# Question 16:

## Part (a):

| Answer/Working | Mark | Guidance |
|---|---|---|
| $A(0, 12)$ | B1 | |
| $2x^2 - 11x + 12 = 0 \Rightarrow x = \ldots$ | M1 | |
| $\left(\frac{3}{2}, 0\right), (4, 0)$ | A1 | |

**(3 marks)**

## Part (b) — WAY 1:

| Answer/Working | Mark | Guidance |
|---|---|---|
| At $D$, $x = \frac{11}{2}$ | B1 | Correct $x$-coordinate at $D$ |
| $\int 2x^2 - 11x + 12\, dx = \frac{2x^3}{3} - \frac{11x^2}{2} + 12x\ (+c)$ | M1A1 | Integrating the curve or curve-line; correct integration |
| Substitutes limits $x = $ "4" to $x = $ "5.5" into integrated expression | dM1 | Must find area completely above $x$-axis |
| $= 10.125$ | A1 | Correct value for area |
| Area of triangle $AOC = \frac{1}{2}\times$"4"$\times$"12" $=$ "24"; Area of $R = 24 + 10.125$ | ddM1 | Correct method to find shaded area |
| $= 34.125$ $\left(=\frac{273}{8}\right)$ | A1 | cso |

## Part (b) — WAY 2:

| Answer/Working | Mark | Guidance |
|---|---|---|
| At $D$, $x = \frac{11}{2}$ | B1 | |
| $\int 2x^2 - 11x + 12\, dx = \frac{2x^3}{3} - \frac{11x^2}{2} + 12x\ (+c)$ | M1A1 | |
| Substitution of limits "4" to "5.5" | dM1 | |
| $= 7.875$ | A1 | |
| Trapezium area $= \frac{1}{2}\times$"12"$\times\left(\frac{11}{2} + \left(\frac{11}{2}-\text{"4"}\right)\right) = 42$ | ddM1 | |
| $= 34.125$ $\left(=\frac{273}{8}\right)$ | A1cso | |

## Part (b) — WAY 3:

| Answer/Working | Mark | Guidance |
|---|---|---|
| At $D$, $x = \frac{11}{2}$ | B1 | |
| $\int 2x^2 - 11x + 12\, dx = \frac{2x^3}{3} - \frac{11x^2}{2} + 12x\ (+c)$ | M1A1 | |
| Substitution of limits "4" to "5.5" | dM1 | |
| $= 7.875$ | A1 | |
| Unshaded triangle area (vertices $(0,12)$, $("4",0)$, $(0,0)$) $= \frac{1}{2}\times$"12"$\times$"4" $= 24$ | ddM1 | |
| Area of $R = 12\times 5.5 - 7.875 - 24 = 34.125$ $\left(=\frac{273}{8}\right)$ | A1cso | |

**(10 marks)**

# Mark Scheme

## Part (a):

| Answer | Mark | Guidance |
|--------|------|----------|
| $(0, 12)$ | B1 | May be indicated on graph. Accept 12 on $y$-axis, or $x=0, y=12$ |
| Solves the quadratic | M1 | Apply general marking principles for solving a quadratic. Award for $\frac{3}{2}$ or $4$ without working shown |
| $\left(\frac{3}{2}, 0\right), (4, 0)$ | A1 | May be indicated on graph. Accept $\frac{3}{2}$ and $4$ on $x$-axis. Can also be written as $x=\frac{3}{2}, y=0$ and $x=4, y=0$. Ignore labelling of coordinates as $A, B$ and $C$ |

## Part (b):

| Answer | Mark | Guidance |
|--------|------|----------|
| At $D$, $x = \frac{11}{2}$ | B1 | May be indicated on the graph |
| Attempts to integrate equation of curve (may be combined with line $y=12$) | M1 | Award for any one term $x^n \rightarrow x^{n+1}$ but powers must have been processed (do not award for $x^{2+1}$) |
| $\frac{2x^3}{3} - \frac{11x^2}{2} + 12x\ (+c)$ or $12x - \frac{2x^3}{3} + \frac{11x^2}{2} - 12x\ (+c)$ | A1 | With or without constant of integration. Ignore spurious notation |
| Substitutes limits and subtracts to find area completely above $x$-axis; integrates between $x=4$ and $x=5.5$, or $x=0$ and $x=\frac{3}{2}$ | dM1 | Dependent on previous method mark |
| $10.125$ or $7.875\ \left(=\frac{63}{8}\right)$ | A1 | |
| Complete method for finding required area (all previous method marks in (b) must be scored) | ddM1 | |
| $34.125$ or exact equivalent $\frac{273}{8}$ | A1 | |