2013 Paper 1 Q8
Year: 2013
Paper: 1
Question Number: 8
Course: LFM Stats And Pure
Section: Functions (Transformations and Inverses)
Difficulty: 1516.0
Banger: 1474.0
functions
composite functions
domain and range
curve sketching
asymptotes
logarithm
square root
inverse functions
Problem
- The functions \(\mathrm{a, b, c}\) and \(\mathrm{d}\) are defined by
- \({\rm a}(x) =x^2 \ \ \ \ (-\infty < x < \infty),\)
- \({\rm b}(x) = \ln x \ \ \ \ (x > 0),\)
- \({\rm c}(x) =2x \ \ \ \ (-\infty < x < \infty),\)
- \({\rm d}(x)= \sqrt x \ \ \ \ (x\ge0) \,.\)
- The functions \(\mathrm{f}\) and \(\mathrm{g}\) are defined by
- \(\f(x)= \sqrt{x^2-1\,} \ \ \ \ (\vert x \vert \ge 1),\)
- \(\g(x) = \sqrt{x^2+1\,} \ \ \ \ (-\infty < x < \infty).\)
- Sketch the graphs of the functions \(\h\) and \({\rm k}\) defined by
- \(\h(x) = x+\sqrt{x^2-1\,}\, \ \ \ \ ( x \ge1)\),
- \({\rm k}(x) = x-\sqrt{x^2-1\,}\, \ \ \ \ (\vert x\vert \ge1),\)
Solution
- \begin{align*} cb(x) &= c(b(x)) \\ &= 2 \ln x \quad (x > 0) \\ ab(x) &= (b(x))^2 \\ &= (\ln x)^2 \quad (x > 0) \\ da(x) &= \sqrt{a(x)} \\ &= \sqrt{x^2} \\ &= |x| \quad (-\infty < x < \infty) \\ ad(x) &= (d(x))^2 \\ &= (\sqrt{x})^2 \\ &= x \quad (x \geq 0) \end{align*} The domains are specified above. The ranges are \(\mathbb{R}, \mathbb{R}_{\geq 0}, \mathbb{R}_{\geq 0}, \mathbb{R}_{\geq 0}\) respectively.
- \begin{align*} fg(x) &= \sqrt{g(x)^2-1} \quad (|g(x)| \geq 1) \\ &= \sqrt{x^2+1-1} \\ &= |x| \end{align*} So \(fg: \mathbb{R} \to \mathbb{R}_{\geq 0}\). \begin{align*} gf(x) &= \sqrt{f(x)^2 + 1} \\ &= \sqrt{\left ( \sqrt{x^2-1} \right)^2+1} \quad (|x| \geq 1) \\ &= |x| \end{align*} So \(gf: \mathbb{R}_{\geq 0} \to \mathbb{R}_{\geq 0}\)
Examiner's report
— 2013 STEP 1, Question 8
Mean: 9.3 / 20
~80% attempted (inferred)
Inferred ~80% from 'almost 1200' out of ~1500; fourth most popular question
From the paper introduction
Around 1500 candidates sat this paper, a significant increase on last year. Overall, responses were good with candidates finding much to occupy them profitably during the three hours of the examination. In hindsight, two or three of the questions lacked sufficient 'punch' in their later parts, but at least most candidates showed sufficient skill to identify them and work on them as part of their chosen selection of questions. On the whole, nearly all candidates managed to attempt 4-6 questions – although there is always a significant minority who attempt 7, 8, 9, … bit and pieces of questions – and most scored well on at least two. Indeed, there were many scripts with 6 question-attempts, most or all of which were fantastically accomplished mathematically, and such excellence is very heart-warming.
Source: Cambridge STEP 2013 Examiner's Report
· 2013-full.pdf
Rating Information
Difficulty Rating: 1516.0
Difficulty Comparisons: 1
Banger Rating: 1474.0
Banger Comparisons: 2
Show LaTeX source
Problem source
\begin{questionparts}
\item
The functions $\mathrm{a, b, c}$ and $\mathrm{d}$ are defined by
\begin{itemize}
\item ${\rm a}(x) =x^2 \ \ \ \ (-\infty < x < \infty),$
\item ${\rm b}(x) = \ln x \ \ \ \ (x > 0),$
\item ${\rm c}(x) =2x \ \ \ \ (-\infty < x < \infty),$
\item ${\rm d}(x)= \sqrt x \ \ \ \ (x\ge0) \,.$
\end{itemize}
Write down the following composite functions, giving the domain and range of each:
\[
\rm
cb, \quad ab, \quad da, \quad ad.
\]
\item The functions $\mathrm{f}$ and $\mathrm{g}$ are defined by
\begin{itemize}
\item $\f(x)= \sqrt{x^2-1\,} \ \ \ \ (\vert x \vert \ge 1),$
\item $\g(x) = \sqrt{x^2+1\,} \ \ \ \ (-\infty < x < \infty).$
\end{itemize}
Determine the composite functions $\mathrm{fg}$ and $\mathrm{gf}$, giving the domain and range of each.
\item Sketch the graphs of the functions $\h$ and ${\rm k}$ defined by
\begin{itemize}
\item $\h(x) = x+\sqrt{x^2-1\,}\, \ \ \ \ ( x \ge1)$,
\item ${\rm k}(x) = x-\sqrt{x^2-1\,}\, \ \ \ \ (\vert x\vert \ge1),$
\end{itemize}
justifying the main features of the graphs, and giving the equations of any asymptotes. Determine the domain and range of the composite function $\mathrm{kh}$.
\end{questionparts}Solution source
\begin{questionparts}
\item \begin{align*}
cb(x) &= c(b(x)) \\
&= 2 \ln x \quad (x > 0) \\
ab(x) &= (b(x))^2 \\
&= (\ln x)^2 \quad (x > 0) \\
da(x) &= \sqrt{a(x)} \\
&= \sqrt{x^2} \\
&= |x| \quad (-\infty < x < \infty) \\
ad(x) &= (d(x))^2 \\
&= (\sqrt{x})^2 \\
&= x \quad (x \geq 0)
\end{align*}
The domains are specified above. The ranges are $\mathbb{R}, \mathbb{R}_{\geq 0}, \mathbb{R}_{\geq 0}, \mathbb{R}_{\geq 0}$ respectively.
\item \begin{align*}
fg(x) &= \sqrt{g(x)^2-1} \quad (|g(x)| \geq 1) \\
&= \sqrt{x^2+1-1} \\
&= |x|
\end{align*}
So $fg: \mathbb{R} \to \mathbb{R}_{\geq 0}$.
\begin{align*}
gf(x) &= \sqrt{f(x)^2 + 1} \\
&= \sqrt{\left ( \sqrt{x^2-1} \right)^2+1} \quad (|x| \geq 1) \\
&= |x|
\end{align*}
So $gf: \mathbb{R}_{\geq 0} \to \mathbb{R}_{\geq 0}$
\item \begin{itemize}
\item \begin{center}
\begin{tikzpicture}
\def\functionf(#1){(#1) + sqrt((#1)^2-1)};
\def\xl{-5};
\def\xu{5};
\def\yl{-10};
\def\yu{10};
% Calculate scaling factors to make the plot square
\pgfmathsetmacro{\xrange}{\xu-\xl}
\pgfmathsetmacro{\yrange}{\yu-\yl}
\pgfmathsetmacro{\xscale}{1}
\pgfmathsetmacro{\yscale}{\xrange/\yrange}
% Define the styles for the axes and grid
\tikzset{
axis/.style={very thick, ->},
grid/.style={thin, gray!30},
x=\xscale cm,
y=\yscale cm
}
% Define the bounding region with clip
\begin{scope}
% You can modify these values to change your plotting region
\clip (\xl,\yl) rectangle (\xu,\yu);
% Draw a grid (optional)
% \draw[grid] (-5,-3) grid (5,3);
\draw[thick, blue, smooth, domain=1:5, samples=100]
plot (\x, {\functionf(\x)});
\draw[thick, red, dashed] (-5, -10) -- (5, 10) node [below left] {$y = 2x$};
\end{scope}
% Set up axes
\draw[axis] (\xl,0) -- (\xu,0) node[right] {$x$};
\draw[axis] (0,\yl) -- (0,\yu) node[above] {$y$};
\end{tikzpicture}
\end{center}
\item \begin{center}
\begin{tikzpicture}
\def\functionf(#1){(#1) - sqrt((#1)^2-1)};
\def\xl{-5};
\def\xu{5};
\def\yl{-10};
\def\yu{10};
% Calculate scaling factors to make the plot square
\pgfmathsetmacro{\xrange}{\xu-\xl}
\pgfmathsetmacro{\yrange}{\yu-\yl}
\pgfmathsetmacro{\xscale}{1}
\pgfmathsetmacro{\yscale}{\xrange/\yrange}
% Define the styles for the axes and grid
\tikzset{
axis/.style={very thick, ->},
grid/.style={thin, gray!30},
x=\xscale cm,
y=\yscale cm
}
% Define the bounding region with clip
\begin{scope}
% You can modify these values to change your plotting region
\clip (\xl,\yl) rectangle (\xu,\yu);
% Draw a grid (optional)
% \draw[grid] (-5,-3) grid (5,3);
\draw[thick, blue, smooth, domain=1:5, samples=100]
plot (\x, {\functionf(\x)});
\draw[thick, blue, smooth, domain=-5:-1, samples=100]
plot (\x, {\functionf(\x)});
\draw[thick, red, dashed] (-5, -10) -- (5, 10) node [below left] {$y = 2x$};
\draw[thick, red, dashed] (-5, 0.1) -- (5, 0.1) node [below left] {$y = 0$};
\end{scope}
% Set up axes
\draw[axis] (\xl,0) -- (\xu,0) node[right] {$x$};
\draw[axis] (0,\yl) -- (0,\yu) node[above] {$y$};
\end{tikzpicture}
\end{center}
\end{itemize}
\begin{align*}
kh(x) &= h(x) - \sqrt{h(x)^2 -1} \quad (|h(x)| \geq 1)\\
&= x + \sqrt{x^2+1} - \sqrt{(x + \sqrt{x^2+1})^2 - 1} \\
&= x + \sqrt{x^2+1} - \sqrt{x^2 + x^2 - 2x} \quad (x \geq 1) \\
&= x + \sqrt{x^2+1} - \sqrt{2x^2-2x} \quad (x \geq 1)
\end{align*}
This has domain $x \geq 1$ and range, $(0, 1]$
\end{questionparts}
Almost 1200 candidates made an attempt at this question, making it fourth favourite, and the mean score on it was 9.3/20 which, if nothing else, suggests that it wasn't quite as easy as folks considered it to be. To begin with, there is a lot to do for the relatively few marks available, and minor slips over domains and ranges subsequently proved quite costly. Apart from the obvious errors from those candidates who thought the order of composition occurred the other way around, and the few who took "ab" to mean the product of the functions a and b, the usual slip-ups were: thinking that √(x²) = x, when it is actually |x|, and not realising that the domain of the composite function fg is just the domain of g. In (ii), although the functions fg and gf look the same (both are |x|), their domains and ranges are different: fg has domain ℝ and range y ≥ 0, while gf has domain |x| ≥ 1 and range y ≥ 1. A lack of a clear grasp of the domains and ranges of h and k in part (iii) was partly responsible for the poor sketches, although the ability to recognise the asymptote y = 2x was also widespread. There were even occasions, when sketching the curve for k, that a correctly drawn asymptote was subsequently labelled as y = –2x simply because of its appearance in the quadrant in which x and y are both negative.