| Exam Board | OCR |
|---|---|
| Module | C1 (Core Mathematics 1) |
| Year | 2012 |
| Session | January |
| Marks | 12 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Curve Sketching |
| Type | Find stationary points of standard polynomial |
| Difficulty | Moderate -0.3 This is a straightforward C1 question requiring standard differentiation to find stationary points and basic algebraic reasoning. The discriminant calculation is routine, and the explanation in part (iii) requires only simple logic about sign of factors. Slightly easier than average due to being methodical rather than requiring problem-solving insight. |
| Spec | 1.02d Quadratic functions: graphs and discriminant conditions1.07n Stationary points: find maxima, minima using derivatives1.07p Points of inflection: using second derivative |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(x^3-3x^2+5x+2x^2-6x+10 = x^3-x^2-x+10\) | M1 | Attempt to multiply out brackets |
| \(\frac{dy}{dx} = 3x^2-2x-1\) | M1 | Attempt to differentiate their cubic |
| \(\frac{dy}{dx}=0\): \((3x+1)(x-1)=0\) | M1* | Sets \(\frac{dy}{dx}=0\); correct method to solve quadratic |
| \(x=-\frac{1}{3}\) or \(x=1\) | A1 | Correct \(x\) values of turning points found www. Any extra values loses all three A marks |
| \(\frac{d^2y}{dx^2}=6x-2\), \(x=1\) gives \(+\)ve \((4)\) | M1dep | Valid method to establish which is min point with a conclusion |
| Min point at \(x=1\) | A1 | Correct conclusion for \(x=1\) found from correct factorisation (even if other root incorrect) |
| \(y=9\) found | A1 | www for \((1,9)\) given as minimum point (ignore other point here). If constant incorrect in initial expansion, max 5/8 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \((-3)^2-4\times1\times5 = -11\) | M1 | Uses \(b^2-4ac\). Note: \(\sqrt{b^2-4ac}\) is M0 |
| \(= -11\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Fully correct argument - no extra incorrect statements | B2 | Award B1 for either: 1) Justifying quadratic factor having no roots so only intersection with \(x\)-axis is at \(x=-2\), stating it's a positive cubic; 2) Sketch of positive cubic with one root at \((-2,0)\) and min point at \((1,9)\) (f/t positive \(y(1)\) from (i)) |
## Question 7(i):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $x^3-3x^2+5x+2x^2-6x+10 = x^3-x^2-x+10$ | M1 | Attempt to multiply out brackets |
| $\frac{dy}{dx} = 3x^2-2x-1$ | M1 | Attempt to differentiate their cubic |
| $\frac{dy}{dx}=0$: $(3x+1)(x-1)=0$ | M1* | Sets $\frac{dy}{dx}=0$; correct method to solve quadratic |
| $x=-\frac{1}{3}$ or $x=1$ | A1 | Correct $x$ values of turning points found www. Any extra values loses all three **A** marks |
| $\frac{d^2y}{dx^2}=6x-2$, $x=1$ gives $+$ve $(4)$ | M1dep | Valid method to establish which is min point with a conclusion |
| Min point at $x=1$ | A1 | Correct conclusion for $x=1$ found from correct factorisation (even if other root incorrect) |
| $y=9$ found | A1 | www for $(1,9)$ given as minimum point (ignore other point here). If constant incorrect in initial expansion, max **5/8** |
**Alternative (product rule):** Attempt to use product rule **M1**; expand brackets of both parts **M1**; then as main scheme
---
## Question 7(ii):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $(-3)^2-4\times1\times5 = -11$ | M1 | Uses $b^2-4ac$. Note: $\sqrt{b^2-4ac}$ is **M0** |
| $= -11$ | A1 | |
---
## Question 7(iii):
| Answer | Marks | Guidance |
|--------|-------|----------|
| Fully correct argument - no extra incorrect statements | B2 | Award **B1** for either: 1) Justifying quadratic factor having no roots so only intersection with $x$-axis is at $x=-2$, stating it's a positive cubic; 2) Sketch of positive cubic with one root at $(-2,0)$ and min point at $(1,9)$ (f/t positive $y(1)$ from (i)) |
---7 A curve has equation $y = ( x + 2 ) \left( x ^ { 2 } - 3 x + 5 \right)$.\\
(i) Find the coordinates of the minimum point, justifying that it is a minimum.\\
(ii) Calculate the discriminant of $x ^ { 2 } - 3 x + 5$.\\
(iii) Explain why $( x + 2 ) \left( x ^ { 2 } - 3 x + 5 \right)$ is always positive for $x > - 2$.
\hfill \mbox{\textit{OCR C1 2012 Q7 [12]}}