| Exam Board | AQA |
|---|---|
| Module | C3 (Core Mathematics 3) |
| Year | 2011 |
| Session | January |
| Marks | 6 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Differentiating Transcendental Functions |
| Type | Differentiate inverse trigonometric functions |
| Difficulty | Moderate -0.3 This is a straightforward application of standard techniques: part (a)(i) requires differentiating tan(3y+1) using chain rule, (a)(ii) uses the reciprocal relationship dy/dx = 1/(dx/dy) with simple substitution, and part (b) is recall of a standard inverse trig graph. All components are routine textbook exercises with no problem-solving or novel insight required, making it slightly easier than average. |
| Spec | 1.05i Inverse trig functions: arcsin, arccos, arctan domains and graphs1.07s Parametric and implicit differentiation |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(\frac{dx}{dy} = 3\sec^2(3y+1)\) | M1 | Differentiating \(\tan\) correctly with chain rule |
| A1 | Correct expression |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| When \(y = -\frac{1}{3}\): \(3y + 1 = 0\), so \(\sec^2(0) = 1\) | M1 | Substituting \(y = -\frac{1}{3}\) |
| \(\frac{dx}{dy} = 3\), so \(\frac{dy}{dx} = \frac{1}{3}\) | A1 | Correct value |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| Correct shape: increasing curve with horizontal asymptotes | B1 | Asymptotes at \(y = \pm\frac{\pi}{2}\) indicated |
| Passes through origin | B1 | Correct asymptotic behaviour shown |
# Question 3:
## Part (a)(i)
| Answer | Mark | Guidance |
|--------|------|----------|
| $\frac{dx}{dy} = 3\sec^2(3y+1)$ | M1 | Differentiating $\tan$ correctly with chain rule |
| A1 | Correct expression |
## Part (a)(ii)
| Answer | Mark | Guidance |
|--------|------|----------|
| When $y = -\frac{1}{3}$: $3y + 1 = 0$, so $\sec^2(0) = 1$ | M1 | Substituting $y = -\frac{1}{3}$ |
| $\frac{dx}{dy} = 3$, so $\frac{dy}{dx} = \frac{1}{3}$ | A1 | Correct value |
## Part (b)
| Answer | Mark | Guidance |
|--------|------|----------|
| Correct shape: increasing curve with horizontal asymptotes | B1 | Asymptotes at $y = \pm\frac{\pi}{2}$ indicated |
| Passes through origin | B1 | Correct asymptotic behaviour shown |
---3
\begin{enumerate}[label=(\alph*)]
\item Given that $x = \tan ( 3 y + 1 )$ :
\begin{enumerate}[label=(\roman*)]
\item find $\frac { \mathrm { d } x } { \mathrm {~d} y }$ in terms of $y$;
\item find the value of $\frac { \mathrm { d } y } { \mathrm {~d} x }$ when $y = - \frac { 1 } { 3 }$.
\end{enumerate}\item Sketch the graph of $y = \tan ^ { - 1 } x$.
\end{enumerate}
\hfill \mbox{\textit{AQA C3 2011 Q3 [6]}}