| 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 straightforward multi-part integration and curve question requiring standard techniques: finding a tangent (gradient substitution), integrating to find the curve equation using a boundary condition, finding roots and minimum via differentiation, and applying a transformation. All steps are routine C2 material with no novel problem-solving required, making it slightly easier than average but not trivial due to the multiple parts and transformation component. |
| Spec | 1.02w 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 differentiation1.08b Integrate x^n: where n != -1 and sums |
| Answer | Marks | Guidance |
|---|---|---|
| 1 | (i) | dy |
| Answer | Marks |
|---|---|
| isw | M1* |
| Answer | Marks |
|---|---|
| [3] | or y − 9 = their (4 × 2 + 3 ) × (x − 2) |
| Answer | Marks | Guidance |
|---|---|---|
| 1 | (ii) | 4x2 |
| Answer | Marks |
|---|---|
| 8 | M1* |
| Answer | Marks |
|---|---|
| [7] | must see “2” and “ + c”; may be earned |
| Answer | Marks | Guidance |
|---|---|---|
| −6.125 or − 6⅛ | B0 for just stating x = 1 and x = −2.5 | |
| 1 | (iii) | substitution to obtain |
| Answer | Marks |
|---|---|
| 8 8 | M1 |
| Answer | Marks |
|---|---|
| [3] | f(x) must be the quadratic in x with linear |
| Answer | Marks |
|---|---|
| linear and constant term obtained in part (ii) | or their x = 1 → their 0.5 and |
Question 1:
1 | (i) | dy
= 4 × 2 + 3 or 11 isw
dx
9 = their (4 × 2 + 3 ) × 2 + c
y = 11x − 13 or y = 11x + c and c = −13
stated
isw | M1*
M1dep*
A1
[3] | or y − 9 = their (4 × 2 + 3 ) × (x − 2)
or y − 9 = 11(x − 2) isw
1 | (ii) | 4x2
+3x
2
[y =] 2x2 + 3x + c
9 = 2×22 + 3×2 + c
y = 2x2 + 3x − 5 cao
(1, 0) and (−2.5, 0) oe cao
3
x=−
4
49
y=−
8 | M1*
A1
M1dep*
A1
B1
B1
B1
[7] | must see “2” and “ + c”; may be earned
later eg after attempt to find c
must include constant, which may be
implied by answer
allow first 4 marks for y = 2x2 + 3x + c and
c = −5 stated
or for x = 1, y = 0 and x = −2.5, y = 0
−6.125 or − 6⅛ | B0 for just stating x = 1 and x = −2.5
1 | (iii) | substitution to obtain
[ y =] f(2x) in polynomial form
y = (2x − 1)(4x + 5) or y = 8x2 + 6x − 5
2
3 49
or y=22x+ −
4 8
3 49
− ,− oe
8 8 | M1
A1FT
B1
[3] | f(x) must be the quadratic in x with linear
and constant term obtained in part (ii), may
be in factorised form
must be simplified to one of these forms,
FT their quadratic in x with linear and
constant term obtained in part (ii)
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) | or their x = 1 → their 0.5 and
their x = −2.5 → their x = −1.25
hence y = (2x − 1)(4x + 5) FT their
x-intercepts from their quadratic in x
with linear and constant term obtained
in part (ii)The gradient of a curve is given by $\frac{dy}{dx} = 4x + 3$. The curve passes through the point $(2, 9)$.
\begin{enumerate}[label=(\roman*)]
\item Find the equation of the tangent to the curve at the point $(2, 9)$. [3]
\item 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. [7]
\item 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. [3]
\end{enumerate}
\hfill \mbox{\textit{OCR MEI C2 Q1 [13]}}