| Exam Board | OCR MEI |
|---|---|
| Module | C2 (Core Mathematics 2) |
| Marks | 13 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Tangents, normals and gradients |
| Type | Find tangent at given point (polynomial/algebraic) |
| Difficulty | Moderate -0.3 This is a multi-part C2 question covering standard techniques: finding tangent equations from derivatives, integration to find curve equations, factorising cubics from roots, and basic transformations. While it has many parts (11 marks total), each individual step uses routine A-level methods without requiring novel insight or complex problem-solving. Slightly easier than average due to the scaffolded nature and straightforward application of techniques. |
| Spec | 1.02j Manipulate polynomials: expanding, factorising, division, factor theorem1.02w Graph transformations: simple transformations of f(x)1.07m Tangents and normals: gradient and equations1.07n Stationary points: find maxima, minima using derivatives1.08a Fundamental theorem of calculus: integration as reverse of differentiation |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\left[\frac{dy}{dx}=\right] 4 \times 2 + 3\) or \(11\) isw | M1* | |
| \(9 = \text{their}(4 \times 2 + 3) \times 2 + c\) | M1dep* | or \(y - 9 = \text{their}(4 \times 2 + 3)(x - 2)\) |
| \(y = 11x - 13\) or \(y = 11x + c\) and \(c = -13\) stated isw | A1 | or \(y - 9 = 11(x-2)\) isw |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\frac{4x^2}{2} + 3x\) | M1* | |
| \([y =] 2x^2 + 3x + c\) | A1 | must see "2" and "\(+ c\)"; may be earned later e.g. after attempt to find \(c\) |
| \(9 = 2 \times 2^2 + 3 \times 2 + c\) | M1dep* | must include constant, which may be implied by answer |
| \(y = 2x^2 + 3x - 5\) cao | A1 | allow first 4 marks for \(y = 2x^2 + 3x + c\) and \(c = -5\) stated |
| \((1, 0)\) and \((-2.5, 0)\) oe cao | B1 | or for \(x = 1, y = 0\) and \(x = -2.5, y = 0\); B0 for just stating \(x = 1\) and \(x = -2.5\) |
| \(x = -\frac{3}{4}\) | B1 | |
| \(y = -\frac{49}{8}\) | B1 | \(-6.125\) or \(-6\frac{1}{8}\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| substitution to obtain \([y =] f(2x)\) in polynomial form | M1 | \(f(x)\) must be the quadratic in \(x\) with linear and constant term obtained in part (ii), may be in factorised form; or their \(x = 1 \rightarrow\) their \(0.5\) and their \(x = -2.5 \rightarrow\) their \(x = -1.25\) |
| \(y = (2x-1)(4x+5)\) or \(y = 8x^2 + 6x - 5\) or \(y = 2\left(2x + \frac{3}{4}\right)^2 - \frac{49}{8}\) | A1FT | must be simplified to one of these forms; FT their quadratic in \(x\) with linear and constant term obtained in part (ii); hence \(y = (2x-1)(4x+5)\) FT their \(x\)-intercepts from their quadratic in \(x\) |
| \(\left(-\frac{3}{8}, -\frac{49}{8}\right)\) oe | B1 | or FT their (both non-zero) co-ordinates for minimum point or their quadratic in \(x\) with linear and constant term obtained in part (ii) |
## Question 2(i):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\left[\frac{dy}{dx}=\right] 4 \times 2 + 3$ or $11$ isw | M1* | |
| $9 = \text{their}(4 \times 2 + 3) \times 2 + c$ | M1dep* | or $y - 9 = \text{their}(4 \times 2 + 3)(x - 2)$ |
| $y = 11x - 13$ or $y = 11x + c$ and $c = -13$ stated isw | A1 | or $y - 9 = 11(x-2)$ isw |
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## Question 2(ii):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\frac{4x^2}{2} + 3x$ | M1* | |
| $[y =] 2x^2 + 3x + c$ | A1 | must see "2" and "$+ c$"; may be earned later e.g. after attempt to find $c$ |
| $9 = 2 \times 2^2 + 3 \times 2 + c$ | M1dep* | must include constant, which may be implied by answer |
| $y = 2x^2 + 3x - 5$ cao | A1 | allow first 4 marks for $y = 2x^2 + 3x + c$ and $c = -5$ stated |
| $(1, 0)$ and $(-2.5, 0)$ oe cao | B1 | or for $x = 1, y = 0$ and $x = -2.5, y = 0$; **B0** for just stating $x = 1$ and $x = -2.5$ |
| $x = -\frac{3}{4}$ | B1 | |
| $y = -\frac{49}{8}$ | B1 | $-6.125$ or $-6\frac{1}{8}$ |
---
## Question 2(iii):
| Answer/Working | Mark | Guidance |
|---|---|---|
| substitution to obtain $[y =] f(2x)$ in polynomial form | M1 | $f(x)$ must be the quadratic in $x$ with linear and constant term obtained in part (ii), may be in factorised form; or their $x = 1 \rightarrow$ their $0.5$ and their $x = -2.5 \rightarrow$ their $x = -1.25$ |
| $y = (2x-1)(4x+5)$ or $y = 8x^2 + 6x - 5$ or $y = 2\left(2x + \frac{3}{4}\right)^2 - \frac{49}{8}$ | A1FT | must be simplified to one of these forms; FT their quadratic in $x$ with linear and constant term obtained in part (ii); hence $y = (2x-1)(4x+5)$ FT their $x$-intercepts from their quadratic in $x$ |
| $\left(-\frac{3}{8}, -\frac{49}{8}\right)$ oe | B1 | or FT their (both non-zero) co-ordinates for minimum point or their quadratic in $x$ with linear and constant term obtained in part (ii) |
---2 The gradient of a curve is given by $\frac { \mathrm { d } y } { \mathrm {~d} x } = 4 x + 3$. The curve passes through the point ( 2,9 ).\\
(i) Find the equation of the tangent to the curve at the point $( 2,9 )$.\\
(ii) Find the equation of the curve and the coordinates of its points of intersection with the $x$-axis. Find also the coordinates of the minimum point of this curve.\\
(iii) Find the equation of the curve after it has been stretched parallel to the $x$-axis with scale factor $\frac { 1 } { 2 }$. Write down the coordinates of the minimum point of the transformed curve.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{4e8d7217-61f7-4ae4-96dd-d34e37c4d623-2_1020_940_244_679}
\captionsetup{labelformat=empty}
\caption{Fig. 11}
\end{center}
\end{figure}
Fig. 11 shows a sketch of the cubic curve $y = \mathrm { f } ( x )$. The values of $x$ where it crosses the $x$-axis are - 5 , - 2 and 2 , and it crosses the $y$-axis at $( 0 , - 20 )$.\\
(i) Express $\mathrm { f } ( x )$ in factorised form.\\
(ii) Show that the equation of the curve may be written as $y = x ^ { 3 } + 5 x ^ { 2 } - 4 x - 20$.\\
(iii) Use calculus to show that, correct to 1 decimal place, the $x$-coordinate of the minimum point on the curve is 0.4 .
Find also the coordinates of the maximum point on the curve, giving your answers correct to 1 decimal place.\\
(iv) State, correct to 1 decimal place, the coordinates of the maximum point on the curve $y = \mathrm { f } ( 2 x )$.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{4e8d7217-61f7-4ae4-96dd-d34e37c4d623-3_768_1023_223_598}
\captionsetup{labelformat=empty}
\caption{Fig. 11}
\end{center}
\end{figure}
Fig. 11 shows the curve $y = x ^ { 3 } - 3 x ^ { 2 } - x + 3$.\\
(i) Use calculus to find $\int _ { 1 } ^ { 3 } \left( x ^ { 3 } - 3 x ^ { 2 } - x + 3 \right) \mathrm { d } x$ and state what this represents.\\
(ii) Find the $x$-coordinates of the turning points of the curve $y = x ^ { 3 } - 3 x ^ { 2 } - x + 3$, giving your answers in surd form. Hence state the set of values of $x$ for which $y = x ^ { 3 } - 3 x ^ { 2 } - x + 3$ is a decreasing function.\\
(i) Differentiate $x ^ { 3 } - 3 x ^ { 2 } - 9 x$. Hence find the $x$-coordinates of the stationary points on the curve $y = x ^ { 3 } - 3 x ^ { 2 } - 9 x$, showing which is the maximum and which the minimum.\\
(ii) Find, in exact form, the coordinates of the points at which the curve crosses the $x$-axis.\\
(iii) Sketch the curve.
A curve has equation $y = x ^ { 3 } - 6 x ^ { 2 } + 12$.\\
(i) Use calculus to find the coordinates of the turning points of this curve. Determine also the nature of these turning points.\\
(ii) Find, in the form $y = m x + c$, the equation of the normal to the curve at the point $( 2 , - 4 )$.
\hfill \mbox{\textit{OCR MEI C2 Q2 [13]}}