| Exam Board | Edexcel |
|---|---|
| Module | P1 (Pure Mathematics 1) |
| Year | 2019 |
| Session | June |
| Marks | 6 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Chain Rule |
| Type | Equation of tangent line |
| Difficulty | Moderate -0.8 This is a straightforward differentiation and tangent line question requiring standard power rule application (rewriting terms as powers first), substitution to find gradient at a point, and using point-slope form. All steps are routine A-level techniques with no problem-solving insight needed, making it easier than average but not trivial due to the algebraic manipulation required. |
| Spec | 1.07i Differentiate x^n: for rational n and sums1.07m Tangents and normals: gradient and equations |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(\frac{dy}{dx} = \frac{1}{8} \times 3x^2 - 24 \times -\frac{1}{2}x^{-\frac{3}{2}}\) | M1 | Reducing a correct power by one on either \(x\) term. Allow \(x^3 \to x^2\) or \(x^{-\frac{1}{2}} \to x^{-\frac{3}{2}}\). Cannot be awarded for \(\frac{24}{x^{0.5}} \to \frac{24}{0.5x^{-0.5}}\) |
| \(\frac{dy}{dx} = \frac{3}{8}x^2 + 12x^{-\frac{3}{2}}\) | A1 | One term correct and simplified |
| \(\frac{dy}{dx} = \frac{3}{8}x^2 + 12x^{-\frac{3}{2}}\) or equivalent e.g. \(\frac{3}{8}x^2 + \frac{12}{x\sqrt{x}}\) | A1 | Fully correct. ISW after correct answer. No need to see \(\frac{dy}{dx}\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(\left.\frac{dy}{dx}\right\ | _{x=4} = \frac{3}{8}\times 4^2 + 12\times 4^{-\frac{3}{2}} = 7.5\) | M1 |
| \(y + 3 = 7.5(x-4) \Rightarrow y = 7.5x - 33\) | M1 | Correct method for finding equation of tangent at \((4,-3)\) using their numerical \(\left.\frac{dy}{dx}\right\ |
| \(y = 7.5x - 33\) or exact equivalent in form \(y = mx + c\) | A1 | ISW after correct answer. If \(y=mx+c\) used must get correct \(m\) and \(c\) |
## Question 1:
### Part (a)
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\frac{dy}{dx} = \frac{1}{8} \times 3x^2 - 24 \times -\frac{1}{2}x^{-\frac{3}{2}}$ | M1 | Reducing a correct power by one on either $x$ term. Allow $x^3 \to x^2$ or $x^{-\frac{1}{2}} \to x^{-\frac{3}{2}}$. Cannot be awarded for $\frac{24}{x^{0.5}} \to \frac{24}{0.5x^{-0.5}}$ |
| $\frac{dy}{dx} = \frac{3}{8}x^2 + 12x^{-\frac{3}{2}}$ | A1 | One term correct and simplified |
| $\frac{dy}{dx} = \frac{3}{8}x^2 + 12x^{-\frac{3}{2}}$ or equivalent e.g. $\frac{3}{8}x^2 + \frac{12}{x\sqrt{x}}$ | A1 | Fully correct. ISW after correct answer. No need to see $\frac{dy}{dx}$ |
**(3 marks)**
### Part (b)
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\left.\frac{dy}{dx}\right\|_{x=4} = \frac{3}{8}\times 4^2 + 12\times 4^{-\frac{3}{2}} = 7.5$ | M1 | Attempting to find value of $\frac{dy}{dx}$ at $x=4$. Score for sight of embedded 4s followed by answer. Condone slips |
| $y + 3 = 7.5(x-4) \Rightarrow y = 7.5x - 33$ | M1 | Correct method for finding equation of tangent at $(4,-3)$ using their numerical $\left.\frac{dy}{dx}\right\|_{x=4}$. Condone one error on sign of 4 and $-3$. Cannot be awarded from made-up gradient |
| $y = 7.5x - 33$ or exact equivalent in form $y = mx + c$ | A1 | ISW after correct answer. If $y=mx+c$ used must get correct $m$ and $c$ |
**(3 marks)**
---\begin{enumerate}
\item The curve $C$ has equation $y = \frac { 1 } { 8 } x ^ { 3 } - \frac { 24 } { \sqrt { x } } + 1$\\
(a) Find $\frac { \mathrm { d } y } { \mathrm {~d} x }$, giving the answer in its simplest form.\\
(3)
\end{enumerate}
The point $P ( 4 , - 3 )$ lies on $C$.\\
(b) Find the equation of the tangent to $C$ at the point $P$. Write your answer in the form $y = m x + c$, where $m$ and $c$ are constants to be found.
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\hfill \mbox{\textit{Edexcel P1 2019 Q1 [6]}}