= 1\left(1 + \tfrac{3}{2}x\right)^{-2} + 3(1 - x)^{-1}\) | |
| \(= 1\left\{\underline{1 + (-2)\left(\tfrac{3x}{2}\right);\ + \dfrac{(-2)(-3)}{2!}\left(\tfrac{3x}{2}\right)^2 + \ldots}\right\}\) \(\quad + 3\left\{\underline{1 + (-1)(-x);\ + \dfrac{(-1)(-2)}{2!}(-x)^2 + \ldots}\right\}\) | dM1; A1 A1 |
| \(= \left\{1 - 3x + \tfrac{27}{4}x^2 + \ldots\right\} + 3\left\{1 + x + x^2 + \ldots\right\}\) | |
| \(= 4 + 0x;\ + \tfrac{39}{4}x^2\) | A1; A1 |
| (6) |
Notes
M1: Moving powers to top on any one of the two expressions
dM1: Either \(1 \pm (-2)\left(\tfrac{3x}{2}\right)\) or \(1 \pm (-1)(-x)\) from either first or second expansions respectively
A1: Ignoring 1 and 3, any one correct \(\underline{\{\ldots\ldots\}}\) expansion.
A1: Both \(\underline{\{\ldots\ldots\}}\) correct.
A1; A1: \(4 + (0x)\); \(\tfrac{39}{4}x^2\)
Aliter 3. (b) Way 2
| Scheme | Marks |
|---|---|
| \(\mathrm{f}(x) = \dfrac{4}{(3x + 2)^2} + \dfrac{3}{(1 - x)}\) | |
| \(= 4(3x + 2)^{-2} + 3(1 - x)^{-1}\) | M1 |
| \(= 4(2 + 3x)^{-2} + 3(1 - x)^{-1}\) | |
| \(= 4\left\{\underline{(2)^{-2} + (-2)(2)^{-3}(3x);\ + \dfrac{(-2)(-3)}{2!}(2)^{-4}(3x)^2 +}\right\}\) \(\quad + 3\left\{\underline{1 + (-1)(-x);\ + \dfrac{(-1)(-2)}{2!}(-x)^2 + \ldots}\right\}\) | dM1; A1 A1 |
| \(= 4\left\{\tfrac{1}{4} - \tfrac{3}{4}x + \tfrac{27}{16}x^2 + \ldots\right\} + 3\left\{1 + x + x^2 + \ldots\right\}\) | |
| \(= 4 + 0x;\ + \tfrac{39}{4}x^2\) | A1; A1 |
| (6) |
M1: Moving powers to top on any one of the two expressions
dM1: Either \((2)^{-2} \pm (-2)(2)^{-3}(3x)\) or \(1 \pm (-1)(-x)\) from either first or second expansions respectively
A1: Ignoring 1 and 3, any one correct \(\underline{\{\ldots\ldots\}}\) expansion. A1: Both \(\underline{\{\ldots\ldots\}}\) correct. A1; A1: \(4 + (0x)\); \(\tfrac{39}{4}x^2\)
| Scheme | Marks |
|---|---|
| Actual \(= \mathrm{f}(0.2) = \dfrac{1.08 + 6.4 + 16}{(6.76)(0.8)}\) \(\qquad = \dfrac{23.48}{5.408} = 4.341715976\ldots = \dfrac{2935}{676}\) | |
| Or Actual \(= \mathrm{f}(0.2) = \dfrac{4}{(3(0.2) + 2)^2} + \dfrac{3}{(1 - 0.2)}\) \(\qquad = \dfrac{4}{6.76} + 3.75 = 4.341715976\ldots = \dfrac{2935}{676}\) | M1 |
| Estimate \(= \mathrm{f}(0.2) = 4 + \tfrac{39}{4}(0.2)^2\) \(\qquad = 4 + 0.39 = 4.39\) | M1ft |
| %age error \(= \dfrac{\left|4.39 - 4.341715976\ldots\right|}{4.341715976\ldots} \times 100\) | M1 |
| \(= 1.112095408\ldots = 1.1\%\) (2sf) | A1 cao |
| (4) | |
| (14 marks) |
Notes
M1: Attempt to find the actual value of f(0.2) or seeing awrt 4.3 and believing it is candidate’s actual f(0.2).
Candidates can also attempt to find the actual value by using \(\dfrac{A}{(3x + 2)} + \dfrac{B}{(3x + 2)^2} + \dfrac{C}{(1 - x)}\) with their \(A\), \(B\) and \(C\).
M1ft: Attempt to find an estimate for f(0.2) using their answer to (b)
M1: \(\left|\dfrac{\text{their estimate} - \text{actual}}{\text{actual}}\right| \times 100\)
A1 cao: 1.1%
Aliter 3. (c) Way 2
| Scheme | Marks |
|---|---|
| Actual \(= \mathrm{f}(0.2) = \dfrac{1.08 + 6.4 + 16}{(6.76)(0.8)}\) \(\qquad = \dfrac{23.48}{5.408} = 4.341715976\ldots = \dfrac{2935}{676}\) | M1 |
| Estimate \(= \mathrm{f}(0.2) = 4 + \tfrac{39}{4}(0.2)^2\) \(\qquad = 4 + 0.39 = 4.39\) | M1ft |
| %age error \(= \left|100 - \left(\dfrac{4.39}{4.341715976\ldots} \times 100\right)\right|\) | M1 |
| \(= \left|100 - 101.1120954\right|\) | |
| \(= \left|-1.112095408\ldots\right| = 1.1\%\) (2sf) | A1 cao |
| (4) |
M1: Attempt to find the actual value of f(0.2). M1ft: Attempt to find an estimate for f(0.2) using their answer to (b)
M1: \(\left|100 - \left(\left(\dfrac{\text{their estimate}}{\text{actual}}\right) \times 100\right)\right|\) A1 cao: 1.1%
3. (c) Note that: %age error \(= \dfrac{\left|4.39 - 4.341715976\ldots\right|}{4.39} \times 100 = 1.0998638\ldots = 1.1\%\) (2sf) — Should be awarded the final marks of M0A0
3. (c) Also note that: %age error \(= \left|100 - \left(\dfrac{4.341715976\ldots}{4.39} \times 100\right)\right| = 1.0998638\ldots = 1.1\%\) (2sf) — Should be awarded the final marks of M0A0
…so be wary of 1.0998638…