## Question 10:

| Answer/Working | Mark | Guidance |
|---|---|---|
| $\int y^{-\frac{1}{3}} \, dy = \int \frac{(12x+9)}{x} \, dx$ | B1 | Separates variables correctly. No need for integral signs |
| $\frac{3}{2}y^{\frac{2}{3}} = 12x + 9\ln x + c$ | M1 | Integrates left hand side $Ay^{-\frac{1}{3}}$ to give $By^{\frac{2}{3}}$ |
| | M1 | Integrates the right hand side $A + \frac{B}{x}$ and obtains $Cx + D\ln x + (c)$ |
| | A1 | Correct answer for lhs and rhs |
| Substitutes $y=8, x=1$: $\frac{3}{2}(8)^{\frac{2}{3}} = 12(1) + 9\ln(1) + c$, giving $c = -6$ | M1 | Substitutes $y=8$ at $x=1$ into their function, finds constant $c$ and attempts to find $y^2 = \cdots$ |
| $y^2 = (8x + 6\ln x - 4)^3$ | A1 | Correct solution only |

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## Question 11a:

| Answer/Working | Mark | Guidance |
|---|---|---|
| $-6x^2 + 24x - 9 \equiv A(x-2)(1-3x) + B(1-3x) + C(x-2)$ | M1 | Uses a correct identity, either full expansion or $10x - 5 \equiv B(1-3x) + C(x-2)$, in a complete method to find $B$ and $C$ |
| $A = 2$ | B1 | |
| Attempts to find $B$ or $C$ from identity | M1 | By substitution or comparing coefficients |
| $B = -3$ and $C = 1$ | A1 | |

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## Question 11b:

| Answer/Working | Mark | Guidance |
|---|---|---|
| $f'(x) = \frac{3}{(x-2)^2} + \frac{3}{(1-3x)^2}$ | M1 | Differentiates $2 - 3(x-2)^{-1} + (1-3x)^{-1}$ to give $A(x-2)^{-2} + B(1-3x)^{-2}$ |
| $3(x-2)^{-2} + 3(1-3x)^{-2}$ | A1 | Correct answer only |

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## Question 11c:

| Answer/Working | Mark | Guidance |
|---|---|---|
| Both terms of $f'(x)$ are positive for all values of $x$, so $f(x)$ is increasing. $f'(x) = +ve + +ve > 0$ | A1 | |

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## Question 12a:

| Answer/Working | Mark | Guidance |
|---|---|---|
| Either splits $\frac{\sec^2 x - 1}{\sec^2 x} = \frac{\sec^2 x}{\sec^2 x} - \frac{1}{\sec^2 x}$ and uses $\frac{1}{\sec^2 x} = \cos^2 x$; or states $\sec^2 x - 1 = \tan^2 x$ and $\sec^2 x = \frac{1}{\cos^2 x}$ | M1 | |
| Either replaces $1 - \cos^2 x = \sin^2 x$, or replaces $\frac{\tan^2 x}{\frac{1}{\cos^2 x}} = \frac{\sin^2 x}{\cos^2 x} \times \cos^2 x$ | M1 | |
| Complete correct proof with all necessary intermediate steps, no errors | A1 | |

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## Question 12b:

| Answer/Working | Mark | Guidance |
|---|---|---|
| Uses $\frac{\sec^2 x - 1}{\sec^2 x} = \sin^2 x$ and replaces $\cos 2x = 1 - 2\sin^2 x$, obtains $A\sin^2 x = 1$ | M1 | |
| $4\sin^2 x = 1$ | A1 | Correct equation |
| Two of $x = 30°, 150°, -210°, -330°, -30°, -150°, 210°, 330°$ | A1 | |
| $x = 30°, 150°, -210°, -330°, -30°, -150°, 210°, 330°$ | A1 | All correct |

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## Question 13a:

| Answer/Working | Mark | Guidance |
|---|---|---|
| $\int 1 \times \ln x = x\ln x - \int x \cdot \frac{1}{x} \, dx$ | M1 | Uses integration by parts the correct way around |
| $x\ln x - \int x \cdot \frac{1}{x} \, dx$ | A1 | Correct expression |
| $\int 1 \, dx = Dx$, $D \neq 0$ | M1 | Integrates the second term |
| $x\ln x - x + c$ | A1 | Correct integration with $+c$ |