| Exam Board | Edexcel |
|---|---|
| Module | C3 (Core Mathematics 3) |
| Year | 2014 |
| Session | June |
| Marks | 6 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Product & Quotient Rules |
| Type | Show derivative equals given algebraic form |
| Difficulty | Moderate -0.8 Part (a) is a routine quotient rule application with simple algebraic simplification to verify a given result. Part (b) requires solving a simple equation. This is a standard textbook exercise testing basic differentiation technique with no problem-solving insight required, making it easier than average. |
| Spec | 1.02k Simplify rational expressions: factorising, cancelling, algebraic division1.07i Differentiate x^n: for rational n and sums1.07m Tangents and normals: gradient and equations1.07q Product and quotient rules: differentiation |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Marks | Guidance |
| Applies \(\frac{vu'-uv'}{v^2}\) to get \(\frac{(x-2)\times4-(4x+1)\times1}{(x-2)^2}\) | M1A1 | M1: quotient rule applied with \(u=4x+1\), \(v=x-2\). If rule quoted it must be correct. May be implied by \(u=4x+1, v=x-2, u'=.., v'=..\) followed by \(\frac{vu'-uv'}{v^2}\). If neither quoted nor implied, only accept form \(\frac{(x-2)\times A-(4x+1)\times B}{(x-2)^2}\), \(A,B>0\) |
| \(=\frac{-9}{(x-2)^2}\) | A1* | All aspects correct including bracketing. Accept \(-9(x-2)^{-2}\) |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Marks | Guidance |
| \(\frac{-9}{(x-2)^2}=-1 \Rightarrow x=\ldots\) | M1 | Sets derivative equal to \(-1\) and proceeds to \(x=\ldots\). Minimum: multiply by \((x-2)^2\) then divide by \(-1\) before square rooting, or multiply out and solve 3TQ |
| \(x=5\) | A1 | |
| \((5,7)\) | A1 | Accept \((5,7)\) or \(x=5, y=7\). Ignore reference to \(x=-1\) (and \(y=1\)). Do not accept \(21/3\) for \(7\). If extra solution with \(x>2\), withhold final mark |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Marks | Guidance |
| \(f(x)=\frac{4x+1}{x-2}=4+\frac{9}{x-2}\), applies chain rule to get \(f'(x)=A(x-2)^{-2}\) | M1 | Must have attempted to divide first |
| \(=-9(x-2)^{-2}=\frac{-9}{(x-2)^2}\) | A1, A1* |
## Question 1:
### Part (a)
| Working | Marks | Guidance |
|---------|-------|----------|
| Applies $\frac{vu'-uv'}{v^2}$ to get $\frac{(x-2)\times4-(4x+1)\times1}{(x-2)^2}$ | M1A1 | M1: quotient rule applied with $u=4x+1$, $v=x-2$. If rule quoted it must be correct. May be implied by $u=4x+1, v=x-2, u'=.., v'=..$ followed by $\frac{vu'-uv'}{v^2}$. If neither quoted nor implied, only accept form $\frac{(x-2)\times A-(4x+1)\times B}{(x-2)^2}$, $A,B>0$ |
| $=\frac{-9}{(x-2)^2}$ | A1* | All aspects correct including bracketing. Accept $-9(x-2)^{-2}$ |
### Part (b)
| Working | Marks | Guidance |
|---------|-------|----------|
| $\frac{-9}{(x-2)^2}=-1 \Rightarrow x=\ldots$ | M1 | Sets derivative equal to $-1$ and proceeds to $x=\ldots$. Minimum: multiply by $(x-2)^2$ then divide by $-1$ before square rooting, or multiply out and solve 3TQ |
| $x=5$ | A1 | |
| $(5,7)$ | A1 | Accept $(5,7)$ or $x=5, y=7$. Ignore reference to $x=-1$ (and $y=1$). Do not accept $21/3$ for $7$. If extra solution with $x>2$, withhold final mark |
### Alt 1(a) — Chain Rule
| Working | Marks | Guidance |
|---------|-------|----------|
| $f(x)=\frac{4x+1}{x-2}=4+\frac{9}{x-2}$, applies chain rule to get $f'(x)=A(x-2)^{-2}$ | M1 | Must have attempted to divide first |
| $=-9(x-2)^{-2}=\frac{-9}{(x-2)^2}$ | A1, A1* | |
---\begin{enumerate}
\item The curve $C$ has equation $y = \mathrm { f } ( x )$ where
\end{enumerate}
$$f ( x ) = \frac { 4 x + 1 } { x - 2 } , \quad x > 2$$
(a) Show that
$$f ^ { \prime } ( x ) = \frac { - 9 } { ( x - 2 ) ^ { 2 } }$$
Given that $P$ is a point on $C$ such that $\mathrm { f } ^ { \prime } ( x ) = - 1$,\\
(b) find the coordinates of $P$.\\
\hfill \mbox{\textit{Edexcel C3 2014 Q1 [6]}}