OCR MEI C1 — Question 4 12 marks

Exam BoardOCR MEI
ModuleC1 (Core Mathematics 1)
Marks12
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicDiscriminant and conditions for roots
TypeProve/show always positive
DifficultyModerate -0.3 This is a multi-part question covering standard C1 techniques: factorising a straightforward quadratic, sketching a parabola, using the discriminant, and finding intersection points via completing the square. While it has four parts, each individual step is routine and requires only direct application of well-practiced methods without novel problem-solving.
Spec1.02d Quadratic functions: graphs and discriminant conditions1.02f Solve quadratic equations: including in a function of unknown1.02n Sketch curves: simple equations including polynomials1.02q Use intersection points: of graphs to solve equations

4
  1. Solve, by factorising, the equation \(2 x ^ { 2 } - x - 3 = 0\).
  2. Sketch the graph of \(y = 2 x ^ { 2 } - x - 3\).
  3. Show that the equation \(x ^ { 2 } - 5 x + 10 = 0\) has no real roots.
  4. Find the \(x\)-coordinates of the points of intersection of the graphs of \(y = 2 x ^ { 2 } - x - 3\) and \(y = x ^ { 2 } - 5 x + 10\). Give your answer in the form \(a \pm \sqrt { b }\).

Question 4(i):
AnswerMarks Guidance
Answer/WorkingMarks Guidance
\((2x-3)(x+1)\)M2 M1 for factors with one sign error or giving two terms correct. Allow M1 for \(2(x - 1.5)(x+1)\) with no better factors seen
\(x = \tfrac{3}{2}\) and \(-1\) obtainedB1 Or ft their factors
Question 4(ii):
AnswerMarks Guidance
Answer/WorkingMarks Guidance
Graph of quadratic the correct way up and crossing both axesB1
Crossing \(x\)-axis only at \(\tfrac{3}{2}\) and \(-1\), or ft from their roots in (i), or their factors if roots not givenB1 For \(x = \tfrac{3}{2}\) condone 1 and 2 marked on axis and crossing roughly halfway between; intersections must be shown labelled or worked out nearby
Crossing \(y\)-axis at \(-3\)B1
Question 4(iii):
AnswerMarks Guidance
Answer/WorkingMarks Guidance
Use of \(b^2 - 4ac\) with numbers substituted (condone one error in substitution) (may be in quadratic formula)M1 May be in formula or \((x - 2.5)^2 = 6.25 - 10\) or \((x-2.5)^2 + 3.75 = 0\) oe (condone one error)
\(25 - 40 < 0\) or \(-15\) obtainedA1 Or \(\sqrt{-15}\) seen in formula, or \((x - 2.5)^2 = -3.75\) oe, or \(x = 2.5 \pm \sqrt{-3.75}\) oe
Question 4(iv):
AnswerMarks Guidance
Answer/WorkingMark Guidance Notes
\(2x^2 - x - 3 = x^2 - 5x + 10\) o.e.M1 attempt at eliminating \(y\) by substitution or subtraction
\(x^2 + 4x - 13 [= 0]\)M1 or \((x+2)^2 = 17\); for rearranging to form \(ax^2 + bx + c [=0]\) or to completing square form; condone one error for each of 2nd and 3rd M1s
use of quadratic formula on resulting equation (do not allow for original quadratics used)M1 or \(x + 2 = \pm\sqrt{17}\) o.e.; 2nd and 3rd M1s may be earned for good attempt at completing square as far as roots obtained
\(-2 \pm \sqrt{17}\) caoA1
## Question 4(i):

| Answer/Working | Marks | Guidance |
|---|---|---|
| $(2x-3)(x+1)$ | M2 | M1 for factors with one sign error or giving two terms correct. Allow M1 for $2(x - 1.5)(x+1)$ with no better factors seen |
| $x = \tfrac{3}{2}$ and $-1$ obtained | B1 | Or ft their factors |

---

## Question 4(ii):

| Answer/Working | Marks | Guidance |
|---|---|---|
| Graph of quadratic the correct way up and crossing both axes | B1 | |
| Crossing $x$-axis only at $\tfrac{3}{2}$ and $-1$, or ft from their roots in (i), or their factors if roots not given | B1 | For $x = \tfrac{3}{2}$ condone 1 and 2 marked on axis and crossing roughly halfway between; intersections must be shown labelled or worked out nearby |
| Crossing $y$-axis at $-3$ | B1 | |

---

## Question 4(iii):

| Answer/Working | Marks | Guidance |
|---|---|---|
| Use of $b^2 - 4ac$ with numbers substituted (condone one error in substitution) (may be in quadratic formula) | M1 | May be in formula or $(x - 2.5)^2 = 6.25 - 10$ or $(x-2.5)^2 + 3.75 = 0$ oe (condone one error) |
| $25 - 40 < 0$ or $-15$ obtained | A1 | Or $\sqrt{-15}$ seen in formula, or $(x - 2.5)^2 = -3.75$ oe, or $x = 2.5 \pm \sqrt{-3.75}$ oe |

## Question 4(iv):

| Answer/Working | Mark | Guidance Notes |
|---|---|---|
| $2x^2 - x - 3 = x^2 - 5x + 10$ o.e. | **M1** | attempt at eliminating $y$ by substitution or subtraction |
| $x^2 + 4x - 13 [= 0]$ | **M1** | or $(x+2)^2 = 17$; for rearranging to form $ax^2 + bx + c [=0]$ or to completing square form; condone one error for each of 2nd and 3rd **M1s** |
| use of quadratic formula on resulting equation (do not allow for original quadratics used) | **M1** | or $x + 2 = \pm\sqrt{17}$ o.e.; 2nd and 3rd **M1s** may be earned for good attempt at completing square as far as roots obtained |
| $-2 \pm \sqrt{17}$ cao | **A1** | |
4 (i) Solve, by factorising, the equation $2 x ^ { 2 } - x - 3 = 0$.\\
(ii) Sketch the graph of $y = 2 x ^ { 2 } - x - 3$.\\
(iii) Show that the equation $x ^ { 2 } - 5 x + 10 = 0$ has no real roots.\\
(iv) Find the $x$-coordinates of the points of intersection of the graphs of $y = 2 x ^ { 2 } - x - 3$ and $y = x ^ { 2 } - 5 x + 10$. Give your answer in the form $a \pm \sqrt { b }$.

\hfill \mbox{\textit{OCR MEI C1  Q4 [12]}}
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