| Exam Board | AQA |
|---|---|
| Module | C3 (Core Mathematics 3) |
| Year | 2014 |
| Session | June |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Product & Quotient Rules |
| Type | Find stationary points coordinates |
| Difficulty | Standard +0.3 This is a straightforward multi-part question testing standard differentiation techniques (chain rule, product rule, quotient rule) with routine algebraic manipulation. Part (a)(i) is basic chain rule, (a)(ii) requires product rule with substitution, and part (b) is a standard quotient rule application to find stationary points. All techniques are textbook exercises with no novel insight required, making it slightly easier than average. |
| Spec | 1.07i Differentiate x^n: for rational n and sums1.07r Chain rule: dy/dx = dy/du * du/dx and connected rates |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\frac{d}{dx}(x^2+1)^{\frac{5}{2}} = \frac{5}{2}(x^2+1)^{\frac{3}{2}} \cdot 2x\) | M1 | Chain rule attempt |
| \(= 5x(x^2+1)^{\frac{3}{2}}\) | A1 | Correct answer |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\frac{dy}{dx} = 2e^{2x}(x^2+1)^{\frac{5}{2}} + e^{2x} \cdot 5x(x^2+1)^{\frac{3}{2}}\) | M1 | Product rule applied correctly |
| At \(x=0\): \(\frac{dy}{dx} = 2e^0(1)^{\frac{5}{2}} + e^0 \cdot 0\) | M1 | Substituting \(x=0\) |
| \(= 2\) | A1 | Correct value |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\frac{dy}{dx} = \frac{4(x^2+1)-(4x-3)(2x)}{(x^2+1)^2}\) | M1 A1 | Quotient rule: M1 for structure, A1 for correct numerator |
| Numerator: \(4x^2+4-8x^2+6x = -4x^2+6x+4\) | A1 | Correct simplified numerator |
| Set numerator \(= 0\): \(-4x^2+6x+4=0\) or \(4x^2-6x-4=0\) | M1 | Setting derivative to zero |
| \(2x^2-3x-2=0 \Rightarrow (2x+1)(x-2)=0\) | ||
| \(x = -\frac{1}{2}\) or \(x = 2\) | A1 | Both correct values |
# Question 3:
## Part (a)(i):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\frac{d}{dx}(x^2+1)^{\frac{5}{2}} = \frac{5}{2}(x^2+1)^{\frac{3}{2}} \cdot 2x$ | M1 | Chain rule attempt |
| $= 5x(x^2+1)^{\frac{3}{2}}$ | A1 | Correct answer |
## Part (a)(ii):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\frac{dy}{dx} = 2e^{2x}(x^2+1)^{\frac{5}{2}} + e^{2x} \cdot 5x(x^2+1)^{\frac{3}{2}}$ | M1 | Product rule applied correctly |
| At $x=0$: $\frac{dy}{dx} = 2e^0(1)^{\frac{5}{2}} + e^0 \cdot 0$ | M1 | Substituting $x=0$ |
| $= 2$ | A1 | Correct value |
## Part (b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\frac{dy}{dx} = \frac{4(x^2+1)-(4x-3)(2x)}{(x^2+1)^2}$ | M1 A1 | Quotient rule: M1 for structure, A1 for correct numerator |
| Numerator: $4x^2+4-8x^2+6x = -4x^2+6x+4$ | A1 | Correct simplified numerator |
| Set numerator $= 0$: $-4x^2+6x+4=0$ or $4x^2-6x-4=0$ | M1 | Setting derivative to zero |
| $2x^2-3x-2=0 \Rightarrow (2x+1)(x-2)=0$ | | |
| $x = -\frac{1}{2}$ or $x = 2$ | A1 | Both correct values |3
\begin{enumerate}[label=(\alph*)]
\item \begin{enumerate}[label=(\roman*)]
\item Differentiate $\left( x ^ { 2 } + 1 \right) ^ { \frac { 5 } { 2 } }$ with respect to $x$.
\item Given that $y = \mathrm { e } ^ { 2 x } \left( x ^ { 2 } + 1 \right) ^ { \frac { 5 } { 2 } }$, find the value of $\frac { \mathrm { d } y } { \mathrm {~d} x }$ when $x = 0$.
\end{enumerate}\item A curve has equation $y = \frac { 4 x - 3 } { x ^ { 2 } + 1 }$. Use the quotient rule to find the $x$-coordinates of the stationary points of the curve.\\[0pt]
[5 marks]
\begin{center}
\includegraphics[max width=\textwidth, alt={}]{57412ec0-ad97-4418-8ba8-93f1f7d8aac1-06_1855_1709_852_153}
\end{center}
\end{enumerate}
\hfill \mbox{\textit{AQA C3 2014 Q3 [10]}}