| Exam Board | Edexcel |
|---|---|
| Module | P1 (Pure Mathematics 1) |
| Year | 2024 |
| Session | January |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Product & Quotient Rules |
| Type | Find equation of tangent |
| Difficulty | Moderate -0.3 This is a straightforward application of the quotient rule followed by finding a tangent equation using point-slope form. Part (a) requires careful algebraic simplification, and part (b) is routine substitution. Slightly easier than average due to being a standard two-part question with no conceptual challenges beyond applying learned techniques. |
| Spec | 1.07i Differentiate x^n: for rational n and sums1.07m Tangents and normals: gradient and equations |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(y = \frac{5x^3-8}{2x^2} = \frac{5}{2}x - 4x^{-2}\) | M1 | Attempts to write \(y\) as a sum of two terms, achieves at least one term with correct index. Award for \(Px + Qx^k\) or \(Px^k + Qx^{-2}\). Condone \(Qx^{-2} \leftrightarrow \frac{Q}{x^2}\) |
| \(\frac{dy}{dx} = \frac{5}{2} + 8x^{-3}\) | dM1 | Requires both: expression written as sum of two terms with both indices correct, AND correct differentiation applied to indices \(Px + Qx^{-2} \rightarrow A + Bx^{-3}\) |
| One correct term (unsimplified acceptable, condone \(\frac{5}{2}x^0\)) | A1 | |
| \(\frac{dy}{dx} = \frac{5}{2} + 8x^{-3}\) or simplified equivalent e.g. \(\frac{1}{2}\left(5 + \frac{16}{x^3}\right)\) | A1 | No need to have \(\frac{dy}{dx}\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Substitutes \(x=2\) into \(\frac{dy}{dx} = "\frac{5}{2} + \frac{8}{2^3}" \Rightarrow \frac{dy}{dx} = "\frac{7}{2}"\) | M1 | \(\frac{dy}{dx}\) cannot be the same as their \(y\). Score for sight of embedded 2's in their \(\frac{dy}{dx}\) followed by a value |
| Uses \(\frac{7}{2}\) and \((2,4) \Rightarrow y-4 = "\frac{7}{2}"(x-2)\) | M1 | Correct method for finding equation of tangent. Look for correct use of \(\frac{dy}{dx}\big |
| \(7x - 2y - 6 = 0\) | A1 | \(7x-2y-6=0\) or any multiple thereof |
## Question 3:
### Part (a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $y = \frac{5x^3-8}{2x^2} = \frac{5}{2}x - 4x^{-2}$ | M1 | Attempts to write $y$ as a sum of two terms, achieves at least one term with correct index. Award for $Px + Qx^k$ or $Px^k + Qx^{-2}$. Condone $Qx^{-2} \leftrightarrow \frac{Q}{x^2}$ |
| $\frac{dy}{dx} = \frac{5}{2} + 8x^{-3}$ | dM1 | Requires both: expression written as sum of two terms with both indices correct, AND correct differentiation applied to indices $Px + Qx^{-2} \rightarrow A + Bx^{-3}$ |
| One correct term (unsimplified acceptable, condone $\frac{5}{2}x^0$) | A1 | |
| $\frac{dy}{dx} = \frac{5}{2} + 8x^{-3}$ or simplified equivalent e.g. $\frac{1}{2}\left(5 + \frac{16}{x^3}\right)$ | A1 | No need to have $\frac{dy}{dx}$ |
### Part (b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Substitutes $x=2$ into $\frac{dy}{dx} = "\frac{5}{2} + \frac{8}{2^3}" \Rightarrow \frac{dy}{dx} = "\frac{7}{2}"$ | M1 | $\frac{dy}{dx}$ cannot be the same as their $y$. Score for sight of embedded 2's in their $\frac{dy}{dx}$ followed by a value |
| Uses $\frac{7}{2}$ and $(2,4) \Rightarrow y-4 = "\frac{7}{2}"(x-2)$ | M1 | Correct method for finding equation of tangent. Look for correct use of $\frac{dy}{dx}\big|_{x=2}$ and point $(2,4)$. If $y=mx+c$ used must proceed to $c=\ldots$ |
| $7x - 2y - 6 = 0$ | A1 | $7x-2y-6=0$ or any multiple thereof |
---\begin{enumerate}
\item The curve $C$ has equation
\end{enumerate}
$$y = \frac { 5 x ^ { 3 } - 8 } { 2 x ^ { 2 } } \quad x > 0$$
(a) Find $\frac { \mathrm { d } y } { \mathrm {~d} x }$ writing your answer in simplest form.
The point $P ( 2,4 )$ lies on $C$.\\
(b) Find an equation for the tangent to $C$ at $P$ writing your answer in the form $a x + b y + c = 0$, where $a , b$ and $c$ are integers.
\hfill \mbox{\textit{Edexcel P1 2024 Q3 [7]}}