| Exam Board | Edexcel |
|---|---|
| Module | Paper 1 (Paper 1) |
| Year | 2019 |
| Session | June |
| Marks | 5 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Product & Quotient Rules |
| Type | Show derivative equals given algebraic form |
| Difficulty | Moderate -0.3 This is a straightforward quotient rule application followed by algebraic simplification to match a given form, then a simple inequality. The quotient rule is standard A-level technique, the algebra is routine, and part (b) requires only basic sign analysis. Slightly easier than average due to being a direct application with clear structure. |
| Spec | 1.07o Increasing/decreasing: functions using sign of dy/dx1.07q Product and quotient rules: differentiation |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Correct method to differentiate \(y = \frac{5x^2+10x}{(x+1)^2}\) | M1 | Use of quotient (& chain) rule or product & chain rule. Expect \(\frac{dy}{dx} = \frac{(x+1)^2(Ax+B)-(5x^2+10x)(Cx+D)}{(x+1)^4}\), \(A,B,C,D>0\) |
| \(\frac{dy}{dx} = \frac{(x+1)^2(10x+10)-(5x^2+10x)\times 2(x+1)}{(x+1)^4}\) | A1 | Correct unsimplified answer |
| Factorises/cancels \((x+1)\): \(\frac{dy}{dx} = \frac{(x+1)(10x+10)-(5x^2+10x)\times 2}{(x+1)^3} = \frac{A}{(x+1)^3}\) | M1 | Valid attempt to proceed to given form. Dependent on quotient rule of \(\pm\frac{v\,du - u\,dv}{v^2}\) and proceeding to \(\frac{A}{(x+1)^3}\) |
| \(\frac{dy}{dx} = \frac{10}{(x+1)^3}\) | A1 | No requirement to see \(\frac{dy}{dx}=\) ; can recover from missing brackets/slips |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Writes \(y = \frac{5x^2+10x}{(x+1)^2}\) in form \(y = A \pm \frac{B}{(x+1)^2}\), \(A,B \neq 0\) | M1 | Strategy of rewriting before differentiating |
| \(y = 5 - \frac{5}{(x+1)^2}\) | A1 | Correct simplified form |
| Uses chain rule \(\Rightarrow \frac{dy}{dx} = \frac{C}{(x+1)^3}\) | M1 | May be scored from \(A=0\) |
| \(\frac{dy}{dx} = \frac{10}{(x+1)^3}\) | A1 | Cannot be awarded from incorrect value of \(A\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| For \(x < -1\) | B1ft | Follow through on their \(\frac{dy}{dx} = \frac{A}{(x+1)^n}\), \(n=1,3\). If \(A>0\) and \(n=1,3 \Rightarrow x<-1\). If \(A<0\) and \(n=1,3\), award for \(x>-1\) |
## Question 3:
### Part (a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Correct method to differentiate $y = \frac{5x^2+10x}{(x+1)^2}$ | M1 | Use of quotient (& chain) rule or product & chain rule. Expect $\frac{dy}{dx} = \frac{(x+1)^2(Ax+B)-(5x^2+10x)(Cx+D)}{(x+1)^4}$, $A,B,C,D>0$ |
| $\frac{dy}{dx} = \frac{(x+1)^2(10x+10)-(5x^2+10x)\times 2(x+1)}{(x+1)^4}$ | A1 | Correct unsimplified answer |
| Factorises/cancels $(x+1)$: $\frac{dy}{dx} = \frac{(x+1)(10x+10)-(5x^2+10x)\times 2}{(x+1)^3} = \frac{A}{(x+1)^3}$ | M1 | Valid attempt to proceed to given form. Dependent on quotient rule of $\pm\frac{v\,du - u\,dv}{v^2}$ and proceeding to $\frac{A}{(x+1)^3}$ |
| $\frac{dy}{dx} = \frac{10}{(x+1)^3}$ | A1 | No requirement to see $\frac{dy}{dx}=$ ; can recover from missing brackets/slips |
### Part (a) — Alternative via division:
| Answer/Working | Mark | Guidance |
|---|---|---|
| Writes $y = \frac{5x^2+10x}{(x+1)^2}$ in form $y = A \pm \frac{B}{(x+1)^2}$, $A,B \neq 0$ | M1 | Strategy of rewriting before differentiating |
| $y = 5 - \frac{5}{(x+1)^2}$ | A1 | Correct simplified form |
| Uses chain rule $\Rightarrow \frac{dy}{dx} = \frac{C}{(x+1)^3}$ | M1 | May be scored from $A=0$ |
| $\frac{dy}{dx} = \frac{10}{(x+1)^3}$ | A1 | Cannot be awarded from incorrect value of $A$ |
### Part (b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| For $x < -1$ | B1ft | Follow through on their $\frac{dy}{dx} = \frac{A}{(x+1)^n}$, $n=1,3$. If $A>0$ and $n=1,3 \Rightarrow x<-1$. If $A<0$ and $n=1,3$, award for $x>-1$ |
---3.
$$y = \frac { 5 x ^ { 2 } + 10 x } { ( x + 1 ) ^ { 2 } } \quad x \neq - 1$$
\begin{enumerate}[label=(\alph*)]
\item Show that $\frac { \mathrm { d } y } { \mathrm {~d} x } = \frac { A } { ( x + 1 ) ^ { n } }$ where $A$ and $n$ are constants to be found.
\item Hence deduce the range of values for $x$ for which $\frac { \mathrm { d } y } { \mathrm {~d} x } < 0$
\end{enumerate}
\hfill \mbox{\textit{Edexcel Paper 1 2019 Q3 [5]}}