| Exam Board | Edexcel |
|---|---|
| Module | P1 (Pure Mathematics 1) |
| Year | 2020 |
| Session | October |
| Marks | 11 |
| 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 P1 differentiation question requiring product rule (or expansion then differentiation), finding a tangent equation by substitution, and using the parallel tangent condition. All steps are routine applications of standard techniques with no novel problem-solving required, making it slightly easier than average. |
| Spec | 1.07i Differentiate x^n: for rational n and sums1.07m Tangents and normals: gradient and equations1.07q Product and quotient rules: differentiation |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(y=(x-2)(x^2-8x+16) \Rightarrow y=x^3-8x^2+16x-2x^2+16x-32 \Rightarrow x^3\pm\ldots x^2\pm\ldots x\pm32\) | M1 | Attempts to multiply out three brackets. Score for expressions of form \(x^3\pm\ldots x^2\pm\ldots x\pm32\) |
| \(y = x^3-10x^2+32x-32\) | A1 | Must have attempted to multiply out brackets |
| \(\frac{\mathrm{d}y}{\mathrm{d}x} = 3x^2-20x+32\) | M1A1* | M1: \(x^n \to x^{n-1}\) correct on one term. A1*: Fully correct with no errors, including \(\frac{\mathrm{d}y}{\mathrm{d}x}\) shown |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(x=6 \Rightarrow y=(6-2)(6-4)^2=16\) | B1 | |
| \(\frac{\mathrm{d}y}{\mathrm{d}x}=3(6)^2-20(6)+32=20\) | B1 | |
| \(y - \text{"}16\text{"} = \text{"}20\text{"}(x-6)\) | M1 | |
| \(y=20x-104\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(3x^2-20x+32=\text{"}20\text{"} \Rightarrow 3x^2-20x+12=0\) | M1 | |
| \(3x^2-20x+12=0 \Rightarrow (3x-2)(x-6)=0 \Rightarrow x=\ldots\) | dM1 | Dependent on previous M1 |
| \(\alpha = \frac{2}{3}\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\frac{dy}{dx} = (x-2) \times A(x-4) \pm B(x-4)^2\) | 2nd M1 | Applies product rule; look for this form or equivalent |
| \(\frac{dy}{dx} = (x-2) \times 2(x-4) + (x-4)^2\) | 1st A1 | |
| \(\frac{dy}{dx} = 2x^2 - 8x - 4x + 16 + x^2 - 8x + 16 \Rightarrow 3x^2 - 20x + 32\) | 1st M1 | Attempts to multiply out and collect terms to form a 3TQ |
| \(\frac{dy}{dx} = 3x^2 - 20x + 32\) | 2nd A1* | With no errors |
| \(y = (x-2)(x^2 - 8x + 16) \Rightarrow y = x^3 \pm ... x^2 \pm ... x \pm 32\) | M1 | Does not require middle terms to score M1 |
| \(y = x^3 - 10x^2 + 32x - 32\) | A1 | oe |
| \(\int(3x^2 - 20x + 32)\,dx = x^3 - 10x^2 + 32x + C\) | M1 | Look for correct index on one term |
| Deduce \(C = -32\) and conclude \(\frac{dy}{dx} = 3x^2 - 20x + 32\) | A1* | With no errors seen |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(y\)-coordinate identified as 16 | B1 | Beware that \(\frac{d^2y}{dx^2} = 16\) when \(x = 6\) |
| Gradient identified as 20; \(\frac{dy}{dx} = 20\), \(m = 20\), \(g = 20\) | B1 | May be used within their equation for the tangent |
| \(y - \text{"16"} = \text{"20"}(x-6)\) | M1 | Using their \(y\) from \(x=6\) into \(y=(x-2)(x^2-8x+16)\); gradient from substituting \(x=6\) into \(\frac{dy}{dx} = 3x^2 - 20x + 32\); cannot be changed gradient |
| \(y = 20x - 104\) | A1 | cao |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Equates \(3x^2 - 20x + 32\) with their 20 and collects terms to obtain a 3TQ | M1 | Condone slips in rearrangement |
| Attempts to solve their 3TQ | dM1 | Dependent on previous M1; if roots just stated, check on calculator |
| \(\alpha = \dfrac{2}{3}\) (allow \(x = ...\)) | A1 | Ignore sight of 6; answer on its own scores full marks. Note: values of 4, \(\frac{8}{3}\) imply solving \(3x^2 - 20x + 32 = 0\) which scores 0 marks |
## Question 8:
### Part (a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $y=(x-2)(x^2-8x+16) \Rightarrow y=x^3-8x^2+16x-2x^2+16x-32 \Rightarrow x^3\pm\ldots x^2\pm\ldots x\pm32$ | M1 | Attempts to multiply out three brackets. Score for expressions of form $x^3\pm\ldots x^2\pm\ldots x\pm32$ |
| $y = x^3-10x^2+32x-32$ | A1 | Must have attempted to multiply out brackets |
| $\frac{\mathrm{d}y}{\mathrm{d}x} = 3x^2-20x+32$ | M1A1* | M1: $x^n \to x^{n-1}$ correct on one term. A1*: Fully correct with no errors, including $\frac{\mathrm{d}y}{\mathrm{d}x}$ shown |
### Part (b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $x=6 \Rightarrow y=(6-2)(6-4)^2=16$ | B1 | |
| $\frac{\mathrm{d}y}{\mathrm{d}x}=3(6)^2-20(6)+32=20$ | B1 | |
| $y - \text{"}16\text{"} = \text{"}20\text{"}(x-6)$ | M1 | |
| $y=20x-104$ | A1 | |
### Part (c):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $3x^2-20x+32=\text{"}20\text{"} \Rightarrow 3x^2-20x+12=0$ | M1 | |
| $3x^2-20x+12=0 \Rightarrow (3x-2)(x-6)=0 \Rightarrow x=\ldots$ | dM1 | Dependent on previous M1 |
| $\alpha = \frac{2}{3}$ | A1 | |
# Mark Scheme Extraction
---
## Question (Part relating to dy/dx and tangent):
### Part (a) - Differentiation via Product Rule (Alternative Method):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\frac{dy}{dx} = (x-2) \times A(x-4) \pm B(x-4)^2$ | 2nd M1 | Applies product rule; look for this form or equivalent |
| $\frac{dy}{dx} = (x-2) \times 2(x-4) + (x-4)^2$ | 1st A1 | |
| $\frac{dy}{dx} = 2x^2 - 8x - 4x + 16 + x^2 - 8x + 16 \Rightarrow 3x^2 - 20x + 32$ | 1st M1 | Attempts to multiply out and collect terms to form a 3TQ |
| $\frac{dy}{dx} = 3x^2 - 20x + 32$ | 2nd A1* | With no errors |
| $y = (x-2)(x^2 - 8x + 16) \Rightarrow y = x^3 \pm ... x^2 \pm ... x \pm 32$ | M1 | Does not require middle terms to score M1 |
| $y = x^3 - 10x^2 + 32x - 32$ | A1 | oe |
| $\int(3x^2 - 20x + 32)\,dx = x^3 - 10x^2 + 32x + C$ | M1 | Look for correct index on one term |
| Deduce $C = -32$ and conclude $\frac{dy}{dx} = 3x^2 - 20x + 32$ | A1* | With no errors seen |
### Part (b) - Tangent:
| Answer/Working | Mark | Guidance |
|---|---|---|
| $y$-coordinate identified as 16 | B1 | Beware that $\frac{d^2y}{dx^2} = 16$ when $x = 6$ |
| Gradient identified as 20; $\frac{dy}{dx} = 20$, $m = 20$, $g = 20$ | B1 | May be used within their equation for the tangent |
| $y - \text{"16"} = \text{"20"}(x-6)$ | M1 | Using their $y$ from $x=6$ into $y=(x-2)(x^2-8x+16)$; gradient from substituting $x=6$ into $\frac{dy}{dx} = 3x^2 - 20x + 32$; cannot be changed gradient |
| $y = 20x - 104$ | A1 | cao |
### Part (c) - Solving for stationary points:
| Answer/Working | Mark | Guidance |
|---|---|---|
| Equates $3x^2 - 20x + 32$ with their 20 and collects terms to obtain a 3TQ | M1 | Condone slips in rearrangement |
| Attempts to solve their 3TQ | dM1 | Dependent on previous M1; if roots just stated, check on calculator |
| $\alpha = \dfrac{2}{3}$ (allow $x = ...$) | A1 | Ignore sight of 6; answer on its own scores full marks. Note: values of 4, $\frac{8}{3}$ imply solving $3x^2 - 20x + 32 = 0$ which scores 0 marks |
---8. The curve $C$ has equation
$$y = ( x - 2 ) ( x - 4 ) ^ { 2 }$$
\begin{enumerate}[label=(\alph*)]
\item Show that
$$\frac { \mathrm { d } y } { \mathrm {~d} x } = 3 x ^ { 2 } - 20 x + 32$$
The line $l _ { 1 }$ is the tangent to $C$ at the point where $x = 6$
\item Find the equation of $l _ { 1 }$, giving your answer in the form $y = m x + c$, where $m$ and $c$ are constants to be found.
The line $l _ { 2 }$ is the tangent to $C$ at the point where $x = \alpha$\\
Given that $l _ { 1 }$ and $l _ { 2 }$ are parallel and distinct,
\item find the value of $\alpha$\\
\begin{center}
\end{center}
\end{enumerate}
\hfill \mbox{\textit{Edexcel P1 2020 Q8 [11]}}