$I = \int_{\frac{5}{2}}^{\frac{5}{2}} \frac{1}{\sqrt{4x^2 - 12x + 13}} dx = \int_{\frac{5}{2}}^{\frac{5}{2}} \frac{1}{\sqrt{(2x-3)^2 + 4}} dx$ | M1 | 2.1 | Completing the square |

Substitute $2x - 3 = 2\sinh\theta$ | M1 | 3.1a | Substitution |

$\Rightarrow dx = \cosh\theta d\theta$, $(2x-3)^2 + 4 = 4\cosh^2\theta$ | | |

When $x = \frac{3}{2}$, $\theta = 0$. When $x = \frac{5}{2}$, $\sinh\theta = 1$ | A1 | 1.1 | Correct reduction (ignore limits) |

$\Rightarrow I = \int_0^{\sinh^{-1}1} \frac{1}{2\cosh\theta} \cosh\theta d\theta = \frac{1}{2}[\theta]_0^{\sinh^{-1}1}$ | A1 | 2.2a | Correct integral (ignore limits) |

$M1$ | 1.1 | Limits |

$= \frac{1}{2}\sinh^{-1}1 = \frac{1}{2}\ln\left(1 + \sqrt{2}\right)$ | A1 | 1.1 |

**OR** | | |

$\int_{\frac{5}{2}}^{\frac{5}{2}} \frac{1}{\sqrt{4(x-\frac{3}{2})^2 - 9 + 13}} dx = \frac{1}{2}\int_{\frac{5}{2}}^{\frac{5}{2}} \frac{1}{\sqrt{(x-\frac{3}{2})^2 + 1}} dx =$ | | | $u = x - \frac{3}{2}$, $du = dx$, $x = \frac{5}{2}, u = 1$, $x = \frac{3}{2}, u = 0$ |

$\frac{1}{4}\int_0^1 \frac{1}{\sqrt{u^2+1}} dx = \frac{1}{4}\left[\ln(u + \sqrt{u^2+1})\right] = \frac{1}{4}\ln(1 + \sqrt{2})$ | | |

| [6] | |

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# Question 6 (i):

$\theta = A\cos 2t + B\sin 2t$ | B1 | 1.2 |

| [1] | |

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# Question 6 (ii):

$\theta = A\cos 2t + B\sin 2t$ | M1 | 3.3 |

$\Rightarrow \frac{d\theta}{dt} = -2A\sin 2t + 2B\cos 2t$ | A1 | 3.4 |

$\frac{d\theta}{dt} = 0$ when $t = 0 \Rightarrow B = 0$ | A1 | 1.1 |

$\theta = \theta_0$ when $t = 0 \Rightarrow A = \theta_0$

$\Rightarrow \theta = \theta_0\cos 2t$ | A1 | 3.4 |

| [4] | |

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# Question 6 (iii):

Angle has to be small otherwise the model is invalid. | B1 | 3.5b |

| [1] | |

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# Question 7 (i):

$\sinh x = \frac{1}{2}(e^x - e^{-x})$, $\sinh 3x = \frac{1}{2}(e^{3x} - e^{-3x})$ | M1 | 1.1a | B1 for one error |

$\sinh^3 x = \frac{1}{8}(e^x - e^{-x})^3 = \frac{1}{8}(e^{3x} - 3e^x + 3e^{-x} - e^{-3x})$ | A1 | 1.1 |

$= \frac{1}{4}\left(\frac{e^{3x} - e^{-3x}}{2} - \frac{3e^x - 3e^{-x}}{2}\right) = \frac{1}{4}\sinh 3x - \frac{3}{4}\sinh x$ | A1 | 2.1 |

$\Rightarrow 4\sinh^3 x = \sinh 3x - 3\sinh x$ | | AG |

| [3] | |

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# Question 7 (ii):

Substitute $u = \sinh x$ | M1 | 3.1a |

$4\sinh^3 x = \sinh 3x - 3\sinh x$ | A1 | 1.1 |

$\Rightarrow 16\sinh^3 x + 12\sinh x = 4\sinh 3x$ | A1 | 1.1 |

$u = \sinh x \Rightarrow 4\sinh 3x = 3 \Rightarrow \sinh 3x = \frac{3}{4}$ | A1 | 1.1 |

$\Rightarrow 3x = \ln\left(\frac{3}{4} + \sqrt{1 + \frac{9}{16}}\right) = \ln 2$ | M1 | 1.1 |

$x = \frac{1}{3}\ln 2 \Rightarrow u = \sinh\left(\frac{1}{3}\ln 2\right) = \frac{2^{\frac{1}{3}} - 2^{-\frac{1}{3}}}{2}$ | A1 | 3.2a | Use of $e^{\ln 2} = 2^1$ |

| [5] | |

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# Question 8 (i):

$f(x) = (1 + a\sin x)e^{bx} = \left[1 + a\left(x - \frac{x^3}{6}\right)\right]\left(1 + bx + \frac{b^2x^2}{2}\right)$ | B1, B1 | 1.2, 1.2 | Sin expansion, Exponential expansion |

$= 1 + bx + \frac{b^2x^2}{2} + ax + abx^2 = 1 + (a + b)x + x^2\left(\frac{b^2}{2} + ab\right)$ | M1 | 1.1a | Multiply out and compare coefficients |

$\approx 1 + 2x + \frac{3}{2}x^2$ | | |

$\Rightarrow a + b = 2, \frac{b^2}{2} + ab = \frac{3}{2}$ | A1 | 1.1 | Both equations |

$\Rightarrow b^2 + 2b(2-b) = 3 \Rightarrow b^2 - 4b + 3 = 0$ | M1 | 1.1 | Solve |

$\Rightarrow b = 1, 3 \Rightarrow a = 1, -1$ | A1 | 1.1 | As $a < 0$ |

$\Rightarrow (-1, 3)$ | |

| [6] | |

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# Question 8 (ii):

There is no restriction on $x$ because there is no restriction on the maclaurin's expansion for $\sin x$ and $e^x$. | B1 | 2.3 |

| [1] | |

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