| Exam Board | OCR |
|---|---|
| Module | C1 (Core Mathematics 1) |
| Year | 2008 |
| Session | June |
| Marks | 14 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Discriminant and conditions for roots |
| Type | Complete the square, then discriminant |
| Difficulty | Moderate -0.8 This is a straightforward multi-part question testing standard C1 techniques: completing the square, identifying vertex, discriminant calculation, and solving simultaneous equations. All parts are routine applications with no problem-solving insight required, making it easier than average but not trivial due to the multiple steps involved. |
| Spec | 1.02c Simultaneous equations: two variables by elimination and substitution1.02d Quadratic functions: graphs and discriminant conditions1.02e Complete the square: quadratic polynomials and turning points |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(2(x^2 - 3x) + 11\) | B1 | \(p = 2\) |
| \(= 2\left[\left(x - \frac{3}{2}\right)^2 - \frac{9}{4}\right] + 11\) | B1 | \(q = -\frac{3}{2}\) |
| \(= 2\left(x - \frac{3}{2}\right)^2 + \frac{13}{2}\) | M1 | \(r = 11 - 2q^2\) or \(\frac{11}{2} - q^2\) |
| A1 | \(r = \frac{13}{2}\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\left(\frac{3}{2}, \frac{13}{2}\right)\) | B1\(\checkmark\) | |
| B1\(\checkmark\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(36 - 4 \times 2 \times 11\) | M1 | Uses \(b^2 - 4ac\) |
| \(= -52\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| 0 real roots | B1 | cao |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(2x^2 - 6x + 11 = 14 - 7x\) | M1* | Substitute for \(x\)/\(y\) or attempt to get an equation in 1 variable only |
| \(2x^2 + x - 3 = 0\) | A1 | Obtain correct 3 term quadratic |
| \((2x+3)(x-1) = 0\) | M1dep | Correct method to solve 3 term quadratic |
| \(x = -\frac{3}{2},\ x = 1\) | A1 | |
| \(y = \frac{49}{2},\ y = 7\) | A1 | SR If A0 A0, one correct pair of values spotted or from correct factorisation www B1 |
## Question 10(i):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $2(x^2 - 3x) + 11$ | B1 | $p = 2$ |
| $= 2\left[\left(x - \frac{3}{2}\right)^2 - \frac{9}{4}\right] + 11$ | B1 | $q = -\frac{3}{2}$ |
| $= 2\left(x - \frac{3}{2}\right)^2 + \frac{13}{2}$ | M1 | $r = 11 - 2q^2$ or $\frac{11}{2} - q^2$ |
| | A1 | $r = \frac{13}{2}$ |
---
## Question 10(ii):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\left(\frac{3}{2}, \frac{13}{2}\right)$ | B1$\checkmark$ | |
| | B1$\checkmark$ | |
---
## Question 10(iii):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $36 - 4 \times 2 \times 11$ | M1 | Uses $b^2 - 4ac$ |
| $= -52$ | A1 | |
---
## Question 10(iv):
| Answer/Working | Mark | Guidance |
|---|---|---|
| 0 real roots | B1 | cao |
---
## Question 10(v):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $2x^2 - 6x + 11 = 14 - 7x$ | M1* | Substitute for $x$/$y$ or attempt to get an equation in 1 variable only |
| $2x^2 + x - 3 = 0$ | A1 | Obtain correct 3 term quadratic |
| $(2x+3)(x-1) = 0$ | M1dep | Correct method to solve 3 term quadratic |
| $x = -\frac{3}{2},\ x = 1$ | A1 | |
| $y = \frac{49}{2},\ y = 7$ | A1 | **SR** If A0 A0, one correct pair of values spotted or from correct factorisation **www B1** |10 (i) Express $2 x ^ { 2 } - 6 x + 11$ in the form $p ( x + q ) ^ { 2 } + r$.\\
(ii) State the coordinates of the vertex of the curve $y = 2 x ^ { 2 } - 6 x + 11$.\\
(iii) Calculate the discriminant of $2 x ^ { 2 } - 6 x + 11$.\\
(iv) State the number of real roots of the equation $2 x ^ { 2 } - 6 x + 11 = 0$.\\
(v) Find the coordinates of the points of intersection of the curve $y = 2 x ^ { 2 } - 6 x + 11$ and the line $7 x + y = 14$.
\hfill \mbox{\textit{OCR C1 2008 Q10 [14]}}