AQA C3 2015 June — Question 4 9 marks

Exam BoardAQA
ModuleC3 (Core Mathematics 3)
Year2015
SessionJune
Marks9
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicComposite & Inverse Functions
TypeFind inverse function
DifficultyModerate -0.3 This is a standard C3 inverse function question with routine techniques: finding range of exponential function, inverting by swapping and rearranging (using logarithms), solving f^(-1)(x)=0, and composing a rational function with itself. All parts follow textbook procedures with no novel insight required, making it slightly easier than average.
Spec1.02u Functions: definition and vocabulary (domain, range, mapping)1.02v Inverse and composite functions: graphs and conditions for existence1.06g Equations with exponentials: solve a^x = b

4 The functions f and g are defined by $$\begin{array} { l l } \mathrm { f } ( x ) = 5 - \mathrm { e } ^ { 3 x } , & \text { for all real values of } x \\ \mathrm {~g} ( x ) = \frac { 1 } { 2 x - 3 } , & \text { for } x \neq 1.5 \end{array}$$
  1. Find the range of f.
  2. The inverse of f is \(\mathrm { f } ^ { - 1 }\).
    1. Find \(\mathrm { f } ^ { - 1 } ( x )\).
    2. Solve the equation \(\mathrm { f } ^ { - 1 } ( x ) = 0\).
  3. Find an expression for \(\operatorname { gg } ( x )\), giving your answer in the form \(\frac { a x + b } { c x + d }\), where \(a , b , c\) and \(d\) are integers.
    [0pt] [3 marks]

Question 4:
Part (a): Range of f
\(f(x) = 5 - e^{3x}\); since \(e^{3x} > 0\), the range is \(f(x) < 5\)
Part (b)(i): Find \(f^{-1}(x)\)
Let \(y = 5 - e^{3x}\) → \(e^{3x} = 5 - y\) → \(x = \frac{\ln(5-y)}{3}\)
\[f^{-1}(x) = \frac{\ln(5-x)}{3}\]
Part (b)(ii): Solve \(f^{-1}(x) = 0\)
\[\frac{\ln(5-x)}{3} = 0 \Rightarrow \ln(5-x) = 0 \Rightarrow x = 4\]
Part (c): Find \(gg(x)\)
\[gg(x) = \frac{1}{2\left(\frac{1}{2x-3}\right)-3} = \frac{2x-3}{2-3(2x-3)} = \frac{2x-3}{11-6x}\]
To obtain the actual mark scheme, you would need the official AQA mark scheme document.
## Question 4:

### Part (a): Range of f
$f(x) = 5 - e^{3x}$; since $e^{3x} > 0$, the range is $f(x) < 5$

### Part (b)(i): Find $f^{-1}(x)$
Let $y = 5 - e^{3x}$ → $e^{3x} = 5 - y$ → $x = \frac{\ln(5-y)}{3}$

$$f^{-1}(x) = \frac{\ln(5-x)}{3}$$

### Part (b)(ii): Solve $f^{-1}(x) = 0$
$$\frac{\ln(5-x)}{3} = 0 \Rightarrow \ln(5-x) = 0 \Rightarrow x = 4$$

### Part (c): Find $gg(x)$
$$gg(x) = \frac{1}{2\left(\frac{1}{2x-3}\right)-3} = \frac{2x-3}{2-3(2x-3)} = \frac{2x-3}{11-6x}$$

To obtain the actual **mark scheme**, you would need the official AQA mark scheme document.
4 The functions f and g are defined by

$$\begin{array} { l l } 
\mathrm { f } ( x ) = 5 - \mathrm { e } ^ { 3 x } , & \text { for all real values of } x \\
\mathrm {~g} ( x ) = \frac { 1 } { 2 x - 3 } , & \text { for } x \neq 1.5
\end{array}$$
\begin{enumerate}[label=(\alph*)]
\item Find the range of f.
\item The inverse of f is $\mathrm { f } ^ { - 1 }$.
\begin{enumerate}[label=(\roman*)]
\item Find $\mathrm { f } ^ { - 1 } ( x )$.
\item Solve the equation $\mathrm { f } ^ { - 1 } ( x ) = 0$.
\end{enumerate}\item Find an expression for $\operatorname { gg } ( x )$, giving your answer in the form $\frac { a x + b } { c x + d }$, where $a , b , c$ and $d$ are integers.\\[0pt]
[3 marks]
\end{enumerate}

\hfill \mbox{\textit{AQA C3 2015 Q4 [9]}}