| Exam Board | OCR |
|---|---|
| Module | H240/03 (Pure Mathematics and Mechanics) |
| Year | 2018 |
| Session | September |
| Marks | 5 |
| Topic | Applied differentiation |
| Type | Stationary points and nature classification |
| Difficulty | Easy -1.8 This is a routine algebraic manipulation (completing the square) followed by immediate reading off of the minimum point from vertex form. Both parts require only standard GCSE/AS techniques with no problem-solving or calculus despite being in a Mechanics module. The question is significantly easier than average A-level work. |
| Spec | 1.02e Complete the square: quadratic polynomials and turning points |
| Answer | Marks | Guidance |
|---|---|---|
| \(4x^2 - 12x + 3 = 4(x^2 - 3x) + 3\) | M1 | Take out a factor of 4 |
| \(x^2 - 3x = \left(x - \frac{3}{2}\right)^2 - \frac{9}{4}\) | A1 | |
| \(4\left(x - \frac{3}{2}\right)^2 - 4 \times \frac{9}{4} + 3 = 4\left(x - \frac{3}{2}\right)^2 - 6\) AG | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| \(4\left(x - \frac{3}{2}\right)^2 - 6 = 4\left[x^2 - 3x + \frac{9}{4}\right] - 6\) | M1 | multiply out square bracket |
| \(= 4x^2 - 4x \times 3 + 4 \times \frac{9}{4} - 6\) | A1 | intermediate step |
| \(= 4x^2 - 12x + 3\) AG | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Minimum point is \(\left(\frac{3}{2}, -6\right)\) | B1, B1 | [2 marks total] |
### (i)
$4x^2 - 12x + 3 = 4(x^2 - 3x) + 3$ | M1 | Take out a factor of 4
$x^2 - 3x = \left(x - \frac{3}{2}\right)^2 - \frac{9}{4}$ | A1 |
$4\left(x - \frac{3}{2}\right)^2 - 4 \times \frac{9}{4} + 3 = 4\left(x - \frac{3}{2}\right)^2 - 6$ AG | A1 |
**Alternative method:**
$4\left(x - \frac{3}{2}\right)^2 - 6 = 4\left[x^2 - 3x + \frac{9}{4}\right] - 6$ | M1 | multiply out square bracket
$= 4x^2 - 4x \times 3 + 4 \times \frac{9}{4} - 6$ | A1 | intermediate step
$= 4x^2 - 12x + 3$ AG | A1 |
**[3 marks total]**
### (ii)
Minimum point is $\left(\frac{3}{2}, -6\right)$ | B1, B1 | **[2 marks total]**
---1 (i) Show that $4 x ^ { 2 } - 12 x + 3 = 4 \left( x - \frac { 3 } { 2 } \right) ^ { 2 } - 6$.\\
(ii) State the coordinates of the minimum point of the curve $y = 4 x ^ { 2 } - 12 x + 3$.
\hfill \mbox{\textit{OCR H240/03 2018 Q1 [5]}}