# Question 9:

## Part (i):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $f(3) = -108 + 81 + 30 - 3 = 0$, hence $(x-3)$ is a factor | B1 | Show that $f(3) = 0$, detail required. Substitute $x = 3$ and confirm $f(3) = 0$. Must show detail of substitution. Allow $f(3) = -4\times3^3 + 9\times3^2 + 10\times3 - 3 = 0$. |
| State $(x-3)$ as factor | B1 | **2 marks** State $(x-3)$ as factor (allow $(3-x)$). Must be seen in (i). |

## Part (ii):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $f(x) = (x-3)(-4x^2 - 3x + 1)$ | M1 | Attempt complete division by $(x-3)$, or equivalent. Must be full attempt to find three term quadratic. |
| Obtain $-4x^2 - 3x + c$ or $-4x^2 + bx + 1$ | A1 | $c, b$ non-zero constants. |
| $(x-3)(-4x^2 - 3x + 1)$ | A1 | **3 marks** Obtain $(x-3)(-4x^2 - 3x + 1)$. Needs to be written as a product. Allow $-(x-3)(4x^2+3x-1)$, but $(x-3)(4x^2+3x-1)$ is A0. |

## Part (iii):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $-4x^2 - 3x + 1 = 0$, $(1-4x)(x+1) = 0$ | M1 | Attempt to solve quadratic. If factorising needs to give two correct terms when brackets expanded. If using formula allow sign slips only. |
| $x = \frac{1}{4}, x = -1$ | A1 | **2 marks** Obtain $\left(\frac{1}{4}, 0\right)$, $(-1, 0)$. Condone only $x$ values given rather than coordinates. Allow if $x = 3$ is still present as well. |

# Question (iv):

## Part (iv) - Integration and Area

**Step 1: Finding the integral**

$\int f(x)dx = -x^4 + 3x^3 + 5x^2 - 3x$ | **B1** | Allow unsimplified coefficients. Condone $+ c$.

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**Step 2: Applying limits**

$F(3) - F(\frac{1}{4}) = (36) - (\frac{-101}{256}) = 36\frac{101}{256}$

$F(\frac{1}{4}) - F(-1) = (\frac{-101}{256}) - (4) = -4\frac{101}{256}$

| **M1*** | Attempt $F(3) - F(\frac{1}{4})$ or $F(\frac{1}{4}) - F(-1)$. Allow use of incorrect limits from their (iii). Limits need to be in correct order, and subtraction. Allow slips when evaluating but clear subtraction attempt must be seen or implied at least once. If minimal method shown then it must appear to be a plausible attempt eg $F(3) = 198$ or even $F(3) - F(\frac{1}{4}) = 198.4$.

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**Step 3: At least one correct area**

| **A1** | Obtain at least one correct area, including decimal equivs. Obtain $36\frac{101}{256}$ or $\frac{9317}{256}$ or $36.4$, or $-4\frac{101}{256}$ or $-\frac{1125}{256}$ or $-4.4$. Can get A1 if both areas attempted and one is correct but the other isn't.

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**Step 4: Full method for total area**

| **M1d*** | Attempt full method to find total area including dealing correctly with negative area. Need to see modulus of negative integral from attempt at $F(\frac{1}{4}) - F(-1)$ (just changing sign from $-$ve to $+$ve is sufficient). If values incorrect in (iii) then can only get this mark if their integral gives negative value. Need to have positive integral from $F(3) - F(\frac{1}{4})$.

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**Step 5: Final answer**

Hence area $= 36\frac{101}{256} + 4\frac{101}{256} = 40\frac{101}{128}$

| **A1** [5 marks, **12** total] | Obtain $40\frac{101}{128}$ or $\frac{5221}{128}$ or $40.8$. Allow exact fraction (including unsimplified ie $\frac{10442}{256}$), or decimal answer to 3dp or better (rounding to 40.8 with no errors seen).

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**SR:** If candidate attempts $F(3) - F(\frac{1}{4})$ **and** $F(-1) - F(\frac{1}{4})$ as an alternative method for dealing with negative area then mark as:
- **B1** correct integral
- **M2** complete method
- **A1** obtain one correct area
- **A1** obtain correct total area

Any attempts using this method must be fully supported by evidence of intention, especially $-1$ as top limit and $\frac{1}{4}$ as bottom limit used consistently throughout integration attempt. It should not be awarded if candidate appears to have simply confused their order of subtraction.