| Exam Board | Edexcel |
|---|---|
| Module | C4 (Core Mathematics 4) |
| Marks | 11 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Product & Quotient Rules |
| Type | Find stationary points and nature |
| Difficulty | Standard +0.3 This is a standard C4 quotient rule question requiring finding stationary points and determining their nature using the second derivative (which is given). The quotient rule application is routine, solving the resulting equation is straightforward, and the second derivative test is mechanical. Slightly above average difficulty only due to the algebraic manipulation required. |
| Spec | 1.02n Sketch curves: simple equations including polynomials1.07n Stationary points: find maxima, minima using derivatives1.07o Increasing/decreasing: functions using sign of dy/dx1.07p Points of inflection: using second derivative |
| Answer | Marks | Guidance |
|---|---|---|
| \(\frac{dy}{dx} = \frac{(4+x^2)(1) - x(2x)}{(4+x^2)^2} = \frac{4-x^2}{(4+x^2)^2}\) | M1 A1 | Quotient rule correctly applied |
| Setting \(\frac{dy}{dx} = 0\): \(4 - x^2 = 0 \Rightarrow x = \pm 2\) | M1 | Setting numerator to zero |
| \(x=2\): \(y = \frac{2}{8} = \frac{1}{4}\); \(x=-2\): \(y = \frac{-2}{8} = -\frac{1}{4}\) | A1 A1 | Both coordinates correct |
| Answer | Marks | Guidance |
|---|---|---|
| At \(x=2\): \(\frac{d^2y}{dx^2} = \frac{2(2)(4-12)}{(8)^3} = \frac{-32}{512} < 0\) → maximum | M1 A1 | Correct substitution and conclusion |
| At \(x=-2\): \(\frac{d^2y}{dx^2} > 0\) → minimum | A1 | Correct conclusion |
| Answer | Marks | Guidance |
|---|---|---|
| Sketch showing odd function, maximum at \((2, \frac{1}{4})\), minimum at \((-2, -\frac{1}{4})\) | B1 | Correct shape |
| Passes through origin, asymptote \(y=0\) as \(x \to \pm\infty\) | B1 B1 | Correct behaviour at origin and asymptotes |
# Question 7:
## Part (a)
| $\frac{dy}{dx} = \frac{(4+x^2)(1) - x(2x)}{(4+x^2)^2} = \frac{4-x^2}{(4+x^2)^2}$ | M1 A1 | Quotient rule correctly applied |
| Setting $\frac{dy}{dx} = 0$: $4 - x^2 = 0 \Rightarrow x = \pm 2$ | M1 | Setting numerator to zero |
| $x=2$: $y = \frac{2}{8} = \frac{1}{4}$; $x=-2$: $y = \frac{-2}{8} = -\frac{1}{4}$ | A1 A1 | Both coordinates correct |
## Part (b)
| At $x=2$: $\frac{d^2y}{dx^2} = \frac{2(2)(4-12)}{(8)^3} = \frac{-32}{512} < 0$ → maximum | M1 A1 | Correct substitution and conclusion |
| At $x=-2$: $\frac{d^2y}{dx^2} > 0$ → minimum | A1 | Correct conclusion |
## Part (c)
| Sketch showing odd function, maximum at $(2, \frac{1}{4})$, minimum at $(-2, -\frac{1}{4})$ | B1 | Correct shape |
| Passes through origin, asymptote $y=0$ as $x \to \pm\infty$ | B1 B1 | Correct behaviour at origin and asymptotes |
---7. The curve $C$ has equation $y = \frac { x } { 4 + x ^ { 2 } }$.
\begin{enumerate}[label=(\alph*)]
\item Use calculus to find the coordinates of the turning points of $C$.
Using the result $\frac { \mathrm { d } ^ { 2 } y } { \mathrm {~d} x ^ { 2 } } = \frac { 2 x \left( x ^ { 2 } - 12 \right) } { \left( 4 + x ^ { 2 } \right) ^ { 3 } }$, or otherwise,
\item determine the nature of each of the turning points.
\item Sketch the curve $C$.
\end{enumerate}
\hfill \mbox{\textit{Edexcel C4 Q7 [11]}}