| Exam Board | Edexcel |
|---|---|
| Module | C3 (Core Mathematics 3) |
| Year | 2018 |
| Session | June |
| Marks | 6 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Product & Quotient Rules |
| Type | Determine increasing/decreasing intervals |
| Difficulty | Moderate -0.3 This is a straightforward application of the product rule followed by routine algebraic manipulation. Part (a) requires differentiating using the product and chain rules, then factorising—a standard C3 skill. Part (b) involves solving a simple inequality from the factorised derivative. While it requires multiple steps, each is routine and commonly practiced, making it slightly easier than average. |
| Spec | 1.07q Product and quotient rules: differentiation |
| Answer | Marks |
|---|---|
| \(y = 2x(3x-1)^5 \Rightarrow \frac{dy}{dx} = 2(3x-1)^5 + 30x(3x-1)^4\) | M1A1 |
| \(\Rightarrow \left(\frac{dy}{dx}\right) = 2(3x-1)^4 \{(3x-1)+15x\} = 2(3x-1)^4(18x-1)\) | M1A1 |
| Answer | Marks |
|---|---|
| \(\frac{dy}{dx} \leq 0 \Rightarrow 2(3x-1)^4(18x-1) \leq 0 \Rightarrow x \leq \frac{1}{18}, x = \frac{1}{3}\) | B1ft, B1 |
**(a)**
| $y = 2x(3x-1)^5 \Rightarrow \frac{dy}{dx} = 2(3x-1)^5 + 30x(3x-1)^4$ | M1A1 |
| $\Rightarrow \left(\frac{dy}{dx}\right) = 2(3x-1)^4 \{(3x-1)+15x\} = 2(3x-1)^4(18x-1)$ | M1A1 |
(4 marks)
**(b)**
| $\frac{dy}{dx} \leq 0 \Rightarrow 2(3x-1)^4(18x-1) \leq 0 \Rightarrow x \leq \frac{1}{18}, x = \frac{1}{3}$ | B1ft, B1 |
(2 marks)
(6 marks total)
---\begin{enumerate}
\item Given $y = 2 x ( 3 x - 1 ) ^ { 5 }$,\\
(a) find $\frac { \mathrm { d } y } { \mathrm {~d} x }$, giving your answer as a single fully factorised expression.\\
(b) Hence find the set of values of $x$ for which $\frac { \mathrm { d } y } { \mathrm {~d} x } \leqslant 0$\\
\end{enumerate}
\hfill \mbox{\textit{Edexcel C3 2018 Q1 [6]}}