C4 June 2018 Q3
3. (i) Given that \[\frac{13 - 4x}{(2x + 1)^2(x + 3)} \equiv \frac{A}{(2x + 1)} + \frac{B}{(2x + 1)^2} + \frac{C}{(x + 3)}\]
| Scheme | Marks |
|---|---|
| \(\dfrac{13 - 4x}{(2x + 1)^2(x + 3)} \equiv \dfrac{A}{(2x + 1)} + \dfrac{B}{(2x + 1)^2} + \dfrac{C}{(x + 3)}\) | |
| \(B = 6,\ C = 1\) At least one of \(B = 6\) or \(C = 1\) Both \(B = 6\) and \(C = 1\) | B1 B1 |
| \(13 - 4x \equiv A(2x + 1)(x + 3) + B(x + 3) + C(2x + 1)^2\) \(x = -3 \Rightarrow 25 = 25C \Rightarrow C = 1\) \(x = -\dfrac{1}{2} \Rightarrow 13 - -2 = \dfrac{5}{2}B \Rightarrow 15 = 2.5B \Rightarrow B = 6\) Writes down a correct identity and attempts to find the value of either one of \(A\) or \(B\) or \(C\) | M1 |
| Either \(x^2: 0 = 2A + 4C\), constant\(: 13 = 3A + 3B + C\), \(x: -4 = 7A + B + 4C\) or \(x = 0 \Rightarrow 13 = 3A + 3B + C\) leading to \(A = -2\) Using a correct identity to find \(A = -2\) | A1 |
| (4) |
Notes
M1: Writes down a correct identity (although this can be implied) and attempts to find the value of at least one of either \(A\) or \(B\) or \(C\). This can be achieved by either substituting values into their identity or comparing coefficients.
Note: The correct partial fraction from no working scores B1B1M1A1
| Scheme | Marks |
|---|---|
| \(\displaystyle\int \dfrac{13 - 4x}{(2x + 1)^2(x + 3)}\,\mathrm{d}x = \int \frac{-2}{(2x + 1)} + \frac{6}{(2x + 1)^2} + \frac{1}{(x + 3)}\,\mathrm{d}x\) | |
| \(= \dfrac{(-2)}{2}\ln(2x + 1) + \dfrac{6(2x + 1)^{-1}}{(-1)(2)} + \ln(x + 3)\ \{+ c\}\) See notes At least two terms correctly integrated o.e. \(\{= -\ln(2x + 1) - 3(2x + 1)^{-1} + \ln(x + 3)\ \{+ c\}\}\) Correct answer, o.e. Simplified or un-simplified. The correct answer must be stated on one line. Ignore the absence of ‘\(+ c\)’ | M1 A1ft A1 |
| (3) |
Notes
M1: At least 2 of either \(\pm\dfrac{P}{(2x + 1)} \to \pm D\ln(2x + 1)\) or \(\pm D\ln(x + \frac{1}{2})\) or \(\pm\dfrac{Q}{(2x + 1)^2} \to \pm E(2x + 1)^{-1}\)
or \(\pm\dfrac{R}{(x + 3)} \to \pm F\ln(x + 3)\) for their constants \(P\), \(Q\), \(R\).
A1ft: At least two terms from any of \(\pm\dfrac{P}{(2x + 1)}\) or \(\pm\dfrac{Q}{(2x + 1)^2}\) or \(\pm\dfrac{R}{(x + 3)}\) correctly integrated.
Note: Can be un-simplified for the A1ft mark.
A1: Correct answer of \(\dfrac{(-2)}{2}\ln(2x + 1) + \dfrac{6(2x + 1)^{-1}}{(-1)(2)} + \ln(x + 3)\ \{+ c\}\) simplified or un-simplified.
with or without ‘\(+ c\)’.
Note: Allow final A1 for equivalent answers, e.g. \(\ln\left(\dfrac{x + 3}{2x + 1}\right) - \dfrac{3}{2x + 1}\ \{+ c\}\) or \(\ln\left(\dfrac{2x + 6}{2x + 1}\right) - \dfrac{3}{2x + 1}\ \{+ c\}\)
Note: Beware that \(\displaystyle\int \frac{-2}{(2x + 1)}\,\mathrm{d}x = \int \frac{-1}{(x + \frac{1}{2})}\,\mathrm{d}x = -\ln(x + \tfrac{1}{2})\ \{+ c\}\) is correct integration
Note: E.g. Allow M1 A1ft A1 for a correct un-simplified \(\ln(x + 3) - \ln(x + \frac{1}{2}) - \frac{3}{2}(x + \frac{1}{2})^{-1}\ \{+ c\}\)
Note: Condone 1st A1ft for poor bracketing, but do not allow poor bracketing for the final A1
E.g. Give final A0 for \(-\ln 2x + 1 - 3(2x + 1)^{-1} + \ln x + 3\ \{+ c\}\) unless recovered
| Scheme | Marks |
|---|---|
| \(\left\{(\mathrm{e}^x + 1)^3 =\right\}\ \mathrm{e}^{3x} + 3\mathrm{e}^{2x} + 3\mathrm{e}^x + 1\) \(\mathrm{e}^{3x} + 3\mathrm{e}^{2x} + 3\mathrm{e}^x + 1\), simplified or un-simplified | B1 |
| \(\left\{\displaystyle\int (\mathrm{e}^x + 1)^3\,\mathrm{d}x\right\} = \dfrac{1}{3}\mathrm{e}^{3x} + \dfrac{3}{2}\mathrm{e}^{2x} + 3\mathrm{e}^x + x\ \{+ c\}\) At least 3 examples (see notes) of correct ft integration \(\dfrac{1}{3}\mathrm{e}^{3x} + \dfrac{3}{2}\mathrm{e}^{2x} + 3\mathrm{e}^x + x\), simplified or un-simplified with or without \(+c\) | M1 A1 |
| (3) |
Notes
Note: Give B1 for an un-simplified \(\mathrm{e}^{3x} + 2\mathrm{e}^{2x} + \mathrm{e}^{2x} + 2\mathrm{e}^x + \mathrm{e}^x + 1\)
M1: At least 3 of either \(\alpha\mathrm{e}^{3x} \to \dfrac{\alpha}{3}\mathrm{e}^{3x}\) or \(\beta\mathrm{e}^{2x} \to \dfrac{\beta}{2}\mathrm{e}^{2x}\) or \(\delta\mathrm{e}^x \to \delta\mathrm{e}^x\) or \(\mu \to \mu x;\ \alpha, \beta, \delta, \mu \neq 0\)
Note: Give A1 for an un-simplified \(\dfrac{1}{3}\mathrm{e}^{3x} + \mathrm{e}^{2x} + \dfrac{1}{2}\mathrm{e}^{2x} + 2\mathrm{e}^x + \mathrm{e}^x + x\), with or without \(+c\)
Alt 1 for part (ii)
| Scheme | Marks |
|---|---|
| \(\displaystyle\int (\mathrm{e}^x + 1)^3\,\mathrm{d}x;\ u = \mathrm{e}^x + 1 \Rightarrow \frac{\mathrm{d}u}{\mathrm{d}x} = \mathrm{e}^x\) | |
| \(\left\{= \displaystyle\int \frac{u^3}{(u - 1)}\,\mathrm{d}u =\right\} \displaystyle\int \left(u^2 + u + 1 + \frac{1}{u - 1}\right)\mathrm{d}u\) \(\displaystyle\int \left(u^2 + u + 1 + \frac{1}{u - 1}\right)\{\mathrm{d}u\}\) where \(u = \mathrm{e}^x + 1\) | B1 |
| \(= \dfrac{1}{3}u^3 + \dfrac{1}{2}u^2 + u + \ln(u - 1)\ \{+ c\}\) At least 3 of either \(\alpha u^2 \to \dfrac{\alpha}{3}u^3\) or \(\beta u \to \dfrac{\beta}{2}u^2\) or \(\delta \to \delta u\) or \(\dfrac{\lambda}{u - 1} \to \lambda\ln(u - 1);\ \alpha, \beta, \delta, \lambda \neq 0\) | M1 |
| \(= \dfrac{1}{3}(\mathrm{e}^x + 1)^3 + \dfrac{1}{2}(\mathrm{e}^x + 1)^2 + (\mathrm{e}^x + 1) + \ln(\mathrm{e}^x + 1 - 1)\ \{+ c\}\) | |
| \(= \dfrac{1}{3}(\mathrm{e}^x + 1)^3 + \dfrac{1}{2}(\mathrm{e}^x + 1)^2 + (\mathrm{e}^x + 1) + x\ \{+ c\}\) \(\dfrac{1}{3}(\mathrm{e}^x + 1)^3 + \dfrac{1}{2}(\mathrm{e}^x + 1)^2 + (\mathrm{e}^x + 1) + x\) or \(\dfrac{1}{3}(\mathrm{e}^x + 1)^3 + \dfrac{1}{2}(\mathrm{e}^x + 1)^2 + \mathrm{e}^x + x\) simplified or un-simplified with or without \(+ c\) Note: \(\ln(\mathrm{e}^x + 1 - 1)\) needs to be simplified to \(x\) for this mark | A1 |
| (3) |
Alt 2 for part (ii)
| Scheme | Marks |
|---|---|
| \(\displaystyle\int (\mathrm{e}^x + 1)^3\,\mathrm{d}x;\ u = \mathrm{e}^x \Rightarrow \frac{\mathrm{d}u}{\mathrm{d}x} = \mathrm{e}^x\) | |
| \(\left\{= \displaystyle\int \frac{(u + 1)^3}{u}\,\mathrm{d}u =\right\} \displaystyle\int \left(u^2 + 3u + 3 + \frac{1}{u}\right)\mathrm{d}u\) \(\displaystyle\int \left(u^2 + 3u + 3 + \frac{1}{u}\right)\{\mathrm{d}u\}\) where \(u = \mathrm{e}^x\) | B1 |
| \(= \dfrac{1}{3}u^3 + \dfrac{3}{2}u^2 + 3u + \ln u\ \{+ c\}\) At least 3 of either \(\alpha u^2 \to \dfrac{\alpha}{3}u^3\) or \(\beta u \to \dfrac{\beta}{2}u^2\) or \(\delta \to \delta u\) or \(\dfrac{\lambda}{u} \to \lambda\ln u;\ \alpha, \beta, \delta, \lambda \neq 0\) | M1 |
| \(= \dfrac{1}{3}\mathrm{e}^{3x} + \dfrac{3}{2}\mathrm{e}^{2x} + 3\mathrm{e}^x + x\ \{+ c\}\) \(\dfrac{1}{3}\mathrm{e}^{3x} + \dfrac{3}{2}\mathrm{e}^{2x} + 3\mathrm{e}^x + x\), simplified or un-simplified with or without \(+ c\) Note: \(\ln(\mathrm{e}^x)\) needs to be simplified to \(x\) for this mark | A1 |
| (3) |
| Scheme | Marks |
|---|---|
| \(\displaystyle\int \frac{1}{4x + 5x^{\frac{1}{3}}}\,\mathrm{d}x,\ x > 0;\ u^3 = x\) | |
| \(3u^2\dfrac{\mathrm{d}u}{\mathrm{d}x} = 1\) \(3u^2\dfrac{\mathrm{d}u}{\mathrm{d}x} = 1\) or \(\dfrac{\mathrm{d}x}{\mathrm{d}u} = 3u^2\) or \(\dfrac{\mathrm{d}u}{\mathrm{d}x} = \dfrac{1}{3}x^{-\frac{2}{3}}\) or \(3u^2\,\mathrm{d}u = \mathrm{d}x\) o.e. | B1 |
| \(= \displaystyle\int \frac{1}{4u^3 + 5u}.3u^2\,\mathrm{d}u\ \left\{= \int \frac{3u}{4u^2 + 5}\,\mathrm{d}u\right\}\) Expression of the form \(\displaystyle\int \frac{\pm ku^2}{4u^3 \pm 5u}\{\mathrm{d}u\},\ k \neq 0\). Does not have to include integral sign or \(\mathrm{d}u\). Can be implied by later working | M1 |
| \(= \dfrac{3}{8}\ln(4u^2 + 5)\ \{+ c\}\) dependent on the previous M mark \(\pm\lambda\ln(4u^2 + 5);\ \lambda\) is a constant\(;\ \lambda \neq 0\) | dM1 |
| \(= \dfrac{3}{8}\ln\left(4x^{\frac{2}{3}} + 5\right)\ \{+ c\}\) Correct answer in \(x\) with or without \(+ c\) | A1 |
| (4) | |
| (14 marks) |
Notes
Note: 1st M1 can be implied by \(\displaystyle\int \frac{\pm ku}{4u^2 \pm 5}\{\mathrm{d}u\},\ k \neq 0\). Does not have to include integral sign or \(\mathrm{d}u\)
Note: Condone 1st M1 for expressions of the form \(\displaystyle\int \left(\frac{\pm 1}{4u^3 \pm 5u}.\frac{\pm k}{u^{-2}}\right)\{\mathrm{d}u\},\ k \neq 0\)
Note: Give 2nd M0 for \(\dfrac{3u}{8u}\ln(4u^2 + 5)\ \{+ c\}\) (\(u\)’s not cancelled) unless recovered in later working
Note: E.g. Give 2nd M0 for integration leading to \(\dfrac{3}{4}u\ln(4u^2 + 5)\) as this is not in the form \(\pm\lambda\ln(4u^2 + 5)\)
Note: Condone 2nd M1 for poor bracketing, but do not allow poor bracketing for the final A1
E.g. Give final A0 for \(\dfrac{3}{8}\ln 4x^{\frac{2}{3}} + 5\ \{+c\}\) unless recovered
Alternative method 1 for part (iii)
| Scheme | Marks |
|---|---|
| \(\left\{\displaystyle\int \frac{1}{4x + 5x^{\frac{1}{3}}}\,\mathrm{d}x\right\} = \displaystyle\int \frac{x^{-\frac{1}{3}}}{4x^{\frac{2}{3}} + 5}\,\mathrm{d}x\) Attempts to multiply numerator and denominator by \(x^{-\frac{1}{3}}\) Expression of the form \(\displaystyle\int \frac{\pm kx^{-\frac{1}{3}}}{4x^{\frac{2}{3}} \pm 5}\,\mathrm{d}x,\ k \neq 0\). Does not have to include integral sign or \(\mathrm{d}u\). Can be implied by later working | M1 M1 |
| \(= \dfrac{3}{8}\ln\left(4x^{\frac{2}{3}} + 5\right)\ \{+ c\}\) \(\pm\lambda\ln(4x^{\frac{2}{3}} + 5);\ \lambda\) is a constant\(;\ \lambda \neq 0\) Correct answer in \(x\) with or without \(+ c\) | dM1 A1 |
| (4) |