| Exam Board | OCR |
|---|---|
| Module | H240/01 (Pure Mathematics) |
| Year | 2019 |
| Session | June |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Topic | Composite & Inverse Functions |
| Type | Find intersection points |
| Difficulty | Standard +0.3 This is a straightforward multi-part question on functions requiring standard techniques: finding domain restrictions for invertibility (vertex of parabola), evaluating composite functions, solving a quadratic equation, and recognizing that f(x)=x implies f(x)=f^(-1)(x). All parts are routine applications with no novel insight required, making it slightly easier than average. |
| Spec | 1.02f Solve quadratic equations: including in a function of unknown1.02v Inverse and composite functions: graphs and conditions for existence |
| Answer | Marks | Guidance |
|---|---|---|
| \(k = 3\) | B1 | State 3. B0 for \(k \geq 3\). Allow B1 for \(x \geq 3\), as this implies \(k = 3\) |
| Answer | Marks | Guidance |
|---|---|---|
| \(f(5) = -13\) | M1 | Attempt \(f(5)\). Could be implied by \(-13\). Could be part of algebraic attempt at \(ff(x)\), with \(x = 5\) used, but does not need to be evaluated for M1 |
| \(-13\) is not in domain so \(f(-13)\), and hence \(ff(5)\), is not defined | A1 | Correct conclusion. Allow equiv, such as 'not possible'. SC Allow A1 for \(f(-13) = 239\) |
| Answer | Marks | Guidance |
|---|---|---|
| \((x-3)^2 - 17 = x\), \(x^2 - 7x - 8 = 0\), \(x = 8,\; x = -1\) | M1 | Equate and attempt to solve. Equate and produce at least one root, not necessarily correct for their equation. Could be implied by sight of 8, or 8 and \(-1\), even if equation not seen |
| Obtain at least \(x = 8\) | A1 | If second root is given, it must also be correct. WWW, e.g. \(x = 8\) given as only root from \((x-8)(x-1)\) is M1A0 |
| \(x = -1\) is not valid as \(x \geq 3\), so \(x = 8\) | A1 | Obtain \(x = 8\) only, having discarded \(x = -1\), with a reason such as 'not in the domain' or 'less than 3'. Must be using \(k = 3\); if referring to 'less than \(k\)' then 3 must have been seen in part (a). Must see some indication that the other root would have been \(-1\), e.g. a factor of \((x+1)\) or a numerical quadratic formula not fully evaluated |
| Answer | Marks | Guidance |
|---|---|---|
| \(f(x)\) and \(f^{-1}(x)\) are reflections in the line \(y = x\) so the point of intersection must be on \(y = x\) | B1 | Correct description. Sufficient to see \(f(x)\) and \(f^{-1}(x)\) intersect on \(y = x\) or reference to reflections in \(y = x\) |
## Question 3:
### Part (a)
$k = 3$ | **B1** | State 3. B0 for $k \geq 3$. Allow B1 for $x \geq 3$, as this implies $k = 3$
### Part (b)
$f(5) = -13$ | **M1** | Attempt $f(5)$. Could be implied by $-13$. Could be part of algebraic attempt at $ff(x)$, with $x = 5$ used, but does not need to be evaluated for M1
$-13$ is not in domain so $f(-13)$, and hence $ff(5)$, is not defined | **A1** | Correct conclusion. Allow equiv, such as 'not possible'. **SC** Allow A1 for $f(-13) = 239$
### Part (c)
$(x-3)^2 - 17 = x$, $x^2 - 7x - 8 = 0$, $x = 8,\; x = -1$ | **M1** | Equate and attempt to solve. Equate and produce at least one root, not necessarily correct for their equation. Could be implied by sight of 8, or 8 and $-1$, even if equation not seen
Obtain at least $x = 8$ | **A1** | If second root is given, it must also be correct. WWW, e.g. $x = 8$ given as only root from $(x-8)(x-1)$ is M1A0
$x = -1$ is not valid as $x \geq 3$, so $x = 8$ | **A1** | Obtain $x = 8$ only, having discarded $x = -1$, with a reason such as 'not in the domain' or 'less than 3'. Must be using $k = 3$; if referring to 'less than $k$' then 3 must have been seen in part **(a)**. Must see some indication that the other root would have been $-1$, e.g. a factor of $(x+1)$ or a numerical quadratic formula not fully evaluated
### Part (d)
$f(x)$ and $f^{-1}(x)$ are reflections in the line $y = x$ so the point of intersection must be on $y = x$ | **B1** | Correct description. Sufficient to see $f(x)$ and $f^{-1}(x)$ intersect on $y = x$ **or** reference to reflections in $y = x$
---3 The function f is defined by $\mathrm { f } ( x ) = ( x - 3 ) ^ { 2 } - 17$ for $x \geqslant k$, where $k$ is a constant.
\begin{enumerate}[label=(\alph*)]
\item Given that $\mathrm { f } ^ { - 1 } ( x )$ exists, state the least possible value of $k$.
\item Evaluate $\mathrm { ff } ( 5 )$.
\item Solve the equation $\mathrm { f } ( x ) = x$.
\item Explain why your solution to part (c) is also the solution to the equation $\mathrm { f } ( x ) = \mathrm { f } ^ { - 1 } ( x )$.
\end{enumerate}
\hfill \mbox{\textit{OCR H240/01 2019 Q3 [7]}}