| Exam Board | AQA |
|---|---|
| Module | C3 (Core Mathematics 3) |
| Year | 2010 |
| Session | June |
| Marks | 12 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Differentiating Transcendental Functions |
| Type | Differentiate logarithmic functions |
| Difficulty | Moderate -0.3 This is a slightly below-average C3 question. Part (a) tests routine application of chain rule for standard functions (ln and sin). Part (b) involves straightforward function composition, finding ranges, solving a simple equation, and finding an inverse—all standard C3 techniques with no novel problem-solving required. The multi-part structure adds length but not conceptual difficulty. |
| Spec | 1.02u Functions: definition and vocabulary (domain, range, mapping)1.02v Inverse and composite functions: graphs and conditions for existence1.07k Differentiate trig: sin(kx), cos(kx), tan(kx)1.07l Derivative of ln(x): and related functions |
| Answer | Marks | Guidance |
|---|---|---|
| \(\frac{dy}{dx} = \frac{5}{5x-2}\) | M1 A1 | M1 for form \(\frac{k}{5x-2}\) |
| Answer | Marks |
|---|---|
| \(\frac{dy}{dx} = 2\cos 2x\) | M1 A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Range of f: \(f(x) \geq \ln\!\left(\frac{3}{2}\right)\) or \(f(x) \geq 0\) when \(x=\frac{1}{2}\): \(f\!\left(\frac{1}{2}\right) = \ln\!\left(\frac{1}{2}\right)\)... range is \(f(x) \geq \ln\!\left(\frac{3}{2}\right)\) i.e. \(\geq 0\) at \(x=\frac{1}{2}\): \(\ln(5\cdot\frac{1}{2}-2)=\ln\frac{1}{2}\), so range \(f \geq \ln\frac{1}{2}\) | M1 A1 | M1 for substituting lower bound |
| Answer | Marks |
|---|---|
| \(\text{gf}(x) = \sin(2\ln(5x-2))\) | B1 |
| Answer | Marks |
|---|---|
| \(\sin(2\ln(5x-2)) = 0 \Rightarrow 2\ln(5x-2) = 0\) | M1 |
| \(\ln(5x-2) = 0 \Rightarrow 5x-2 = 1 \Rightarrow x = \frac{3}{5}\) | M1 A1 |
| Answer | Marks |
|---|---|
| \(y = \sin 2x \Rightarrow x = \frac{1}{2}\arcsin(y) \Rightarrow g^{-1}(x) = \frac{1}{2}\arcsin(x)\) | M1 A1 |
# Question 3:
## Part (a)(i)
| $\frac{dy}{dx} = \frac{5}{5x-2}$ | M1 A1 | M1 for form $\frac{k}{5x-2}$ |
## Part (a)(ii)
| $\frac{dy}{dx} = 2\cos 2x$ | M1 A1 | |
## Part (b)(i)
| Range of f: $f(x) \geq \ln\!\left(\frac{3}{2}\right)$ or $f(x) \geq 0$ when $x=\frac{1}{2}$: $f\!\left(\frac{1}{2}\right) = \ln\!\left(\frac{1}{2}\right)$... range is $f(x) \geq \ln\!\left(\frac{3}{2}\right)$ i.e. $\geq 0$ at $x=\frac{1}{2}$: $\ln(5\cdot\frac{1}{2}-2)=\ln\frac{1}{2}$, so range $f \geq \ln\frac{1}{2}$ | M1 A1 | M1 for substituting lower bound |
## Part (b)(ii)
| $\text{gf}(x) = \sin(2\ln(5x-2))$ | B1 | |
## Part (b)(iii)
| $\sin(2\ln(5x-2)) = 0 \Rightarrow 2\ln(5x-2) = 0$ | M1 | |
| $\ln(5x-2) = 0 \Rightarrow 5x-2 = 1 \Rightarrow x = \frac{3}{5}$ | M1 A1 | |
## Part (b)(iv)
| $y = \sin 2x \Rightarrow x = \frac{1}{2}\arcsin(y) \Rightarrow g^{-1}(x) = \frac{1}{2}\arcsin(x)$ | M1 A1 | |
---3
\begin{enumerate}[label=(\alph*)]
\item Find $\frac { \mathrm { d } y } { \mathrm {~d} x }$ when:
\begin{enumerate}[label=(\roman*)]
\item $\quad y = \ln ( 5 x - 2 )$;
\item $y = \sin 2 x$.
\end{enumerate}\item The functions f and g are defined with their respective domains by
$$\begin{array} { l l }
\mathrm { f } ( x ) = \ln ( 5 x - 2 ) , & \text { for real values of } x \text { such that } x \geqslant \frac { 1 } { 2 } \\
\mathrm {~g} ( x ) = \sin 2 x , & \text { for real values of } x \text { in the interval } - \frac { \pi } { 4 } \leqslant x \leqslant \frac { \pi } { 4 }
\end{array}$$
\begin{enumerate}[label=(\roman*)]
\item Find the range of f .
\item Find an expression for $\operatorname { gf } ( x )$.
\item Solve the equation $\operatorname { gf } ( x ) = 0$.
\item The inverse of g is $\mathrm { g } ^ { - 1 }$. Find $\mathrm { g } ^ { - 1 } ( x )$.
\end{enumerate}\end{enumerate}
\hfill \mbox{\textit{AQA C3 2010 Q3 [12]}}