**(a)**

(i) Shape (either W-shape, N-shape or similar): shape can show cubic with turning points. Max. at (0, 0). (2, 0), (or 2 shown on x-axis). | B1, B1, B1 | (3 marks)

(ii) Shape (parabola, curve bending same way throughout): (It need not go below x-axis). Through origin. (6, 0), (or 6 shown on x-axis). | B1, B1, B1 | (3 marks)

**Total: 6 marks**

**(b)** $x^2(x-2) = x(6-x)$ | | M1 |

$x^3 - x^2 - 6x = 0$ | Expand to form 3-term cubic (or 3-term quadratic if divided by x), with all terms on one side. The "= 0" may be implied. | M1 |

$x(x-3)(x+2) = 0$ | $x = \ldots$ | Factor x (or divide by x), and solve quadratic. | M1 |

$x = 3$ and $x = -2$ | || A1 |

$x = -2:$ | $y = -16$ | Attempt y value for a non-zero x value by substituting back into $x^2(x-2)$ or $x(6-x)$. | M1 |

$x = 3:$ | $y = 9$ | Both y values are needed for A1. | A1 |

$(-2, -16)$ and $(3, 9)$ | | | |

$(0, 0)$ | This can just be written down. Ignore any 'method' shown. (But must be seen in part (b)). | | B1 | (7 marks)

**Total: 13 marks**

**Guidance:**
- (a) (i) For the third 'shape' shown above, where a section of the graph coincides with the x-axis, the B1 for (2, 0) can still be awarded if the 2 is shown on the x-axis. For the final B1 in (i), and similarly for (6, 0) in (ii): There must be a sketch. If, for example (2, 0) is written separately from the sketch, the sketch must not clearly contradict this. If (0, 2) instead of (2, 0) is shown on the sketch, allow the mark. Ignore extra intersections with the x-axis.
  - (ii) 2nd B is dependent on 1st B.

- Separate sketches can score all marks.

- (b) Note the dependence of the first three M marks. A common wrong solution is (−2, 0), (3, 0), (0, 0), which scores M0 A0 B1 as the last 3 marks. A solution using no algebra (e.g. trial and error), can score up to 3 marks: M0 M0 M0 A0 M1 A1 B1. (The final A1 requires both y values). Also, if the cubic is found but not solved algebraically, up to 5 marks: M1 M1 M0 A0 M1 A1 B1. (The final A1 requires both y values).

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# GENERAL PRINCIPLES FOR C1 MARKING

## Method mark for solving 3 term quadratic:

1. **Factorisation**
   - $(x^2 + bx + c) = (x + p)(x + q)$, where $|pq| = |c|$, leading to $x = \ldots$
   - $(ax^2 + bx + c) = (mx + p)(nx + q)$, where $|pq| = |c|$ and $|mn| = |a|$, leading to $x = \ldots$

2. **Formula**
   - Attempt to use correct formula (with values for a, b and c).

3. **Completing the square**
   - Solving $x^2 + bx + c = 0$: $(x \pm p)^2 = q + c, p \neq 0, q \neq 0$, leading to $x = \ldots$

## Method marks for differentiation and integration:

1. **Differentiation**
   - Power of at least one term decreased by 1. ($x^n \to x^{n-1}$)

2. **Integration**
   - Power of at least one term increased by 1. ($x^n \to x^{n+1}$)

## Use of a formula

Where a method involves using a formula that has been learnt, the advice given in recent examiners' reports is that the formula should be quoted first.

Normal marking procedure is as follows:

**Method mark for quoting a correct formula and attempting to use it, even if there are mistakes in the substitution of values.**

Where the formula is not quoted, the method mark can be gained by implication from correct working with values, but will be lost if there is any mistake in the working.

## Exact answers

Examiners' reports have emphasised that where, for example, an exact answer is asked for, or working with surds is clearly required, marks will normally be lost if the candidate resorts to using rounded decimals.

## Answers without working

The rubric says that these may gain no credit. Individual mark schemes will give details of what happens in particular cases. General policy is that if it could be done "in your head", detailed working would not be required. Most candidates do show working, but there are occasional awkward cases and if the mark scheme does not cover this, please contact your team leader for advice.

## Misreads

(See the next sheet for a simple example).

A misread must be consistent for the whole question to be interpreted as such.

These are not common. In clear cases, please deduct the first 2 A (or B) marks which would have been lost by following the scheme. (Note that 2 marks is the maximum misread penalty, but that misreads which alter the nature of the difficulty of the question cannot be treated so generously and it will usually be necessary here to follow the scheme as written).

Sometimes following the scheme as written is more generous to the candidate than applying the misread rule, so in this case use the scheme as written.

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# MISREADS

**Question 7.** $3x^2$ misread as $3x^3$

(a) $f(x) = \frac{3x^4}{4} - 6x - \frac{8x^{-1}}{-1}$ | | M1 A1 A0 |

$1 = 12 - 12 + 4 + C$, $C = -3$ | | M1 A0 |

(b) $m = 3 \times 2^3 - 6 - \frac{8}{2^2} = 16$ | | M1 A1 |

Eqn. of tangent: $y - 1 = 16(x - 2)$ | | M1 |

$y = 16x - 31$ | | A1 |