# Question 16:

## Part (a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\frac{1}{2}x + 1 = x^2 - 4x + 3$ | M1 | |
| $2x^2 - 9x + 4 = 0 \Rightarrow x = \frac{1}{2}$ or $x = 4$ | dM1 A1 | |
| $y = \frac{5}{4}$ or $y = 3$ | dM1 A1 | |

## Part (b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Curve meets $x$-axis at $x=3$ and $x=1$ | M1 A1 | No need to see $y=0$ |

## Part (c) — Way 1:
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\int x^2 - 4x + 3\, dx = \frac{1}{3}x^3 - 2x^2 + 3x$ | M1 A1 | |
| Use limits 1 and $\frac{1}{2}$ for $A_1$ | M1 | |
| Use limits 4 and 3 for $A_2$ | M1 | |
| Area of trapezium $= \frac{1}{2}(a+b)\times h = \frac{1}{2}(\frac{5}{4}+3)\times(4-\frac{1}{2})$ | M1 | |
| $7.4375$ $\left(7\frac{7}{16}\right)$ $\left(\frac{119}{16}\right)$ | A1 | May be implied by correct final answer |
| Area of region $=$ Area of trapezium $- A_1 - A_2$ | ddddM1 | Dependent on all previous method marks |
| $= 7.4375 - \frac{7}{24} - \frac{4}{3} = \frac{93}{16}$ or $5.8125$ | A1 | |

## Part (c) — Way 2:
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\int -x^2 + \frac{9}{2}x - 2\, dx = -\frac{x^3}{3} + \frac{9}{4}x^2 - 2x$ | M1A1 | |
| Use limits $\frac{1}{2}$ and 4 | M1 | |
| $\pm\int x^2 - 4x + 3\, dx = \pm\left(\frac{1}{3}x^3 - 2x^2 + 3x\right)$ | M1 | |
| Use limits 1 and 3 to obtain $A_6 = \pm\frac{4}{3}$ | M1A1 | |
| Area $= (A_3 + A_4 + A_5 + A_6) - A_6$ | ddddM1 | Dependent on all previous method marks |
| $6\frac{2}{3} + \frac{23}{48} - \frac{4}{3} = \frac{93}{16}$ | A1 | |

## Part (c) — Way 3:
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\int -x^2 + \frac{9}{2}x - 2\, dx = -\frac{x^3}{3} + \frac{9}{4}x^2 - 2x$ | M1A1 | |
| Use limits 1 and $\frac{1}{2}$ for $A_3$ | M1 | |
| Use limits 4 and 3 for $A_4$ | M1 | |
| Area of trapezium $= \frac{1}{2}(\frac{3}{2}+\frac{5}{2})\times(3-1) = 4$ | M1 A1 | |
| Area $= A_3 + A_4 + A_5$ | ddddM1 | Dependent on all previous method marks |
| $\frac{19}{48} + \frac{17}{12} + 4 = \frac{93}{16}$ | A1 | |

## Question 16:

### Part (c) – Way 4: Alternative method (larger trapezium minus areas)

| Answer/Working | Mark | Guidance |
|---|---|---|
| $\int x^2 - 4x + 3 \, dx = \frac{1}{3}x^3 - 2x^2 + 3x$ | M1 A1 | Attempt at integration; correct integration |
| Use limits 4 and $\frac{1}{2}$: $[(\frac{1}{3}(4)^3 - 2(4)^2 + 3\times4) - (\frac{1}{3}(\frac{1}{2})^3 - 2(\frac{1}{2})^2 + 3\times(\frac{1}{2}))]$ giving $A_1 + A_2 - A_6$ **AND** use limits 3 and 1: $\pm[(\frac{1}{3}(3)^3 - 2(3)^2 + 3\times3) - (\frac{1}{3}(1)^3 - 2(1)^2 + 3\times(1))] \pm A_6$ | M2 | Both sets of limits applied correctly |
| Area of trapezium $= \frac{1}{2}(a+b)\times h = \frac{1}{2}(\frac{5}{4}+3)\times(4-\frac{1}{2}) = \ldots$ or $\int_{\frac{1}{2}}^{4}(\frac{1}{2}x+1)dx = [\frac{1}{4}x^2+x]_{\frac{1}{2}}^{4} = (4+4) - (\frac{1}{16}+\frac{1}{2}) = \ldots$ | M1 | Finds area of appropriate trapezium |
| $7.4375 \quad (7\frac{7}{16})$ | A1 | Correct area of trapezium (may be implied by correct final answer) |
| Uses correct combination of correct areas: Area of region $= 7.4375 - (A_1 + A_2 - A_6 + A_6)$ | ddddM1 | Dependent on all previous method marks; correct final combination |
| $= 7.4375 - \left(\frac{7}{24} + \frac{4}{3}\right) = \frac{93}{16}$ | A1 | Any correct form of this exact answer |

**Total: [8] marks; 15 marks for whole question**

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### Notes by Part:

**Part (a):**
| Guidance |
|---|
| **M1:** Puts equations equal or finds $x$ in terms of $y$ and substitutes, or substitutes for $x$ |
| **dM1:** Solves three-term quadratic in $x$ to obtain $x = \ldots$ or in $y$ to obtain $y = \ldots$ (Dependent on **first** M) |
| **A1:** Both answers correct |
| **dM1:** Obtains at least one value for $y$ or $x$ (Dependent on **first** M) |
| **A1:** Both correct |
| **Note:** Allow candidates to obtain $x^2 - \frac{9}{2}x + 2 = 0$ and solve as $(2x-1)(x-4)=0 \Rightarrow x = \frac{1}{2}, 4$ |
| The coordinates do not need to be 'paired' |

**Part (b):**
| Guidance |
|---|
| **M1:** Attempts to solve $0 = x^2 - 4x + 3$ according to usual rules |
| **A1:** cao |
| Attempts by T&I can score both marks for $x=1$ and $x=3$. If one solution obtained, score M1A0 |
| For (c) do not allow 'mixed' methods — apply the method giving most credit to candidate |

**Part (c) – Way 1:**
| Guidance |
|---|
| **M1:** Attempt at integration of given quadratic $(x^n \to x^{n+1}$ at least once$)$ |
| **A1:** Correct integration of given quadratic expression |
| **M1:** Finds area of $A_1$; **M1:** Finds area of $A_2$ |
| **M1:** Finds area of appropriate trapezium |
| **A1:** Correct area of trapezium $7.4375\ (7\frac{7}{16})$ |
| **ddddM1:** Correct final combination; **A1:** Any correct form of exact answer |

**Part (c) – Way 2:**
| Guidance |
|---|
| **M1:** Attempt at integration of $\pm$(quadratic expression $-$ given line) $(x^n \to x^{n+1}$ at least once$)$ |
| **A1:** Correct integration as in mark scheme (allow if terms not collected/simplified; sign errors before valid attempt score M1A0) |
| **M1:** Uses limits $\frac{1}{2}$ and 4 on *subtracted* integration |
| **M1:** Attempts to integrate curve; **M1:** Uses limits 1 and 3 on integrated curve $C$ |
| **A1:** Obtains $A_6 = \pm\frac{4}{3}$ |
| **ddddM1:** Correct final combination; **A1:** Any correct form of exact answer |
| Note: Common error — using limits $\frac{1}{2}$ and 4 on *subtracted* integration and stopping gives area $\frac{343}{48}$, usually scoring 3/8 |

**Part (c) – Way 3:**
| Guidance |
|---|
| **M1:** Attempt at integration of $\pm$(quadratic $-$ line) $(x^n \to x^{n+1}$ at least once$)$ |
| **A1:** Correct integration (allow if not simplified; sign errors before valid attempt score M1A0) |
| **M1:** Uses limits $\frac{1}{2}$ and 1 on *subtracted* integration |
| **M1:** Uses limits 4 and 3 on *subtracted* integration |
| **M1:** Finds area of appropriate trapezium; **A1:** Correct area of trapezium $= 4$ |
| **ddddM1:** Correct final combination; **A1:** Any correct form of exact answer |

**Part (c) – Way 4:**
| Guidance |
|---|
| **M1:** Attempt at integration of given quadratic $(x^n \to x^{n+1}$ at least once$)$ |
| **A1:** Correct integration of given quadratic |
| **M2:** Finds area of $A_1 + A_2 - A_6$ using limits $\frac{1}{2}$ and 4, **and** finds area of $A_6$ using limits 1 and 3 |
| **M1:** Finds area of appropriate trapezium; **A1:** Correct area of trapezium $7.4375\ (7\frac{7}{16})$ |
| **ddddM1:** Correct final combination; **A1:** Any correct form of exact answer |