## Question 10(i):

| Answer | Marks | Guidance |
|--------|-------|----------|
| $\frac{d}{dx}(\cot^{n+1}x) = -(n+1)\cot^n x\csc^2 x$ | M1 A1 | Differentiates using chain rule |
| $= -(n+1)\cot^n x(\cot^2 x + 1) = -(n+1)\cot^{n+2}x - (n+1)\cot^n x$ | M1 | Uses $\csc^2 x = \cot^2 x + 1$. OE |
| $\left[\cot^{n+1}x\right]_{\frac{1}{4}\pi}^{\frac{1}{2}\pi} = -(n+1)(I_{n+2}+I_n)$ | M1 | Integrates both sides |
| $-1 = -(n+1)(I_{n+2}+I_n) \Rightarrow I_{n+2} = \frac{1}{n+1} - I_n$ | A1 | AG |
| **Alternative:** $\frac{d}{dx}\left(\frac{\cos^{n+1}x}{\sin^{n+1}x}\right) = \frac{-(n+1)\sin^{n+2}x\cos^n x - (n+1)\sin^n x\cos^{n+2}x}{\sin^{2n+2}x}$ | M1 A1 | Differentiates using quotient/chain rule |
| $= -(n+1)\cot^n x - (n+1)\cot^{n+2}x$ | M1 | Separates |
| $\left[\cot^{n+1}x\right]_{\frac{1}{4}\pi}^{\frac{1}{2}\pi} = -(n+1)(I_{n+2}+I_n)$ | M1 | Integrates both sides |
| $-1 = -(n+1)(I_{n+2}+I_n) \Rightarrow I_{n+2} = \frac{1}{n+1} - I_n$ | A1 | AG |
| **Total: 5** | | |

## Question 10(ii):

| Answer | Marks | Guidance |
|--------|-------|----------|
| $\bar{y} = \frac{1}{2A}\int_{\frac{1}{4}\pi}^{\frac{1}{2}\pi} \cot^2 x\, dx = \frac{I_2}{2A}$ | M1 | Uses correct formula for $\bar{y}$. Allow incorrect or missing limits. |
| $A = \int_{\frac{1}{4}\pi}^{\frac{1}{2}\pi} \cot x\, dx = [\ln\sin x]_{\frac{1}{4}\pi}^{\frac{1}{2}\pi} = \ln\sqrt{2}$ | M1 A1 | Integrates $\cot x$ to find area under curve with correct limits. |
| $I_2 = 1 - I_0 = 1 - \frac{1}{4}\pi$ | M1 A1 | Finds $I_2$ |
| $\bar{y} = \frac{1}{\ln 2}\left(1 - \frac{1}{4}\pi\right)$ | A1 | AEF. Must be exact. |
| **Total: 6** | | |

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## Question 11E(i):

| Answer | Marks | Guidance |
|--------|-------|----------|
| $\mathbf{P} = \begin{pmatrix} 1 & -1 & 0 \\ 1 & 0 & 1 \\ 0 & b & -1 \end{pmatrix}$, $\mathbf{D} = \begin{pmatrix} 2 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 3 \end{pmatrix}$ | B1 B1 | Writes **P** and **D** (accept correctly matched permutations of columns) |
| $\det\mathbf{P} = -b - 1$ | B1 | |
| $\mathbf{P}^{-1} = \frac{1}{b+1}\begin{pmatrix} b & 1 & 1 \\ -1 & 1 & 1 \\ -b & b & -1 \end{pmatrix}$ | M1 A1 | Finds inverse of **P**. (Adj $\div$ Det). |
| $\mathbf{A} = \mathbf{PDP}^{-1}$ | M1 | Applies $\mathbf{A} = \mathbf{PDP}^{-1}$ |
| $\mathbf{A} = \frac{1}{b+1}\begin{pmatrix} 1 & -1 & 0 \\ 1 & 0 & 1 \\ 0 & b & -1 \end{pmatrix}\begin{pmatrix} 2b & 2 & 2 \\ -1 & 1 & 1 \\ -3b & 3b & -3 \end{pmatrix}$ | M1 A1 | Multiplies two adjacent matrices. $\mathbf{PD} = \begin{pmatrix} 2 & -1 & 0 \\ 2 & 0 & 3 \\ 0 & b & -3 \end{pmatrix}$ |
| $\frac{1}{b+1}\begin{pmatrix} 1+2b & 1 & 1 \\ -b & 3b+2 & -1 \\ 2b & -2b & 3+b \end{pmatrix}$ | A1 | |
| **Total: 9** | | |

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## Question 11E(i) Alternative Method:

| Answer | Marks | Guidance |
|--------|-------|----------|
| Matching eigenvalues with eigenvectors to form systems of equations | B3 | B1 for each correctly matched pair |
| Three values found correctly from solving systems of equations; another three values found correctly; attempting to find final three values | M1 A1, M1 A1, M1 | |
| $\mathbf{A} = \frac{1}{b+1}\begin{pmatrix} 1+2b & 1 & 1 \\ -b & 3b+2 & -1 \\ 2b & -2b & 3+b \end{pmatrix}$ | A1 | Final answer given in matrix form. |

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## Question 11E(ii):

| Answer | Marks | Guidance |
|--------|-------|----------|
| $\mathbf{A}^{-1}\begin{pmatrix} 0 \\ 2 \\ -2 \end{pmatrix} = 2\mathbf{A}^{-1}\begin{pmatrix} 0 \\ 1 \\ -1 \end{pmatrix} = \frac{2}{3}\begin{pmatrix} 0 \\ 1 \\ -1 \end{pmatrix}$ | M1 A1 | M1 for $\mathbf{A}^{-1}\begin{pmatrix} 0 \\ 2 \\ -2 \end{pmatrix} = 2\mathbf{A}^{-1}\begin{pmatrix} 0 \\ 1 \\ -1 \end{pmatrix}$ |
| **Total: 2** | | |

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## Question 11E(iii):

| Answer | Marks | Guidance |
|--------|-------|----------|
| $\mathbf{A}^n\begin{pmatrix} 1 \\ 1 \\ 0 \end{pmatrix} = 2^2\begin{pmatrix} 1 \\ 1 \\ 0 \end{pmatrix} \Rightarrow n = 2$ | B1 | Uses $\mathbf{A}^n\begin{pmatrix} 1 \\ 1 \\ 0 \end{pmatrix} = 2^n\begin{pmatrix} 1 \\ 1 \\ 0 \end{pmatrix}$ |
| $\mathbf{A}^n\begin{pmatrix} -1 \\ 0 \\ b \end{pmatrix} = \begin{pmatrix} -1 \\ 0 \\ b^{-1} \end{pmatrix} \Rightarrow b = b^{-1} \Rightarrow b = 1$ | M1 A1 | M1 for stating $b = b^{-1}$. Must be only one answer ($b=1$) for A1. |
| **Total: 3** | | |

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## Question 11O(i)(a):

| Answer | Marks | Guidance |
|--------|-------|----------|
| $\ln y = t\ln a$ | M1 A1 | Differentiates both sides. |
| $\frac{1}{y}\frac{dy}{dt} = \ln a$ | | |
| $\frac{dy}{dt} = y\ln a = a^t \ln a$ | A1 | AG |
| **Total: 3** | | |

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## Question 11O(i)(b):

| Answer | Marks | Guidance |
|--------|-------|----------|
| $\frac{d^2y}{dt^2} = a^t(\ln a)^2$ | B1 | |

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## Question 11O(ii):

| Answer | Marks | Guidance |
|--------|-------|----------|
| $r = \ln a$ | B1 | States common ratio. |
| $-1 < \ln a < 1 \Rightarrow e^{-1} < a < e$ | M1 A1 | Uses convergence condition for a geometric series. |
| **Total: 3** | | |

## Question 11O(iii):

| Answer | Marks | Guidance |
|--------|-------|----------|
| $\frac{dy}{dx} = \frac{dy}{dt} \times \frac{dt}{dx} = \frac{a^t \ln a}{at^{a-1}} = \left(\frac{\ln a}{a}\right)\left(\frac{a^t}{t^{a-1}}\right)$ | **M1 A1** | Uses chain rule to find $\frac{dy}{dx}$ |
| $\frac{d}{dt}\left(\frac{dy}{dx}\right) = \left(\frac{\ln a}{a}\right)\left(\frac{t^{a-1}a^t \ln a - (a-1)t^{a-2}a^t}{t^{2(a-1)}}\right)$ | **M1 A1** | Uses quotient or product rule to find $\frac{d}{dt}\left(\frac{dy}{dx}\right)$ |
| $\frac{d^2y}{dx^2} = \left(\frac{\ln a}{a^2}\right)\left(\frac{t^{a-1}a^t \ln a - (a-1)t^{a-2}a^t}{t^{3(a-1)}}\right)$ $= \left(\frac{\ln a}{a^2}\right)\left(\frac{\ln a - (a-1)t^{-1}}{t^{2(a-1)}}\right)a^t$ | **M1 A1** | Uses chain rule to find $\frac{d^2y}{dx^2}$ |
| $t = 2 \Rightarrow \frac{d^2y}{dx^2} = 2^{1-2a}\ln a\left(2\ln a - a + 1\right)$ | **A1** | Substitutes $t = 2$, AG |
| | **7** | |