| Exam Board | OCR |
|---|---|
| Module | H240/02 (Pure Mathematics and Statistics) |
| Year | 2018 |
| Session | June |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Topic | Discriminant and conditions for roots |
| Type | Find k for equal roots |
| Difficulty | Moderate -0.8 This is a straightforward multi-part question testing completing the square (routine algebraic manipulation), interpreting the completed square form to determine number of roots (direct observation), and using the discriminant condition for repeated roots (standard application of b²-4ac=0). All parts are textbook exercises requiring only recall and basic application of standard techniques with no problem-solving insight needed. |
| Spec | 1.02d Quadratic functions: graphs and discriminant conditions1.02e Complete the square: quadratic polynomials and turning points1.02f Solve quadratic equations: including in a function of unknown |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(2x^2 - 6x + 11.5\) | B1 | or \(a = 2\) |
| \(2((x-3)^2 + 11.5 - 9)\) | B1 | or \(b = -3\) |
| M1 | \(23 - 2(\text{their } b)^2\) | |
| \(2(x-3)^2 + 5\) | A1 | or \(c = 5\) |
| [4] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(2(x+3)^2 + 5\) is always \(+\)ve, or \(2(x+3)^2 + 5 > 0\), or \(2(x+3)^2 + 5 \geq 5\), hence no real roots | B1f [1] | or \(2(x+3)^2 = -5\) which is impossible; or "\(+\)ve quadratic" and min on \(y=5\); or "\(+\)ve" quadratic; TP at \((3,5)\). Must use (i), not use D. ft their (i) (\(a\) & \(c > 0\)) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(2(x-3)^2 = 2(2x^2 - 6x + 9)\) | M1 | or \(12^2 - 8k = 0\) |
| \(k = 18\) | A1 [2] |
## Question 1:
### Part (i)
| Answer | Marks | Guidance |
|--------|-------|----------|
| $2x^2 - 6x + 11.5$ | B1 | or $a = 2$ |
| $2((x-3)^2 + 11.5 - 9)$ | B1 | or $b = -3$ |
| | M1 | $23 - 2(\text{their } b)^2$ |
| $2(x-3)^2 + 5$ | A1 | or $c = 5$ |
| **[4]** | | |
### Part (ii)
| Answer | Marks | Guidance |
|--------|-------|----------|
| $2(x+3)^2 + 5$ is always $+$ve, or $2(x+3)^2 + 5 > 0$, or $2(x+3)^2 + 5 \geq 5$, hence no real roots | B1f [1] | or $2(x+3)^2 = -5$ which is impossible; or "$+$ve quadratic" and min on $y=5$; or "$+$ve" quadratic; TP at $(3,5)$. Must use (i), not use D. ft their (i) ($a$ & $c > 0$) |
### Part (iii)
| Answer | Marks | Guidance |
|--------|-------|----------|
| $2(x-3)^2 = 2(2x^2 - 6x + 9)$ | M1 | or $12^2 - 8k = 0$ |
| $k = 18$ | A1 **[2]** | |
---1 (i) Express $2 x ^ { 2 } - 12 x + 23$ in the form $a ( x + b ) ^ { 2 } + c$.\\
(ii) Use your result to show that the equation $2 x ^ { 2 } - 12 x + 23 = 0$ has no real roots.\\
(iii) Given that the equation $2 x ^ { 2 } - 12 x + k = 0$ has repeated roots, find the value of the constant $k$.
\hfill \mbox{\textit{OCR H240/02 2018 Q1 [7]}}