| Exam Board | OCR |
|---|---|
| Module | C1 (Core Mathematics 1) |
| Year | 2010 |
| Session | June |
| Marks | 14 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Stationary points and optimisation |
| Type | Find range where function increasing/decreasing |
| Difficulty | Standard +0.3 This is a standard C1 differentiation question with routine stationary point finding, determining decreasing intervals, and tangent equation verification. Part (iv) adds mild problem-solving by connecting the tangent to root-counting via a sketch, but all techniques are straightforward applications of basic calculus with no novel insight required—slightly above average due to the multi-part nature and final synthesis step. |
| Spec | 1.02p Interpret algebraic solutions: graphically1.07m Tangents and normals: gradient and equations1.07n Stationary points: find maxima, minima using derivatives1.07o Increasing/decreasing: functions using sign of dy/dx |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(\frac{dy}{dx} = 6x^2 + 10x - 4\) | B1 | 1 term correct |
| B1 | Completely correct (no \(+c\)) | |
| \(6x^2 + 10x - 4 = 0\) | M1* | Sets \(\frac{dy}{dx} = 0\) |
| \(2(3x^2 + 5x - 2) = 0\) | ||
| \((3x-1)(x+2) = 0\) | M1 dep* | Correct method to solve quadratic |
| \(x = \frac{1}{3}\) or \(x = -2\) | A1 | SC If A0 A0, one correct pair of values spotted or from correct factorisation www B1 |
| \(y = -\frac{19}{27}\) or \(y = 12\) | A1 | 6 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(-2 < x < \frac{1}{3}\) | M1 | Any inequality (or inequalities) involving both their \(x\) values from part (i) |
| A1 | 2 Allow \(\leq\) and \(\geq\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| When \(x = \frac{1}{2}\), \(6x^2 + 10x - 4 = \frac{5}{2}\) | M1 | Substitute \(x = \frac{1}{2}\) into their \(\frac{dy}{dx}\) |
| and \(2x^3 + 5x^2 - 4x = -\frac{1}{2}\) | B1 | Correct \(y\) coordinate |
| \(y + \frac{1}{2} = \frac{5}{2}\left(x - \frac{1}{2}\right)\) | M1 | Correct equation of straight line using their values. Must use their \(\frac{dy}{dx}\) value not e.g. the negative reciprocal |
| \(10x - 4y - 7 = 0\) | A1 | 4 Shows rearrangement to given equation. CWO throughout for A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| Sketch of a cubic with a tangent which meets it at 2 points only | B1 | |
| +ve cubic with max/min points and line with +ve gradient as tangent to the curve to the right of the min | B1 | 2 |
| SC1: B1 Convincing algebra to show that \(8x^3 + 20x^2 - 26x + 7 = 0\) factorises into \((2x-1)(2x-1)(x+7)\); B1 Correct argument to say there are 2 distinct roots | ||
| SC2: B1 Recognising \(y = 2.5x - \frac{7}{4}\) is tangent from part (iii); B1 As second B1 on main scheme |
# Question 10:
## Part (i):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\frac{dy}{dx} = 6x^2 + 10x - 4$ | B1 | 1 term correct |
| | B1 | Completely correct (no $+c$) |
| $6x^2 + 10x - 4 = 0$ | M1* | Sets $\frac{dy}{dx} = 0$ |
| $2(3x^2 + 5x - 2) = 0$ | | |
| $(3x-1)(x+2) = 0$ | M1 dep* | Correct method to solve quadratic |
| $x = \frac{1}{3}$ or $x = -2$ | A1 | SC If A0 A0, one correct pair of values spotted or from correct factorisation **www** B1 |
| $y = -\frac{19}{27}$ or $y = 12$ | A1 | **6** |
## Part (ii):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $-2 < x < \frac{1}{3}$ | M1 | Any inequality (or inequalities) involving both their $x$ values from part (i) |
| | A1 | **2** Allow $\leq$ and $\geq$ |
## Part (iii):
| Answer/Working | Marks | Guidance |
|---|---|---|
| When $x = \frac{1}{2}$, $6x^2 + 10x - 4 = \frac{5}{2}$ | M1 | Substitute $x = \frac{1}{2}$ into their $\frac{dy}{dx}$ |
| and $2x^3 + 5x^2 - 4x = -\frac{1}{2}$ | B1 | Correct $y$ coordinate |
| $y + \frac{1}{2} = \frac{5}{2}\left(x - \frac{1}{2}\right)$ | M1 | Correct equation of straight line using their values. Must use their $\frac{dy}{dx}$ value not e.g. the negative reciprocal |
| $10x - 4y - 7 = 0$ | A1 | **4** Shows rearrangement to given equation. CWO throughout for A1 |
## Part (iv):
| Answer/Working | Marks | Guidance |
|---|---|---|
| Sketch of a cubic with a tangent which meets it at 2 points only | B1 | |
| +ve cubic with max/min points and line with +ve gradient as tangent to the curve to the right of the min | B1 | **2** |
| **SC1:** B1 Convincing algebra to show that $8x^3 + 20x^2 - 26x + 7 = 0$ factorises into $(2x-1)(2x-1)(x+7)$; B1 Correct argument to say there are 2 distinct roots | | |
| **SC2:** B1 Recognising $y = 2.5x - \frac{7}{4}$ is tangent from part (iii); B1 As second B1 on main scheme | | |10 (i) Find the coordinates of the stationary points of the curve $y = 2 x ^ { 3 } + 5 x ^ { 2 } - 4 x$.\\
(ii) State the set of values for $x$ for which $2 x ^ { 3 } + 5 x ^ { 2 } - 4 x$ is a decreasing function.\\
(iii) Show that the equation of the tangent to the curve at the point where $x = \frac { 1 } { 2 }$ is $10 x - 4 y - 7 = 0$.\\
(iv) Hence, with the aid of a sketch, show that the equation $2 x ^ { 3 } + 5 x ^ { 2 } - 4 x = \frac { 5 } { 2 } x - \frac { 7 } { 4 }$ has two distinct real roots.
\hfill \mbox{\textit{OCR C1 2010 Q10 [14]}}