| Exam Board | OCR |
|---|---|
| Module | C1 (Core Mathematics 1) |
| Year | 2011 |
| Session | January |
| Marks | 13 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Tangents, normals and gradients |
| Type | Tangent meets curve/axis — further geometry |
| Difficulty | Moderate -0.8 This is a straightforward C1 differentiation question with routine steps: find derivative, evaluate at a point, write tangent equation, find intercept and midpoint, and identify parabola properties. All parts use standard techniques with no problem-solving insight required, making it easier than average but not trivial due to the multi-part nature. |
| Spec | 1.07m 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 | Mark | Guidance |
| \(\frac{dy}{dx} = 6 - 2x\) | M1, A1 | Attempt to differentiate \(\pm y\); correct expression cao; one correct non-zero term |
| When \(x=5\), \(6-2x = -4\) | M1 | Substitute \(x=5\) into their \(\frac{dy}{dx}\) |
| When \(x=5\), \(y=12\) | B1 | Correct \(y\) coordinate |
| \(y - 12 = -4(x-5)\) | M1 | Correct equation of straight line through \((5, \text{their } y)\), non-zero numerical gradient; allow \(\frac{y-12}{x-5} = \text{their gradient}\) |
| \(4x + y - 32 = 0\) | A1 | Shows rearrangement to correct form; allow any correct form e.g. \(0 = 2y + 8x - 64\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(Q\) is point \((8, 0)\) | B1ft | ft from line in (i) |
| Midpoint of \(PQ = \left(\frac{5+8}{2}, \frac{12+0}{2}\right)\) | M1 | Uses \(\left(\frac{x_1+x_2}{2}, \frac{y_1+y_2}{2}\right)\) o.e. for their P, Q |
| \(= \left(\frac{13}{2}, 6\right)\) | A1 | Do not accept \(\left(\frac{13}{2}, \frac{12}{2}\right)\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(6 - 2x = 0\) | M1 | Solution of their \(\frac{dy}{dx} = 0\); alternatives: a) completing square with \(\pm(x-3)^2\); b) solve quadratic and find midpoint of roots; c) use \(x = -\frac{b}{2a}\) |
| Line of symmetry is \(x = 3\) | A1 | Allow from \(\pm[16-(x-3)^2]\), \(\pm[6-2x=0]\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(x < 3\) | M1 | \(x <\) their 3 or \(x >\) their 3, OR attempt to solve \(\frac{dy}{dx} > 0\) |
| A1 | Allow from \(\pm[16-(x-3)^2]\), \(\pm[6-2x=0]\) in (iii); may solve \(\frac{dy}{dx}=0\) then use \(\frac{d^2y}{dx^2} < 0\) implies maximum; allow \(x \leq 3\) |
## Question 8(i):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\frac{dy}{dx} = 6 - 2x$ | M1, A1 | Attempt to differentiate $\pm y$; correct expression cao; one correct non-zero term |
| When $x=5$, $6-2x = -4$ | M1 | Substitute $x=5$ into their $\frac{dy}{dx}$ |
| When $x=5$, $y=12$ | B1 | Correct $y$ coordinate |
| $y - 12 = -4(x-5)$ | M1 | Correct equation of straight line through $(5, \text{their } y)$, non-zero numerical gradient; allow $\frac{y-12}{x-5} = \text{their gradient}$ |
| $4x + y - 32 = 0$ | A1 | Shows rearrangement to correct form; allow any correct form e.g. $0 = 2y + 8x - 64$ |
---
## Question 8(ii):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $Q$ is point $(8, 0)$ | B1ft | ft from line in (i) |
| Midpoint of $PQ = \left(\frac{5+8}{2}, \frac{12+0}{2}\right)$ | M1 | Uses $\left(\frac{x_1+x_2}{2}, \frac{y_1+y_2}{2}\right)$ o.e. for their P, Q |
| $= \left(\frac{13}{2}, 6\right)$ | A1 | Do not accept $\left(\frac{13}{2}, \frac{12}{2}\right)$ |
---
## Question 8(iii):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $6 - 2x = 0$ | M1 | Solution of their $\frac{dy}{dx} = 0$; alternatives: a) completing square with $\pm(x-3)^2$; b) solve quadratic and find midpoint of roots; c) use $x = -\frac{b}{2a}$ |
| Line of symmetry is $x = 3$ | A1 | Allow from $\pm[16-(x-3)^2]$, $\pm[6-2x=0]$ |
---
## Question 8(iv):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $x < 3$ | M1 | $x <$ their 3 or $x >$ their 3, OR attempt to solve $\frac{dy}{dx} > 0$ |
| | A1 | Allow from $\pm[16-(x-3)^2]$, $\pm[6-2x=0]$ in (iii); may solve $\frac{dy}{dx}=0$ then use $\frac{d^2y}{dx^2} < 0$ implies maximum; allow $x \leq 3$ |
---8 (i) Find the equation of the tangent to the curve $y = 7 + 6 x - x ^ { 2 }$ at the point $P$ where $x = 5$, giving your answer in the form $a x + b y + c = 0$.\\
(ii) This tangent meets the $x$-axis at $Q$. Find the coordinates of the mid-point of $P Q$.\\
(iii) Find the equation of the line of symmetry of the curve $y = 7 + 6 x - x ^ { 2 }$.\\
(iv) State the set of values of $x$ for which $7 + 6 x - x ^ { 2 }$ is an increasing function.
\hfill \mbox{\textit{OCR C1 2011 Q8 [13]}}