| Exam Board | Edexcel |
|---|---|
| Module | C34 (Core Mathematics 3 & 4) |
| Year | 2019 |
| Session | June |
| Marks | 11 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Product & Quotient Rules |
| Type | Find range using calculus |
| Difficulty | Standard +0.3 This is a standard multi-part calculus question requiring quotient rule differentiation, solving f'(x)=0 for turning points, applying transformations, and determining range from critical values. While it has multiple parts (7 marks typical), each step follows routine A-level procedures with no novel problem-solving required. The quotient rule application and algebraic simplification are straightforward, making this slightly easier than average. |
| Spec | 1.02w Graph transformations: simple transformations of f(x)1.07n Stationary points: find maxima, minima using derivatives1.07q Product and quotient rules: differentiation |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(f(x) = \frac{6x+2}{3x^2+5} \Rightarrow f'(x) = \frac{6(3x^2+5) - 6x(6x+2)}{(3x^2+5)^2}\) | M1, A1 | M1 for \(\frac{\alpha(3x^2+5) - \beta x(6x+2)}{(3x^2+5)^2}\) or equivalent. Condone obvious slips and bracketing errors. If product/quotient rule quoted it must be correct. A1: Fully correct derivative in any form |
| \(f'(x) = \frac{30 - 12x - 18x^2}{(3x^2+5)^2}\) | ||
| \(f'(x) = \frac{-6(3x^2+2x-5)}{(3x^2+5)^2}\) | A1 | Correct simplified expression. Apply isw once correct expression seen |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(f'(x) = 0 \Rightarrow 30 - 12x - 18x^2 = 0 \Rightarrow -6(3x+5)(x-1) = 0 \Rightarrow x = \ldots\) | M1 | Sets numerator \(= 0\) and attempts to solve quadratic |
| \(x = -\frac{5}{3},\ 1\) | A1 | Correct values |
| \(x = -\frac{5}{3} \Rightarrow y = \frac{6\left(-\frac{5}{3}\right)+2}{3\left(-\frac{5}{3}\right)^2+5}\) or \(x = 1 \Rightarrow y = \frac{6(1)+2}{3(1)^2+5}\) | dM1 | Finds \(y\) coordinate from \(x\) coordinate for one of their values. Dependent on previous M mark |
| \(\Rightarrow \left(-\frac{5}{3}, -\frac{3}{5}\right),\ (1, 1)\) | A1 | Correct coordinates; allow equivalent exact fractions/decimals for \(-\frac{5}{3}\) and/or \(-\frac{3}{5}\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Either \(\left(\frac{1}{2} \times \text{their } 1,\ \right)\) or \(\left(\ ,\ \text{their } 1 + 4\right)\) | B1ft | One correct or correct follow through coordinate; allow \(x = \ldots,\ y = \ldots\) |
| \(\left(\frac{1}{2} \times \text{their } 1,\ \text{their } 1 + 4\right)\) | B1ft | Both correct or correct follow through coordinates (allow \(x=\ldots, y=\ldots\)) but there should be no other points that have clearly not been discarded unless their point is clearly indicated as being the maximum |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(-\frac{3}{5} \leqslant y \leqslant \frac{2}{5}\) | M1 | For either end of inequality including \(\leqslant\) or \(\geqslant\) but allow \(<\) and/or \(>\); or e.g. max \(= \frac{2}{5}\), min \(= -\frac{3}{5}\) but not just values |
| \(-\frac{3}{5} \leqslant y \leqslant \frac{2}{5}\) | A1ft | Both ends fully correct with \(\leqslant\) and \(\geqslant\). Allow alternative notation: \(\left[-\frac{3}{5}, \frac{2}{5}\right]\), \(\left\{y: y \geqslant -\frac{3}{5} \cap y \leqslant \frac{2}{5}\right\}\) etc. Do not allow \(x\) for the range but allow \(g\) or \(g(x)\) but not \(f\) or \(f(x)\) |
## Question 8:
### Part (a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $f(x) = \frac{6x+2}{3x^2+5} \Rightarrow f'(x) = \frac{6(3x^2+5) - 6x(6x+2)}{(3x^2+5)^2}$ | M1, A1 | M1 for $\frac{\alpha(3x^2+5) - \beta x(6x+2)}{(3x^2+5)^2}$ or equivalent. Condone obvious slips and bracketing errors. If product/quotient rule quoted it must be correct. A1: Fully correct derivative in any form |
| $f'(x) = \frac{30 - 12x - 18x^2}{(3x^2+5)^2}$ | | |
| $f'(x) = \frac{-6(3x^2+2x-5)}{(3x^2+5)^2}$ | A1 | Correct simplified expression. Apply isw once correct expression seen |
### Part (b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $f'(x) = 0 \Rightarrow 30 - 12x - 18x^2 = 0 \Rightarrow -6(3x+5)(x-1) = 0 \Rightarrow x = \ldots$ | M1 | Sets numerator $= 0$ and attempts to solve quadratic |
| $x = -\frac{5}{3},\ 1$ | A1 | Correct values |
| $x = -\frac{5}{3} \Rightarrow y = \frac{6\left(-\frac{5}{3}\right)+2}{3\left(-\frac{5}{3}\right)^2+5}$ or $x = 1 \Rightarrow y = \frac{6(1)+2}{3(1)^2+5}$ | dM1 | Finds $y$ coordinate from $x$ coordinate for one of their values. **Dependent on previous M mark** |
| $\Rightarrow \left(-\frac{5}{3}, -\frac{3}{5}\right),\ (1, 1)$ | A1 | Correct coordinates; allow equivalent exact fractions/decimals for $-\frac{5}{3}$ and/or $-\frac{3}{5}$ |
### Part (c):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Either $\left(\frac{1}{2} \times \text{their } 1,\ \right)$ or $\left(\ ,\ \text{their } 1 + 4\right)$ | B1ft | One correct or correct follow through coordinate; allow $x = \ldots,\ y = \ldots$ |
| $\left(\frac{1}{2} \times \text{their } 1,\ \text{their } 1 + 4\right)$ | B1ft | Both correct or correct follow through coordinates (allow $x=\ldots, y=\ldots$) **but there should be no other points that have clearly not been discarded unless their point is clearly indicated as being the maximum** |
### Part (d):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $-\frac{3}{5} \leqslant y \leqslant \frac{2}{5}$ | M1 | For either end of inequality including $\leqslant$ or $\geqslant$ but allow $<$ and/or $>$; or e.g. max $= \frac{2}{5}$, min $= -\frac{3}{5}$ but not just values |
| $-\frac{3}{5} \leqslant y \leqslant \frac{2}{5}$ | A1ft | Both ends fully correct with $\leqslant$ and $\geqslant$. Allow alternative notation: $\left[-\frac{3}{5}, \frac{2}{5}\right]$, $\left\{y: y \geqslant -\frac{3}{5} \cap y \leqslant \frac{2}{5}\right\}$ etc. Do not allow $x$ for the range but allow $g$ or $g(x)$ but **not** $f$ or $f(x)$ |
---8.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{a9870c94-0910-46ec-a54a-44a431cb324e-22_524_1443_260_246}
\captionsetup{labelformat=empty}
\caption{Figure 2}
\end{center}
\end{figure}
Figure 2 shows a sketch of part of the curve $y = \mathrm { f } ( x )$, where
$$\mathrm { f } ( x ) = \frac { 6 x + 2 } { 3 x ^ { 2 } + 5 } , \quad x \in \mathbb { R }$$
\begin{enumerate}[label=(\alph*)]
\item Find $\mathrm { f } ^ { \prime } ( x )$, writing your answer as a single fraction in its simplest form.
The curve has two turning points, a maximum at point $A$ and a minimum at point $B$, as shown in Figure 2.
\item Using part (a), find the coordinates of point $A$ and the coordinates of point $B$.
\item State the coordinates of the maximum turning point of the function with equation
$$y = \mathrm { f } ( 2 x ) + 4 \quad x \in \mathbb { R }$$
\item Find the range of the function
$$\operatorname { g } ( x ) = \frac { 6 x + 2 } { 3 x ^ { 2 } + 5 } , \quad x \leqslant 0$$
\end{enumerate}
\hfill \mbox{\textit{Edexcel C34 2019 Q8 [11]}}