| Exam Board | OCR MEI |
|---|---|
| Module | C1 (Core Mathematics 1) |
| Year | 2015 |
| Session | June |
| Marks | 5 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Solving quadratics and applications |
| Type | Quadratic with surd roots, exact form |
| Difficulty | Moderate -0.8 This is a straightforward application of Pythagoras' theorem requiring basic algebraic manipulation to form a quadratic equation, then solving it using the quadratic formula. The steps are routine and clearly signposted, making it easier than average for A-level. |
| Spec | 1.01a Proof: structure of mathematical proof and logical steps1.01b Logical connectives: congruence, if-then, if and only if |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \((3x)^2 = h^2 + (2x+1)^2\) oe | B1 | For a correct Pythagoras statement for this triangle, in terms of \(x\), with correct brackets; condone another letter instead of \(h\) for one mark but not both unless recovered at some point |
| \(9x^2 = h^2 + 4x^2 + 4x + 1\) and completion to given answer \(h^2 = 5x^2 - 4x - 1\) | B1 | For correct expansion, with brackets or correct signs; must complete to the given answer with no errors in any interim working; may follow \(3x^2 = h^2 + (2x+1)^2\) oe for B0 B1; eg B1 for \(h^2 = 9x^2 - (4x^2 + 4x + 1)\) and completion to correct answer but B0 for \(h^2 = 9x^2 - 4x^2 + 4x + 1\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \([0=]\ 5x^2 - 4x - 8\) | B1 | For subst and correctly rearranging to zero |
| \(\frac{4 \pm \sqrt{(-4)^2 - 4 \times 5 \times -8}}{2 \times 5}\) or ft | M1 | For use of formula in their eqn rearranged to zero, condoning one error; ft only if their rearranged eqn is a 3-term quadratic; no ft from \(5x^2 - 4x - 1\ [=0]\); or M1 for \(\left(x - \frac{2}{5}\right)^2 = \left(\frac{2}{5}\right)^2 + \frac{8}{5}\) oe (condoning one error), which also implies first M1 if not previously earned; M0 for factorising ft |
| \(\frac{4 + \sqrt{176}}{10}\) or \(\frac{2}{5} + \frac{\sqrt{44}}{5}\) oe | A1 | isw wrong simplification; A0 if negative root also included |
## Question 8(i):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $(3x)^2 = h^2 + (2x+1)^2$ oe | B1 | For a correct Pythagoras statement for this triangle, in terms of $x$, with correct brackets; condone another letter instead of $h$ for one mark but not both unless recovered at some point |
| $9x^2 = h^2 + 4x^2 + 4x + 1$ and completion to given answer $h^2 = 5x^2 - 4x - 1$ | B1 | For correct expansion, with brackets or correct signs; must complete to the given answer with no errors in any interim working; may follow $3x^2 = h^2 + (2x+1)^2$ oe for B0 B1; eg B1 for $h^2 = 9x^2 - (4x^2 + 4x + 1)$ and completion to correct answer but B0 for $h^2 = 9x^2 - 4x^2 + 4x + 1$ |
## Question 8(ii):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $[0=]\ 5x^2 - 4x - 8$ | B1 | For subst and correctly rearranging to zero |
| $\frac{4 \pm \sqrt{(-4)^2 - 4 \times 5 \times -8}}{2 \times 5}$ or ft | M1 | For use of formula in their eqn rearranged to zero, condoning one error; ft only if their rearranged eqn is a 3-term quadratic; no ft from $5x^2 - 4x - 1\ [=0]$; or M1 for $\left(x - \frac{2}{5}\right)^2 = \left(\frac{2}{5}\right)^2 + \frac{8}{5}$ oe (condoning one error), which also implies first M1 if not previously earned; M0 for factorising ft |
| $\frac{4 + \sqrt{176}}{10}$ or $\frac{2}{5} + \frac{\sqrt{44}}{5}$ oe | A1 | isw wrong simplification; A0 if negative root also included |8 Fig. 8 shows a right-angled triangle with base $2 x + 1$, height $h$ and hypotenuse $3 x$.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{c55e1f96-670a-4bc3-9e77-92d28769b7f5-2_317_593_1653_543}
\captionsetup{labelformat=empty}
\caption{Not to scale}
\end{center}
\end{figure}
Fig. 8\\
(i) Show that $h ^ { 2 } = 5 x ^ { 2 } - 4 x - 1$.\\
(ii) Given that $h = \sqrt { 7 }$, find the value of $x$, giving your answer in surd form.
\hfill \mbox{\textit{OCR MEI C1 2015 Q8 [5]}}