OCR MEI AS Paper 1 2019 June — Question 1 3 marks

Exam BoardOCR MEI
ModuleAS Paper 1 (AS Paper 1)
Year2019
SessionJune
Marks3
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicDiscriminant and conditions for roots
TypeProve/show always positive
DifficultyEasy -1.2 This is a straightforward discriminant question requiring rearrangement to standard form (2x² - x + 7 = 0), then showing b² - 4ac < 0. It's simpler than average A-level questions as it only requires one standard technique with no problem-solving insight, though the 'show detailed reasoning' requirement adds minimal complexity.
Spec1.02d Quadratic functions: graphs and discriminant conditions

1 In this question you must show detailed reasoning. Show that the equation \(x = 7 + 2 x ^ { 2 }\) has no real roots.

Question 1:
DR (Detailed Reasoning required)
AnswerMarks Guidance
AnswerMarks Guidance
Rearrange as \(2x^2 - x + 7 = 0\)M1 (AO 1.1a)
Discriminant is \((-1)^2 - 4 \times 2 \times 7\)M1 dep (AO 1.1a)
\(= -55 < 0\) so no real rootsA1 (AO 2.2a) Must be clearly argued from a correct discriminant which need not be evaluated if clearly negative. Expression for discriminant must be precise if not evaluated.
[3]
Alternative method:
AnswerMarks Guidance
AnswerMarks Guidance
Rearrange as \(2x^2 - x + 7 = 0\)M1
Attempt to complete the squareM1 dep Allow for \(2(x-0.25)^2 + \ldots\) soi
\(2(x - 0.25)^2 + 6.875 = 0\)
\(x - 0.25 = \pm\sqrt{-3.4375}\) so no real rootsA1 Must be clearly argued from correct working
[3]
Second alternative method:
AnswerMarks Guidance
AnswerMarks Guidance
Rearrange as \([y =]\, 2x^2 - x + 7 = 0\)M1 Must equate to zero
Differentiate \(\dfrac{\mathrm{d}y}{\mathrm{d}x} = 4x - 1 = 0\)M1 dep
Stationary point at \((0.25,\ 6.875)\)
Stationary point is minimum so \(y \geq 6.875\) so is never zeroA1 Must be clearly argued from correct working
[3]
# Question 1:

**DR** (Detailed Reasoning required)

| Answer | Marks | Guidance |
|--------|-------|----------|
| Rearrange as $2x^2 - x + 7 = 0$ | M1 (AO 1.1a) | |
| Discriminant is $(-1)^2 - 4 \times 2 \times 7$ | M1 dep (AO 1.1a) | |
| $= -55 < 0$ so no real roots | A1 (AO 2.2a) | Must be clearly argued from a correct discriminant which need not be evaluated if clearly negative. Expression for discriminant must be precise if not evaluated. |
| **[3]** | | |

**Alternative method:**

| Answer | Marks | Guidance |
|--------|-------|----------|
| Rearrange as $2x^2 - x + 7 = 0$ | M1 | |
| Attempt to complete the square | M1 dep | Allow for $2(x-0.25)^2 + \ldots$ soi |
| $2(x - 0.25)^2 + 6.875 = 0$ | | |
| $x - 0.25 = \pm\sqrt{-3.4375}$ so no real roots | A1 | Must be clearly argued from correct working |
| **[3]** | | |

**Second alternative method:**

| Answer | Marks | Guidance |
|--------|-------|----------|
| Rearrange as $[y =]\, 2x^2 - x + 7 = 0$ | M1 | Must equate to zero |
| Differentiate $\dfrac{\mathrm{d}y}{\mathrm{d}x} = 4x - 1 = 0$ | M1 dep | |
| Stationary point at $(0.25,\ 6.875)$ | | |
| Stationary point is minimum so $y \geq 6.875$ so is never zero | A1 | Must be clearly argued from correct working |
| **[3]** | | |
1 In this question you must show detailed reasoning.
Show that the equation $x = 7 + 2 x ^ { 2 }$ has no real roots.

\hfill \mbox{\textit{OCR MEI AS Paper 1 2019 Q1 [3]}}