2019 Paper 2 Q2
Year: 2019
Paper: 2
Question Number: 2
Course: LFM Pure
Section: Differentiation
Difficulty: 1500.0
Banger: 1500.0
integration
inverse function
geometric proof
curve sketching
implicit differentiation
definite integral
area between curves
Problem
The function f satisfies \(f(0) = 0\) and \(f'(t) > 0\) for \(t > 0\). Show by means of a sketch that, for \(x > 0\),
$$\int_0^x f(t) \, dt + \int_0^{f(x)} f^{-1}(y) \, dy = xf(x).$$
- The (real) function g is defined, for all \(t\), by $$(g(t))^3 + g(t) = t.$$ Prove that \(g(0) = 0\), and that \(g'(t) > 0\) for all \(t\). Evaluate \(\int_0^2 g(t) \, dt\).
- The (real) function h is defined, for all \(t\), by $$(h(t))^3 + h(t) = t + 2.$$ Evaluate \(\int_0^8 h(t) \, dt\).
Solution
- Suppose \((g(t))^3 + g(t) = t\). Notice that \((g(0))^3 + g(0) =0 \Rightarrow g(0)((g(0))^2 + 1) = 0 \Rightarrow g(0) = 0\).
\begin{align*}
&& t &= (g(t))^3 + g(t) \\
\Rightarrow && 1 &= 3(g(t))^2 g'(t) + g'(t) \\
\Rightarrow && g'(t) &= \frac{1}{1 + 3(g(t))^2} > 0
\end{align*}
From our sketch, we can see we are interested in: \begin{align*} && \int_0^2 g(t) \d t &= 2 - \int_0^1 (x^3 + x) \d x \\ &&&= 2 - \frac14 - \frac12 = \frac54 \end{align*}
- \(\,\)
From our second sketch, we can see that: \begin{align*} && \int_0^8 h(t) \d t &= 16 - \int_1^2 (x^3+x-2) \d x \\ &&&= 16 - \left ( \frac{8}{4} + \frac{2^2}{2} - 2 \cdot 2 \right)+ \left ( \frac{1}{4} + \frac{1}{2} - 2 \right) \\ &&&= \frac{59}{4} \end{align*}
Examiner's report
— 2019 STEP 2, Question 2
~80% attempted (inferred)
Inferred ~80%: 'another popular question', Pure (≥50%), described alongside Q1 as popular; second-best average mark of all questions.
From the paper introduction
The Pure questions were again the most popular of the paper, with only one of the questions attempted by fewer than half of the candidates (of the remaining four questions, only question 9 was attempted by more than a quarter of the candidates). In many of the questions candidates were often unable to make good use of the results shown in the earlier parts of the question in order to solve the more complex later parts. Nevertheless, some good solutions were seen to all of the questions. For many of the questions, solutions were seen in which the results were reached, but without sufficient justification of some of the steps.
Source: Cambridge STEP 2019 Examiner's Report
· 2019-p2.pdf
Rating Information
Difficulty Rating: 1500.0
Difficulty Comparisons: 0
Banger Rating: 1500.0
Banger Comparisons: 0
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Problem source
The function f satisfies $f(0) = 0$ and $f'(t) > 0$ for $t > 0$. Show by means of a sketch that, for $x > 0$,
$$\int_0^x f(t) \, dt + \int_0^{f(x)} f^{-1}(y) \, dy = xf(x).$$
\begin{questionparts}
\item The (real) function g is defined, for all $t$, by
$$(g(t))^3 + g(t) = t.$$
Prove that $g(0) = 0$, and that $g'(t) > 0$ for all $t$.
Evaluate $\int_0^2 g(t) \, dt$.
\item The (real) function h is defined, for all $t$, by
$$(h(t))^3 + h(t) = t + 2.$$
Evaluate $\int_0^8 h(t) \, dt$.
\end{questionparts}Solution source
\begin{center}
\begin{tikzpicture}
\def\functionf(#1){1.1*(#1) - sin(180*#1/pi) };
\def\xl{-2};
\def\xu{15};
\def\yl{-2};
\def\yu{15};
% Calculate scaling factors to make the plot square
\pgfmathsetmacro{\xrange}{\xu-\xl}
\pgfmathsetmacro{\yrange}{\yu-\yl}
\pgfmathsetmacro{\xscale}{10/\xrange}
\pgfmathsetmacro{\yscale}{10/\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);
% \filldraw (-0.5, 0) circle (1.5pt) node[below]{$p$};
\filldraw (10, 0) circle (1.5pt) node[below]{$x$};
\filldraw (10, {\functionf(10)}) circle (1.5pt) node[right]{$(x, f(x))$};
\filldraw (0, {\functionf(10)}) circle (1.5pt) node[left]{$f(x)$};
% \filldraw (2.1, 0) circle (1.5pt) node[below]{$r$};
\draw[thick, blue, smooth, domain=0:\xu, samples=100]
plot ({\x}, {\functionf(\x)});
\filldraw[thick, blue, opacity=0.25, smooth, domain=0:10, samples=100]
plot ({\x}, {\functionf(\x)}) -- (10,0);
\filldraw[thick, red, opacity=0.25, smooth, domain=0:10, samples=100]
plot ({\x}, {\functionf(\x)}) -- (0,{\functionf(10)});
\node[blue, above, rotate=10] at ({12}, {\functionf(12)}) {\tiny $y=f(t)$};
\node at ({7}, {0.5*\functionf(5)}) {$\displaystyle \int_0^x f(t) dt$};
\node at ({0.5*\functionf(5)}, 8) {$\displaystyle \int_0^{f(x)} f^{-1}(y) dy$};
\end{scope}
% Set up axes
\draw[axis] (\xl,0) -- (\xu,0) node[right] {$t$};
\draw[axis] (0,\yl) -- (0,\yu) node[above] {$y$};
\end{tikzpicture}
\end{center}
Notice the total area is $xf(x)$ and it is made up of the sum of the two integrals.
\begin{questionparts}
\item Suppose $(g(t))^3 + g(t) = t$. Notice that $(g(0))^3 + g(0) =0 \Rightarrow g(0)((g(0))^2 + 1) = 0 \Rightarrow g(0) = 0$.
\begin{align*}
&& t &= (g(t))^3 + g(t) \\
\Rightarrow && 1 &= 3(g(t))^2 g'(t) + g'(t) \\
\Rightarrow && g'(t) &= \frac{1}{1 + 3(g(t))^2} > 0
\end{align*}
\begin{center}
\begin{tikzpicture}
\def\functionf(#1){(#1)^3+(#1)};
\def\xl{-0.5};
\def\xu{3};
\def\yl{-0.5};
\def\yu{2};
% Calculate scaling factors to make the plot square
\pgfmathsetmacro{\xrange}{\xu-\xl}
\pgfmathsetmacro{\yrange}{\yu-\yl}
\pgfmathsetmacro{\xscale}{10/\xrange}
\pgfmathsetmacro{\yscale}{10/\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);
% \filldraw (-0.5, 0) circle (1.5pt) node[below]{$p$};
\filldraw (2, 0) circle (1.5pt) node[below]{$2$};
\filldraw (2, 1) circle (1.5pt) node[above]{$(2,1)$};
% \filldraw (2.1, 0) circle (1.5pt) node[below]{$r$};
\draw[thick, blue, smooth, domain=0:\yu, samples=100]
plot ({\functionf(\x)}, \x);
\filldraw[thick, blue, opacity=0.25, smooth, domain=0:1, samples=100]
plot ({\functionf(\x)}, \x) -- (2,0);
\filldraw[thick, red, opacity=0.25, smooth, domain=0:1, samples=100]
plot ({\functionf(\x)}, \x) -- (0,1);
\node[blue, above, rotate=10] at ({\functionf(1.15)}, 1.15) {\tiny $y^3+y=x$};
\node at (1.2, 0.25) {$\displaystyle \int_0^2 g(t) dt$};
\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}
From our sketch, we can see we are interested in:
\begin{align*}
&& \int_0^2 g(t) \d t &= 2 - \int_0^1 (x^3 + x) \d x \\
&&&= 2 - \frac14 - \frac12 = \frac54
\end{align*}
\item $\,$
\begin{center}
\begin{tikzpicture}
\def\functionf(#1){(#1)^3+(#1)-2};
\def\xl{-0.5};
\def\xu{11};
\def\yl{-0.5};
\def\yu{3};
% Calculate scaling factors to make the plot square
\pgfmathsetmacro{\xrange}{\xu-\xl}
\pgfmathsetmacro{\yrange}{\yu-\yl}
\pgfmathsetmacro{\xscale}{10/\xrange}
\pgfmathsetmacro{\yscale}{10/\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);
% \filldraw (-0.5, 0) circle (1.5pt) node[below]{$p$};
\filldraw (8, 0) circle (1.5pt) node[below]{$8$};
\filldraw (8, 2) circle (1.5pt) node[above]{$(8,2)$};
\filldraw (0, 1) circle (1.5pt) node[left]{$1$};
\draw[thick, blue, smooth, domain=0:\yu, samples=100]
plot ({\functionf(\x)}, \x);
\filldraw[thick, blue, opacity=0.25, smooth, domain=1:2, samples=100]
plot ({\functionf(\x)}, \x) -- (8,0) -- (0,0);
\filldraw[thick, red, opacity=0.25, smooth, domain=1:2, samples=100]
plot ({\functionf(\x)}, \x) -- (0,2);
\node[blue, above, rotate=15] at ({\functionf(2.1)}, 2.1) {\tiny $y^3+y-2=x$};
\node at (4, .8) {$\displaystyle \int_0^8 h(t) dt$};
\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}
From our second sketch, we can see that:
\begin{align*}
&& \int_0^8 h(t) \d t &= 16 - \int_1^2 (x^3+x-2) \d x \\
&&&= 16 - \left ( \frac{8}{4} + \frac{2^2}{2} - 2 \cdot 2 \right)+ \left ( \frac{1}{4} + \frac{1}{2} - 2 \right) \\
&&&= \frac{59}{4}
\end{align*}
\end{questionparts}
This question was another popular question that was generally well answered, achieving the second-best average mark of all of the questions and was also the question for which the largest number of solutions received full marks. Most candidates drew a convincing sketch to demonstrate that the two integrals make a rectangle. Arguments from sketches showing the inverse function and reflective symmetry were less successful and often candidates' diagrams assumed x to be a fixed point of f(t). By far the most common mistake in the first part was to notice the solution g(2) = 1 but not to factorise and use the quadratic discriminant to show that no other solutions were possible. The conceptually difficult part was to use g(y) = y + y, and many candidates stopped just before this point. In the final part, many candidates tried to apply the stem identity in its original form, without noticing that h(0) ≠ 0. This was the most difficult part, and those who modified it correctly generally did well. Candidates sometimes failed to check that h′(t) > 0, but this was not necessary for those who used h(t) = g(t−2).