| Exam Board | OCR |
|---|---|
| Module | C1 (Core Mathematics 1) |
| Year | 2012 |
| Session | January |
| Marks | 12 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Completing the square and sketching |
| Type | Sketch quadratic curve |
| Difficulty | Moderate -0.3 This is a straightforward multi-part question on quadratic functions requiring standard techniques: factorising to find intercepts, sketching a parabola, solving a quadratic inequality, and finding intersection points with a line. All parts are routine C1 exercises with no novel problem-solving required, making it slightly easier than average but not trivial due to the multiple steps involved. |
| Spec | 1.02g Inequalities: linear and quadratic in single variable1.02n Sketch curves: simple equations including polynomials1.02q Use intersection points: of graphs to solve equations |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \((x-3)(x+4) = 0\) | M1 | Correct method to find roots |
| \(x = 3\) or \(x = -4\) | A1 | Correct roots |
| Negative quadratic curve sketch | B1 | Negative quadratic curve |
| \(y\) intercept \((0, 12)\) | B1 | \(y\) intercept \((0, 12)\) |
| Good curve with correct roots \(3\) and \(-4\) indicated, max point in \(2^{nd}\) quadrant | B1 | i.e. max at \((0, 12)\) B0; curve must go below \(x\)-axis for final mark |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(-4 < x < 3\) | M1 | Correct method to solve quadratic inequality; allow \(\leq\) for method mark but not accuracy mark |
| A1 | Allow "\(x > -4\), \(x < 3\)"; allow "\(x > -4\) and \(x < 3\)"; do not allow "\(x > -4\) or \(x < 3\)" |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(y = 4 - 3x\) | ||
| \(12 - x - x^2 = 4 - 3x\) | M1 | Substitute for \(x/y\) or attempt to get an equation in 1 variable only; e.g. \(3x + 12 - x - x^2 = 4\), or \(y = 12 - \left(\frac{4-y}{3}\right) - \left(\frac{4-y}{3}\right)^2\) |
| \(x^2 - 2x - 8 = 0\) | A1 | Obtain correct 3 term quadratic |
| \((x-4)(x+2) = 0\) | M1 | Correct method to solve 3 term quadratic; (this leads to \(y^2 - 2y - 80 = 0\)); condone poor algebra for this mark |
| \(x = 4\) or \(x = -2\) | A1 | SC If A0 A0, give B1 for one correct pair of values spotted or from correct factorisation |
| \(y = -8\) or \(y = 10\) | A1 |
## Question 9:
### Part (i):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $(x-3)(x+4) = 0$ | M1 | Correct method to find roots |
| $x = 3$ or $x = -4$ | A1 | Correct roots |
| Negative quadratic curve sketch | B1 | Negative quadratic curve |
| $y$ intercept $(0, 12)$ | B1 | $y$ intercept $(0, 12)$ |
| Good curve with correct roots $3$ and $-4$ indicated, max point in $2^{nd}$ quadrant | B1 | i.e. max at $(0, 12)$ **B0**; curve must go below $x$-axis for final mark |
**Total: [5]**
### Part (ii):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $-4 < x < 3$ | M1 | Correct method to solve quadratic inequality; allow $\leq$ for method mark but not accuracy mark |
| | A1 | Allow "$x > -4$, $x < 3$"; allow "$x > -4$ and $x < 3$"; do not allow "$x > -4$ **or** $x < 3$" |
**Total: [2]**
### Part (iii):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $y = 4 - 3x$ | | |
| $12 - x - x^2 = 4 - 3x$ | M1 | Substitute for $x/y$ or attempt to get an equation in 1 variable only; e.g. $3x + 12 - x - x^2 = 4$, or $y = 12 - \left(\frac{4-y}{3}\right) - \left(\frac{4-y}{3}\right)^2$ |
| $x^2 - 2x - 8 = 0$ | A1 | Obtain correct 3 term quadratic |
| $(x-4)(x+2) = 0$ | M1 | Correct method to solve 3 term quadratic; (this leads to $y^2 - 2y - 80 = 0$); condone poor algebra for this mark |
| $x = 4$ or $x = -2$ | A1 | **SC** If **A0 A0**, give **B1** for one correct pair of values spotted or from correct factorisation |
| $y = -8$ or $y = 10$ | A1 | |
**Total: [5]**
---9 (i) Sketch the curve $y = 12 - x - x ^ { 2 }$, giving the coordinates of all intercepts with the axes.\\
(ii) Solve the inequality $12 - x - x ^ { 2 } > 0$.\\
(iii) Find the coordinates of the points of intersection of the curve $y = 12 - x - x ^ { 2 }$ and the line $3 x + y = 4$.
\hfill \mbox{\textit{OCR C1 2012 Q9 [12]}}