| Exam Board | OCR |
|---|---|
| Module | C1 (Core Mathematics 1) |
| Year | 2013 |
| Session | June |
| Marks | 12 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Curve Sketching |
| Type | Sketch with inequalities or regions |
| Difficulty | Moderate -0.8 This is a straightforward C1 question testing basic quadratic curve sketching, differentiation for decreasing functions, and solving a quadratic equation. All parts use standard techniques with no problem-solving insight required—factorising to find intercepts, setting dy/dx < 0, and finding distance between intersection points. Easier than average A-level content. |
| Spec | 1.02n Sketch curves: simple equations including polynomials1.02q Use intersection points: of graphs to solve equations1.07o Increasing/decreasing: functions using sign of dy/dx |
| Answer | Marks | Guidance |
|---|---|---|
| Marks: M1 | A1 | B1 |
| Guidance: Correct method to find roots | Correct roots | Reasonably symmetrical positive quadratic curve, must cross x axis |
| Answer | Marks | Guidance |
|---|---|---|
| Marks: M1 | A1 | A1 FT [3] |
| Guidance: Attempt to find x coordinate of vertex by differentiating and equating/comparing to zero, completing the square, finding the mid-point of their roots oe | cao | \(x <\) their vertex, allow \(\leq\) |
| Answer | Marks | Guidance |
|---|---|---|
| Marks: M1 | M1 | M1 |
| Guidance: Set quadratic expression equal to 4 | Correct method to solve resulting three term quadratic | Must have both solutions – no mark for one spotted root |
## (i)
**Answer:** $(2x+3)(x-2) = 0$; $x = -\frac{3}{2}, x = 2$; [Graph of quadratic]
**Marks:** M1 | A1 | B1 | B1 | B1 [5]
**Guidance:** Correct method to find roots | Correct roots | Reasonably symmetrical positive quadratic curve, must cross x axis | y intercept $(0, -6)$ only | Good curve, with correct roots indicated and min point in 4th quadrant (not on axis)
**Additional notes:** Indicated on graph or clearly stated, but there must be a curve; Only allow final B1 if curve is clearly intended to be a quadratic symmetrical about min point in 4th quadrant
## (ii)
**Answer:** $\frac{dy}{dx} = 4x - 1 = 0$; Vertex when $x = \frac{1}{4}$; $x < \frac{1}{4}$
**Marks:** M1 | A1 | A1 FT [3]
**Guidance:** Attempt to find x coordinate of vertex by differentiating and equating/comparing to zero, completing the square, finding the mid-point of their roots oe | cao | $x <$ their vertex, allow $\leq$
**Additional notes:** SC Award B1 (FT) for $x < 0$ if clearly from their graph; NB Look for solution to 9ii done in the space below 9i graph
## (iii)
**Answer:** $2x^2 - x - 6 = 4$; $2x^2 - x - 10 = 0$; $(2x - 5)(x + 2) = 0$; $x = \frac{5}{2}, x = -2$; Distance $PQ = 4\frac{1}{2}$
**Marks:** M1 | M1 | M1 | A1 | B1FT [4]
**Guidance:** Set quadratic expression equal to 4 | Correct method to solve resulting three term quadratic | Must have both solutions – no mark for one spotted root | FT from their x values found from their resulting quadratic, provided $y = 4$
**Additional notes:** Not $2x^2 - x - 6 = 0$ with no use of 4; Allow $\frac{9}{2}$ oe, but do not accept unsimplified expressions like $\sqrt{\frac{81}{4}}$
---\begin{enumerate}[label=(\roman*)]
\item Sketch the curve $y = 2x^2 - x - 6$, giving the coordinates of all points of intersection with the axes. [5]
\item Find the set of values of $x$ for which $2x^2 - x - 6$ is a decreasing function. [3]
\item The line $y = 4$ meets the curve $y = 2x^2 - x - 6$ at the points $P$ and $Q$. Calculate the distance $PQ$. [4]
\end{enumerate}
\hfill \mbox{\textit{OCR C1 2013 Q9 [12]}}