| Exam Board | OCR |
|---|---|
| Module | C1 (Core Mathematics 1) |
| Year | 2011 |
| Session | January |
| Marks | 11 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Completing the square and sketching |
| Type | Quadratic with equal roots |
| Difficulty | Moderate -0.3 This is a standard completing-the-square question with routine follow-up parts. Part (i) requires factoring out 4 and completing the square (textbook procedure), part (ii) applies this to solve an equation (straightforward once completed), and part (iii) uses the equal roots condition (discriminant = 0 or vertex form). All techniques are core C1 material with no problem-solving insight required, making it slightly easier than average. |
| 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 | Mark | Guidance |
| \(p = 4\) | B1 | If \(p\), \(q\), \(r\) found correctly, then ISW slips in format |
| \(q = \frac{3}{2}\) | B1 | |
| \(r = -3 - 4q^2\) or \(r = -\frac{3}{4} - q^2\) | M1 | |
| \(= 4\left(x + \frac{3}{2}\right)^2 - 12\) | A1 | 4 marks; \(r = -12\) (from \(q = \pm 1.5\)) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(\frac{-12 \pm \sqrt{12^2 - 4 \times 4 \times -3}}{2 \times 4}\) | M1 | Correct method to solve quadratic |
| \(= \frac{-12 \pm \sqrt{192}}{8}\) | A1 | \(\frac{-12 \pm \sqrt{192}}{8}\) or \(\frac{-3 \pm \sqrt{12}}{2}\) |
| \(= \frac{-12 \pm 8\sqrt{3}}{8}\) | B1 | \(\sqrt{192} = 8\sqrt{3}\) or \(\sqrt{12} = 2\sqrt{3}\) from correct \(b^2 - 4ac\) |
| \(= -\frac{3}{2} \pm \sqrt{3}\) | A1 | 4 marks; \(\frac{-3 \pm 2\sqrt{3}}{2}\) or \(-\frac{12}{8} \pm \sqrt{3}\), \(-\frac{6}{4} \pm \sqrt{3}\); Do not ISW; SR One correct root www B1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(4\left(x + \frac{3}{2}\right)^2 - 12 = 0\) | ||
| \(x + \frac{3}{2} = \pm\sqrt{3}\) | M1 | Must have \(\pm\) for method mark |
| A1ft | \(x + 1.5\) ft \(x + q\) from part (i) www in LHS in part (ii); Not for \(2(x+q) = \ldots\) | |
| \(x = -\frac{3}{2} \pm \sqrt{3}\) | A1 | \(\pm\sqrt{3}\) |
| A1 | Do not ISW |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(12^2 - 4 \times 4 \times (-k) = 0\) | M1 | Attempts \(b^2 - 4ac = 0\) or \(\sqrt{b^2 - 4ac} = 0\) involving \(k\). If \(b^2 - 4ac\) not quoted then expression must be correct. |
| \(144 + 16k = 0\) | A1 | Correct, unsimplified expression |
| \(k = -9\) | A1 | 3 marks |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(4x^2 + 12x = k\) | Must involve \(k\) in working | |
| \(4(x + \frac{3}{2})^2 - 9 = k\) | M1 | Attempts completing the square, or factorises to \((2x+3)^2 - 9 = k\) |
| Equal roots when \(x = -\frac{3}{2}\) | M1 | Substitutes \(x = -\frac{3}{2}\) |
| \(k = -9\) | A1 |
# Question 7:
## Part (i)
| Answer | Mark | Guidance |
|--------|------|----------|
| $p = 4$ | B1 | If $p$, $q$, $r$ found correctly, then **ISW** slips in format |
| $q = \frac{3}{2}$ | B1 | |
| $r = -3 - 4q^2$ or $r = -\frac{3}{4} - q^2$ | M1 | |
| $= 4\left(x + \frac{3}{2}\right)^2 - 12$ | A1 | **4 marks**; $r = -12$ (from $q = \pm 1.5$) |
## Part (ii)
| Answer | Mark | Guidance |
|--------|------|----------|
| $\frac{-12 \pm \sqrt{12^2 - 4 \times 4 \times -3}}{2 \times 4}$ | M1 | Correct method to solve quadratic |
| $= \frac{-12 \pm \sqrt{192}}{8}$ | A1 | $\frac{-12 \pm \sqrt{192}}{8}$ or $\frac{-3 \pm \sqrt{12}}{2}$ |
| $= \frac{-12 \pm 8\sqrt{3}}{8}$ | B1 | $\sqrt{192} = 8\sqrt{3}$ or $\sqrt{12} = 2\sqrt{3}$ from correct $b^2 - 4ac$ |
| $= -\frac{3}{2} \pm \sqrt{3}$ | A1 | **4 marks**; $\frac{-3 \pm 2\sqrt{3}}{2}$ or $-\frac{12}{8} \pm \sqrt{3}$, $-\frac{6}{4} \pm \sqrt{3}$; Do not **ISW**; **SR** One correct root **www B1** |
*OR using completed square:*
| Answer | Mark | Guidance |
|--------|------|----------|
| $4\left(x + \frac{3}{2}\right)^2 - 12 = 0$ | | |
| $x + \frac{3}{2} = \pm\sqrt{3}$ | M1 | Must have $\pm$ for method mark |
| | A1ft | $x + 1.5$ **ft** $x + q$ from part (i) **www** in LHS in part (ii); Not for $2(x+q) = \ldots$ |
| $x = -\frac{3}{2} \pm \sqrt{3}$ | A1 | $\pm\sqrt{3}$ |
| | A1 | Do not **ISW** |
## Part (iii)
| Answer | Mark | Guidance |
|--------|------|----------|
| $12^2 - 4 \times 4 \times (-k) = 0$ | M1 | Attempts $b^2 - 4ac = 0$ or $\sqrt{b^2 - 4ac} = 0$ involving $k$. If $b^2 - 4ac$ not quoted then expression must be correct. |
| $144 + 16k = 0$ | A1 | Correct, unsimplified expression |
| $k = -9$ | A1 | **3 marks** |
*Other alternative methods:*
- a) Attempt to factorise into two equal brackets (may divide by 4 first – must be correct) **M1**; Equate coefficient of $x$ to 12 (or 3) **A1** $k = -9$ **A1**
- b) Uses differentiation to find $x$ ordinate of turning point and uses this to form equation in $k$ **M1**; Correct equation in $k$ **A1** $k = -9$ **A1**
## Question 7(iii):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $4x^2 + 12x = k$ | | Must involve $k$ in working |
| $4(x + \frac{3}{2})^2 - 9 = k$ | M1 | Attempts completing the square, or factorises to $(2x+3)^2 - 9 = k$ |
| Equal roots when $x = -\frac{3}{2}$ | M1 | Substitutes $x = -\frac{3}{2}$ |
| $k = -9$ | A1 | |
---7 (i) Express $4 x ^ { 2 } + 12 x - 3$ in the form $p ( x + q ) ^ { 2 } + r$.\\
(ii) Solve the equation $4 x ^ { 2 } + 12 x - 3 = 0$, giving your answers in simplified surd form.\\
(iii) The quadratic equation $4 x ^ { 2 } + 12 x - k = 0$ has equal roots. Find the value of $k$.
\hfill \mbox{\textit{OCR C1 2011 Q7 [11]}}