CAIE P1 2012 June — Question 11 10 marks

Exam BoardCAIE
ModuleP1 (Pure Mathematics 1)
Year2012
SessionJune
Marks10
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicComposite & Inverse Functions
TypeFind minimum domain for inverse
DifficultyModerate -0.8 This is a straightforward multi-part question on standard P1 topics. Part (i) requires routine differentiation to find a stationary point. Part (ii) tests understanding that a function needs to be one-to-one for an inverse (requiring k=2 for the vertex). Parts (iii) and (iv) involve standard inverse function manipulation and sketching reflections in y=x. All techniques are textbook exercises with no novel problem-solving required, making this easier than average.
Spec1.02e Complete the square: quadratic polynomials and turning points1.02m Graphs of functions: difference between plotting and sketching1.02v Inverse and composite functions: graphs and conditions for existence1.07n Stationary points: find maxima, minima using derivatives1.07o Increasing/decreasing: functions using sign of dy/dx

The function \(f\) is such that \(f(x) = 8 - (x - 2)^2\), for \(x \in \mathbb{R}\).
  1. Find the coordinates and the nature of the stationary point on the curve \(y = f(x)\). [3]
The function \(g\) is such that \(g(x) = 8 - (x - 2)^2\), for \(k \leqslant x \leqslant 4\), where \(k\) is a constant.
  1. State the smallest value of \(k\) for which \(g\) has an inverse. [1]
For this value of \(k\),
  1. find an expression for \(g^{-1}(x)\), [3]
  2. sketch, on the same diagram, the graphs of \(y = g(x)\) and \(y = g^{-1}(x)\). [3]

\(f(x) = 8 - (x-2)^2\)
AnswerMarks Guidance
(i) Stationary point at \(x = 2\); \(y\)-coordinate \(= 8\); Nature: Maximum; (or \(y = -x^2 + 4x + 4 \to -2x + 4 = 0 \to (2, 8)\) Max)B1, B1, B1 co; co; co independent of first two marks
(ii) \(k = 2\)B1 ✓ on "x-value"
(iii) \(y = 8 - (x-2)^2 \to (x-2)^2 + y = 8 \to (x-2) = \pm\sqrt{8-y} \to g^{-1} = 2 + \sqrt{8-x}\)M1, M1, A1 Attempt to make x the subject; Order of operations correct; Must be f(x)
(iv)B1, B1, B1 B1 arc 1st quad (no tp, no axes); B1 Evidence of symmetry about y = x; B1 all correct as shown left
$f(x) = 8 - (x-2)^2$

**(i)** Stationary point at $x = 2$; $y$-coordinate $= 8$; Nature: Maximum; (or $y = -x^2 + 4x + 4 \to -2x + 4 = 0 \to (2, 8)$ Max) | B1, B1, B1 | co; co; co independent of first two marks

**(ii)** $k = 2$ | B1 | ✓ on "x-value"

**(iii)** $y = 8 - (x-2)^2 \to (x-2)^2 + y = 8 \to (x-2) = \pm\sqrt{8-y} \to g^{-1} = 2 + \sqrt{8-x}$ | M1, M1, A1 | Attempt to make x the subject; Order of operations correct; Must be f(x)

**(iv)** | B1, B1, B1 | B1 arc 1st quad (no tp, no axes); B1 Evidence of symmetry about y = x; B1 all correct as shown left
The function $f$ is such that $f(x) = 8 - (x - 2)^2$, for $x \in \mathbb{R}$.
\begin{enumerate}[label=(\roman*)]
\item Find the coordinates and the nature of the stationary point on the curve $y = f(x)$. [3]
\end{enumerate}

The function $g$ is such that $g(x) = 8 - (x - 2)^2$, for $k \leqslant x \leqslant 4$, where $k$ is a constant.
\begin{enumerate}[label=(\roman*)]
\setcounter{enumi}{1}
\item State the smallest value of $k$ for which $g$ has an inverse. [1]
\end{enumerate}

For this value of $k$,
\begin{enumerate}[label=(\roman*)]
\setcounter{enumi}{2}
\item find an expression for $g^{-1}(x)$, [3]

\item sketch, on the same diagram, the graphs of $y = g(x)$ and $y = g^{-1}(x)$. [3]
\end{enumerate}

\hfill \mbox{\textit{CAIE P1 2012 Q11 [10]}}