| Exam Board | CAIE |
|---|---|
| Module | P1 (Pure Mathematics 1) |
| Year | 2017 |
| Session | March |
| Marks | 11 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Chain Rule |
| Type | Equation of tangent line |
| Difficulty | Standard +0.8 This is a multi-part question requiring differentiation, finding tangent/normal equations, solving a cubic equation (when the normal intersects the curve again), and finding the intersection of two tangents. While each individual step uses standard techniques, the question requires sustained problem-solving across three connected parts with increasing complexity, placing it moderately above average difficulty. |
| Spec | 1.07i Differentiate x^n: for rational n and sums1.07m Tangents and normals: gradient and equations |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\dfrac{dy}{dx} = 2x-2\). At \(x=2,\ m=2\) | B1B1 | Numerical \(m\) |
| Equation of tangent is \(y - 2 = 2(x-2)\) | B1 | Expect \(y = 2x-2\) |
| Total: | 3 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Equation of normal \(y - 2 = -\frac{1}{2}(x-2)\) | M1 | Through \((2,2)\) with gradient \(= -1/m\). Expect \(y = -\frac{1}{2}x+3\) |
| \(x^2 - 2x + 2 = -\frac{1}{2}x + 3 \rightarrow 2x^2 - 3x - 2 = 0\) | M1 | Equate and simplify to 3-term quadratic |
| \(x = -\frac{1}{2},\ y = 3\frac{1}{4}\) | A1A1 | Ignore answer of \((2,2)\) |
| Total: | 4 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| At \(x = -\frac{1}{2}\), grad \(= 2(-\frac{1}{2}) - 2 = -3\) | B1 | Follow through *their* \(-\frac{1}{2}\) |
| Equation of tangent is \(y - 3\frac{1}{4} = -3(x + \frac{1}{2})\) | \*M1 | Through *their* B with grad *their* \(-3\) (not \(m_1\) or \(m_2\)). Expect \(y = -3x + \frac{7}{4}\) |
| \(2x - 2 = -3x + \frac{7}{4}\) | DM1 | Equate *their* tangents or attempt to solve simultaneous equations |
| \(x = \frac{3}{4}\), \(y = -\frac{1}{2}\) | A1 | Both required |
## Question 9(i):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\dfrac{dy}{dx} = 2x-2$. At $x=2,\ m=2$ | B1B1 | Numerical $m$ |
| Equation of tangent is $y - 2 = 2(x-2)$ | B1 | Expect $y = 2x-2$ |
| **Total:** | **3** | |
## Question 9(ii):
| Answer | Marks | Guidance |
|--------|-------|----------|
| Equation of normal $y - 2 = -\frac{1}{2}(x-2)$ | M1 | Through $(2,2)$ with gradient $= -1/m$. Expect $y = -\frac{1}{2}x+3$ |
| $x^2 - 2x + 2 = -\frac{1}{2}x + 3 \rightarrow 2x^2 - 3x - 2 = 0$ | M1 | Equate and simplify to 3-term quadratic |
| $x = -\frac{1}{2},\ y = 3\frac{1}{4}$ | A1A1 | Ignore answer of $(2,2)$ |
| **Total:** | **4** | |
## Question 9(iii):
| Answer | Marks | Guidance |
|--------|-------|----------|
| At $x = -\frac{1}{2}$, grad $= 2(-\frac{1}{2}) - 2 = -3$ | **B1** | Follow through *their* $-\frac{1}{2}$ |
| Equation of tangent is $y - 3\frac{1}{4} = -3(x + \frac{1}{2})$ | **\*M1** | Through *their* B with grad *their* $-3$ (not $m_1$ or $m_2$). Expect $y = -3x + \frac{7}{4}$ |
| $2x - 2 = -3x + \frac{7}{4}$ | **DM1** | Equate *their* tangents or attempt to solve simultaneous equations |
| $x = \frac{3}{4}$, $y = -\frac{1}{2}$ | **A1** | Both required |
---9 The point $A ( 2,2 )$ lies on the curve $y = x ^ { 2 } - 2 x + 2$.\\
(i) Find the equation of the tangent to the curve at $A$.\\
The normal to the curve at $A$ intersects the curve again at $B$.\\
(ii) Find the coordinates of $B$.\\
The tangents at $A$ and $B$ intersect each other at $C$.\\
(iii) Find the coordinates of $C$.\\
\hfill \mbox{\textit{CAIE P1 2017 Q9 [11]}}