Standard +0.3 Part (a) is routine inverse function finding for a simple square root. Part (b) requires understanding of range and inverse existence but is standard bookwork. Part (c) involves composition and equation solving with some algebraic manipulation to reach the required form, but follows predictable steps without requiring novel insight. Overall slightly easier than average A-level.
The function f is defined as follows:
$$f(x) = \sqrt{x-1} \text{ for } x > 1.$$
\begin{enumerate}[label=(\alph*)]
\item Find an expression for \(f^{-1}(x)\). [1]
\end enumerate}
\includegraphics{figure_4}
The diagram shows the graph of \(y = g(x)\) where \(g(x) = \frac{1}{x^2+2}\) for \(x \in \mathbb{R}\).
\begin{enumerate}[label=(\alph*)]
\setcounter{enumi}{1}
\item State the range of g and explain whether \(g^{-1}\) exists. [2]
\end enumerate}
The function h is defined by \(h(x) = \frac{1}{x^2+2}\) for \(x \geqslant 0\).
\begin{enumerate}[label=(\alph*)]
\setcounter{enumi}{2}
\item Solve the equation \(hf(x) = f\left(\frac{25}{16}\right)\). Give your answer in the form \(a + b\sqrt{c}\), where \(a\), \(b\) and \(c\) are integers. [4]
\end enumerate}
g–1 does not exist because g fails the horizontal line test.
B1
g–1 can be replaced by ‘It’ throughout.
A correct statement followed by any further incorrect
explanation can be awarded B1.
2
Answer
Marks
Guidance
Question
Answer
Marks
Answer
Marks
4(c)
25 1
f
Answer
Marks
Guidance
16 4
B1
SOI
1 1
2 4
Answer
Marks
Guidance
x 1 2
M1
1
Equating , or their ‘simplified’
2
x 1 2
25
version, to their f .
16
2 2
x 1 24 leading to x 1 2 leading to x 1 2
Or
x2 x 124 leading to x2 x 10 leading to x 12 2
Or
6 364
x12 x leading to x2 6x10 leading to x
Answer
Marks
Guidance
2
A1
Simplification as far as x =…
Allow just + in the results because can be
disregarded at this stage.
Can be implied by the final answer.
Note: x1 2 scores A0.
Answer
Marks
Guidance
32 2
A1
Must discount the solution 3 2 2.
4
Answer
Marks
Guidance
Question
Answer
Marks
Question 4:
--- 4(a) ---
4(a) | f1x x12
| B1 | ISW
Condone ‘y =’.
1
--- 4(b) ---
4(b) | 1 1 1
0gx or gx0 and gx or 0,
2 2 2 | B1 | 1
Do not allow gx0, gx .
2
1
Do not allow gx0 or gx .
2
Condone g or y in place of gx.
g–1 does not exist because it is one to many or
g–1 does not exist because it is not one to one.
Or
g–1 does not exist because g is not one to one or
g–1 does not exist because g is many to one or
g–1 does not exist because g fails the horizontal line test. | B1 | g–1 can be replaced by ‘It’ throughout.
A correct statement followed by any further incorrect
explanation can be awarded B1.
2
Question | Answer | Marks | Guidance
--- 4(c) ---
4(c) | 25 1
f
16 4 | B1 | SOI
1 1
2 4
x 1 2 | M1 | 1
Equating , or their ‘simplified’
2
x 1 2
25
version, to their f .
16
2 2
x 1 24 leading to x 1 2 leading to x 1 2
Or
x2 x 124 leading to x2 x 10 leading to x 12 2
Or
6 364
x12 x leading to x2 6x10 leading to x
2 | A1 | Simplification as far as x =…
Allow just + in the results because can be
disregarded at this stage.
Can be implied by the final answer.
Note: x1 2 scores A0.
32 2 | A1 | Must discount the solution 3 2 2.
4
Question | Answer | Marks | Guidance
The function f is defined as follows:
$$f(x) = \sqrt{x-1} \text{ for } x > 1.$$
\begin{enumerate}[label=(\alph*)]
\item Find an expression for $f^{-1}(x)$. [1]
\end enumerate}
\includegraphics{figure_4}
The diagram shows the graph of $y = g(x)$ where $g(x) = \frac{1}{x^2+2}$ for $x \in \mathbb{R}$.
\begin{enumerate}[label=(\alph*)]
\setcounter{enumi}{1}
\item State the range of g and explain whether $g^{-1}$ exists. [2]
\end enumerate}
The function h is defined by $h(x) = \frac{1}{x^2+2}$ for $x \geqslant 0$.
\begin{enumerate}[label=(\alph*)]
\setcounter{enumi}{2}
\item Solve the equation $hf(x) = f\left(\frac{25}{16}\right)$. Give your answer in the form $a + b\sqrt{c}$, where $a$, $b$ and $c$ are integers. [4]
\end enumerate}
\hfill \mbox{\textit{CAIE P1 2024 Q4 [7]}}