| Exam Board | OCR |
|---|---|
| Module | H240/01 (Pure Mathematics) |
| Year | 2023 |
| Session | June |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Topic | Modulus function |
| Type | Graph y = a|bx+c| + d: identify vertex and intercepts |
| Difficulty | Standard +0.3 This is a straightforward modulus function question requiring standard techniques: using vertex and point information to find constants, explaining why a V-shaped graph isn't one-to-one, finding an inverse for a restricted domain, and analyzing when g(x) = g^(-1)(x) has no solutions. All parts follow predictable patterns with no novel insight required, making it slightly easier than average. |
| Spec | 1.02l Modulus function: notation, relations, equations and inequalities1.02u Functions: definition and vocabulary (domain, range, mapping)1.02v Inverse and composite functions: graphs and conditions for existence |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(a = 2\) | B1 | Either stated or embedded in equation. eg \( |
| \(b = 6\) | B1 | Either stated or embedded in equation. eg \( |
| \(c = 1\) | B1 | Either stated or embedded in equation. eg \( |
| [3] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Because f is a many to one function eg \(f(0) = f(6)\) | B1 | Any correct reason. Condone no explicit example. Could also say 'because f is not one to one'. B1 BOD for 'it is not one to one'. If referring to 'one to many' or 'many to one' it must be clear whether this is f or \(f^{-1}\) (just 'it' or 'the function' is not enough). Allow implication of function eg 'as it is a many to one function there is no inverse function'. May also refer to the 'horizontal line test', but need to state outcome eg horizontal line would cross graph of \(y = f(x)\) twice |
| [1] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(y = px - q\), \(px = y + q\), \(x = \frac{1}{p}(y + q)\) | M1 | Complete attempt to find inverse function of \(f(x) = px - q\); correct order of operations, allow sign error only |
| \(g^{-1}(x) = \frac{1}{p}x + \frac{q}{p}\) | A1 | Obtain correct inverse in terms of \(x\); could be single term \(g^{-1}(x) = \frac{x+q}{p}\); A1 for just \(\frac{1}{p}x + \frac{q}{p}\); if LHS seen must be \(g^{-1}(x)\) or \(y\) |
| \(x \geq 0\) | B1 | Correct domain; B0 for \(x > 0\); B1 for 'any non-negative \(x\)' but B0 for 'any positive \(x\)'; \(g^{-1}(x) \geq 0\) is B0 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(0 < p \leq 1\) | B1 | Correct set of values, any notation; no need for \(0 < p\) as specified in question, so B1 for \(p \leq 1\); B0 for \(p < 1\); B0 for any additional incorrect values |
# Question 5:
## Part (a)(i)
| Answer | Marks | Guidance |
|--------|-------|----------|
| $a = 2$ | **B1** | Either stated or embedded in equation. eg $|2x - b|$ seen. Ignore any other values seen. **B0** for $a = -2$, unless subsequently corrected |
| $b = 6$ | **B1** | Either stated or embedded in equation. eg $|ax - 6|$ seen. Ignore any other values seen |
| $c = 1$ | **B1** | Either stated or embedded in equation. eg $|ax - b| + 1$ seen. Ignore any other values seen |
| **[3]** | | |
## Part (a)(ii)
| Answer | Marks | Guidance |
|--------|-------|----------|
| Because f is a many to one function eg $f(0) = f(6)$ | **B1** | Any correct reason. Condone no explicit example. Could also say 'because f is not one to one'. **B1** BOD for 'it is not one to one'. If referring to 'one to many' or 'many to one' it must be clear whether this is f or $f^{-1}$ (just 'it' or 'the function' is not enough). Allow implication of function eg 'as it is a many to one function there is no inverse function'. May also refer to the 'horizontal line test', but need to state outcome eg horizontal line would cross graph of $y = f(x)$ twice |
| **[1]** | | |
## Question 5(b)(i):
**Finding inverse function of $g(x) = px - q$**
| Answer | Mark | Guidance |
|--------|------|----------|
| $y = px - q$, $px = y + q$, $x = \frac{1}{p}(y + q)$ | M1 | Complete attempt to find inverse function of $f(x) = px - q$; correct order of operations, allow sign error only |
| $g^{-1}(x) = \frac{1}{p}x + \frac{q}{p}$ | A1 | Obtain correct inverse in terms of $x$; could be single term $g^{-1}(x) = \frac{x+q}{p}$; A1 for just $\frac{1}{p}x + \frac{q}{p}$; if LHS seen must be $g^{-1}(x)$ or $y$ |
| $x \geq 0$ | B1 | Correct domain; B0 for $x > 0$; B1 for 'any non-negative $x$' but B0 for 'any positive $x$'; $g^{-1}(x) \geq 0$ is B0 |
---
## Question 5(b)(ii):
| Answer | Mark | Guidance |
|--------|------|----------|
| $0 < p \leq 1$ | B1 | Correct set of values, any notation; no need for $0 < p$ as specified in question, so B1 for $p \leq 1$; B0 for $p < 1$; B0 for any additional incorrect values |
---5
\begin{enumerate}[label=(\alph*)]
\item The function $\mathrm { f } ( x )$ is defined for all values of $x$ as $\mathrm { f } ( x ) = | a x - b |$, where $a$ and $b$ are positive constants.
\begin{enumerate}[label=(\roman*)]
\item The graph of $y = \mathrm { f } ( x ) + c$, where $c$ is a constant, has a vertex at $( 3,1 )$ and crosses the $y$-axis at $( 0,7 )$.
Find the values of $a , b$ and $c$.
\item Explain why $\mathrm { f } ^ { - 1 } ( x )$ does not exist.
\end{enumerate}\item The function $\mathrm { g } ( x )$ is defined for $x \geqslant \frac { q } { p }$ as $\mathrm { g } ( x ) = | p x - q |$, where $p$ and $q$ are positive constants.
\begin{enumerate}[label=(\roman*)]
\item Find, in terms of $p$ and $q$, an expression for $\mathrm { g } ^ { - 1 } ( x )$, stating the domain of $\mathrm { g } ^ { - 1 } ( x )$.
\item State the set of values of $p$ for which the equation $\mathrm { g } ( x ) = \mathrm { g } ^ { - 1 } ( x )$ has no solutions.
\end{enumerate}\end{enumerate}
\hfill \mbox{\textit{OCR H240/01 2023 Q5 [8]}}