| Exam Board | OCR |
|---|---|
| Module | C1 (Core Mathematics 1) |
| Marks | 14 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Curve Sketching |
| Type | Find tangent to polynomial curve |
| Difficulty | Moderate -0.3 This is a straightforward C1 question involving standard techniques: sketching a cubic transformation, differentiating using chain rule, finding a tangent equation, and using the derivative to find a parallel tangent. All steps are routine applications of basic calculus with no novel problem-solving required, making it slightly easier than average. |
| Spec | 1.02n Sketch curves: simple equations including polynomials1.07i Differentiate x^n: for rational n and sums1.07m Tangents and normals: gradient and equations |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Notes |
| Graph with \((0,8)\) and \((-2,0)\) marked, correct cubic shape | B3 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Notes |
| \(f(x) = (x+2)(x^2+4x+4)\) | ||
| \(f(x) = x^3 + 4x^2 + 4x + 2x^2 + 8x + 8\) | M1 | |
| \(f(x) = x^3 + 6x^2 + 12x + 8\) | A1 | |
| \(f'(x) = 3x^2 + 12x + 12\) | M1 A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Notes |
| \(\text{grad} = 3 - 12 + 12 = 3\) | B1 | |
| \(\therefore y-1 = 3(x+1) \quad [y = 3x+4]\) | M1 A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Notes |
| \(\text{grad}\, m = 3\) | ||
| \(\therefore 3x^2 + 12x + 12 = 3\) | ||
| \(x^2 + 4x + 3 = 0\) | ||
| \((x+1)(x+3) = 0\) | M1 | |
| \(x = -1\) (at \(P\)), \(-3\) | A1 | |
| \(x = -3 \therefore y = -1\) | ||
| \(\therefore y+1 = 3(x+3)\) | M1 | |
| \(y = 3x+8\) | A1 | (14) |
# Question 10:
## Part (i):
| Answer | Mark | Notes |
|--------|------|-------|
| Graph with $(0,8)$ and $(-2,0)$ marked, correct cubic shape | B3 | |
## Part (ii):
| Answer | Mark | Notes |
|--------|------|-------|
| $f(x) = (x+2)(x^2+4x+4)$ | | |
| $f(x) = x^3 + 4x^2 + 4x + 2x^2 + 8x + 8$ | M1 | |
| $f(x) = x^3 + 6x^2 + 12x + 8$ | A1 | |
| $f'(x) = 3x^2 + 12x + 12$ | M1 A1 | |
## Part (iii):
| Answer | Mark | Notes |
|--------|------|-------|
| $\text{grad} = 3 - 12 + 12 = 3$ | B1 | |
| $\therefore y-1 = 3(x+1) \quad [y = 3x+4]$ | M1 A1 | |
## Part (iv):
| Answer | Mark | Notes |
|--------|------|-------|
| $\text{grad}\, m = 3$ | | |
| $\therefore 3x^2 + 12x + 12 = 3$ | | |
| $x^2 + 4x + 3 = 0$ | | |
| $(x+1)(x+3) = 0$ | M1 | |
| $x = -1$ (at $P$), $-3$ | A1 | |
| $x = -3 \therefore y = -1$ | | |
| $\therefore y+1 = 3(x+3)$ | M1 | |
| $y = 3x+8$ | A1 | **(14)** |
**Total: (72)**10. The curve $C$ has the equation $y = \mathrm { f } ( x )$ where
$$\mathrm { f } ( x ) = ( x + 2 ) ^ { 3 }$$
(i) Sketch the curve $C$, showing the coordinates of any points of intersection with the coordinate axes.\\
(ii) Find $\mathrm { f } ^ { \prime } ( x )$.
The straight line $l$ is the tangent to $C$ at the point $P ( - 1,1 )$.\\
(iii) Find an equation for $l$.
The straight line $m$ is parallel to $l$ and is also a tangent to $C$.\\
(iv) Show that $m$ has the equation $y = 3 x + 8$.
\hfill \mbox{\textit{OCR C1 Q10 [14]}}