2012 Paper 1 Q3
Year: 2012
Paper: 1
Question Number: 3
Course: LFM Pure
Section: Integration
Difficulty: 1516.0
Banger: 1484.0
integration
curve sketching
inequality
area comparison
trigonometric
exponential
logarithm
geometric proof
Problem
- Sketch the curve \(y=\sin x\) for \(0\le x \le \tfrac12 \pi\) and add to your diagram the tangent to the curve at the origin and the chord joining the origin to the point \((b, \sin b)\), where \(0 < b < \frac12\pi\). By considering areas, show that \[ 1-\tfrac12 b^2 <\cos b < 1-\tfrac 12 b \sin b\,. \]
- By considering the curve \(y=a^x\), where \(a>1\), show that \[ \frac{2(a-1)}{a+1} < \ln a < -1 + \sqrt{2a-1\,}\,. \] [Hint: You may wish to write \(a^x\) as \(\e^{x\ln a}\).]
Solution
- \(\,\)
The area under the blue curve is \(1-\cos b\). The area under the green line is \(\frac12 b \sin b\) The area under the red line is \(\frac12 b^2\) Therefore \(\frac12 b \sin b < 1- \cos b < \frac12 b^2 \Rightarrow 1- \frac12 b^2 < \cos b < 1 - \frac12 b \sin b\)
- \(\,\)
\begin{align*} &&\text{Area under blue curve}: &= \int_0^1 a^x \d x\\ &&&= \left [ \frac{1}{\ln a}e^{x \ln a} \right]_0^1 \\ &&&= \frac{a-1}{\ln a} \\ \\ &&\text{Area under green line}: &=\frac12 \cdot 1 \cdot (a + 1)\\ &&&= \frac{a+1}{2} \\ \\ &&\text{Area under tangent}: &=\frac12 \cdot 1 \cdot (1+\ln a + 1)\\ &&&= \frac{\ln a+2}{2} \\ \\ \Rightarrow && \frac{a+1}{2} & > \frac{a-1}{\ln a} \\ \Rightarrow && \ln a& > \frac{2(a-1)}{a+1} \\ \\ \Rightarrow && \frac{a-1}{\ln a} &> \frac{\ln a +2}{2} \\ \Rightarrow && 2(a-1) -2\ln a - (\ln a)^2 &> 0 \\ \Rightarrow && \ln a & < -1 + \sqrt{2a-1} \end{align*}
Examiner's report
— 2012 STEP 1, Question 3
Source: Cambridge STEP 2012 Examiner's Report
· 2012-full.pdf
Rating Information
Difficulty Rating: 1516.0
Difficulty Comparisons: 1
Banger Rating: 1484.0
Banger Comparisons: 1
Show LaTeX source
Problem source
\begin{questionparts}
\item
Sketch the curve $y=\sin x$ for $0\le x \le \tfrac12 \pi$ and add to your diagram the tangent to the curve at the origin and the chord joining the origin to the point $(b, \sin b)$, where $0 < b < \frac12\pi$.
By considering areas, show that
\[
1-\tfrac12 b^2 <\cos b < 1-\tfrac 12 b \sin b\,.
\]
\item By considering the curve $y=a^x$, where $a>1$, show
that
\[
\frac{2(a-1)}{a+1} < \ln a < -1 + \sqrt{2a-1\,}\,.
\]
[\textbf{Hint}: You may wish to write $a^x$ as $\e^{x\ln a}$.]
\end{questionparts}Solution source
\begin{questionparts}
\item $\,$
\begin{center}
\begin{tikzpicture}
\def\functionf(#1){sin(180/pi * (#1))};
\def\xl{-0.2};
\def\xu{2};
\def\yl{-0.2};
\def\yu{1.5};
% 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);
\draw[thick, blue, smooth, domain=0:{pi/2}, samples=100]
plot (\x, {\functionf(\x)});
\filldraw ({pi/2},0) circle (1.5pt) node[below] {$\frac{\pi}{2}$};
\filldraw ({pi/3}, {sin(60)}) circle (1.5pt) node[above] {\tiny $(b, \sin b)$};
\filldraw ({pi/3}, {pi/3}) circle (1.5pt) node[above] {\tiny $(b, b)$};
\draw[red, dashed] (\xl, \xl) -- (\xu, \xu) node[pos = 0.5, above, sloped] {\tiny $y = x$};
% \draw[red, dashed] (\xl, {cos(60)*(\xl - pi/3) + sin(60)}) -- (\xu, {cos(60)*(\xu - pi/3) + sin(60)}) node[pos = 0.8, above, sloped] {\tiny $y = \cos b(x-b)+\sin b$};
\draw[green] (0,0) -- ({pi/3}, {sin(60)});
\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}
The area under the blue curve is $1-\cos b$.
The area under the green line is $\frac12 b \sin b$
The area under the red line is $\frac12 b^2$
Therefore $\frac12 b \sin b < 1- \cos b < \frac12 b^2 \Rightarrow 1- \frac12 b^2 < \cos b < 1 - \frac12 b \sin b$
\item $\,$
\begin{center}
\begin{tikzpicture}
\def\functionf(#1){exp(ln(4)*(#1))};
\def\xl{-0.4};
\def\xu{2};
\def\yl{-0.4};
\def\yu{10};
% 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);
\draw[thick, blue, smooth, domain=\xl:\xu, samples=100]
plot (\x, {\functionf(\x)});
\node[blue, below, rotate=50] at (2.4, {\functionf(2.4)}) {\tiny $y = a^x$};
\filldraw (1, {\functionf(1)}) circle (1.5pt) node[above] {$(1, a)$};
\filldraw (0,1) circle (1.5pt) node[left] {$1$};
\filldraw (1,{ln(4)+1}) circle (1.5pt) node[below] {$(1, 1+\ln a)$};
\draw[green] (0,1) -- (1, {\functionf(1)});
\draw[red, dashed] (\xl,{ln(4)*\xl+1}) -- (\xu,{ln(4)*\xu+1}) node[pos = 0.75, above, sloped] {\tiny $y = x \ln a + 1$};
% \draw[red, dashed] (\xl, {cos(60)*(\xl - pi/3) + sin(60)}) -- (\xu, {cos(60)*(\xu - pi/3) + sin(60)}) node[pos = 0.8, above, sloped] {\tiny $y = \cos b(x-b)+\sin b$};
\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}
\begin{align*}
&&\text{Area under blue curve}: &= \int_0^1 a^x \d x\\
&&&= \left [ \frac{1}{\ln a}e^{x \ln a} \right]_0^1 \\
&&&= \frac{a-1}{\ln a} \\
\\
&&\text{Area under green line}: &=\frac12 \cdot 1 \cdot (a + 1)\\
&&&= \frac{a+1}{2} \\
\\
&&\text{Area under tangent}: &=\frac12 \cdot 1 \cdot (1+\ln a + 1)\\
&&&= \frac{\ln a+2}{2} \\
\\
\Rightarrow && \frac{a+1}{2} & > \frac{a-1}{\ln a} \\
\Rightarrow && \ln a& > \frac{2(a-1)}{a+1} \\
\\
\Rightarrow && \frac{a-1}{\ln a} &> \frac{\ln a +2}{2} \\
\Rightarrow && 2(a-1) -2\ln a - (\ln a)^2 &> 0 \\
\Rightarrow && \ln a & < -1 + \sqrt{2a-1}
\end{align*}
\end{questionparts}
The first part of this question was generally well answered by those candidates who recognised the concept of placing the required area between two triangles. There were a number of answers where the graph was sketched, but no further progress was made however. Of those who solved the first part of the question, many were able to do some work on the second part of the question, but took the upper limit of the integral as (consistent with the method for the first part of the question, but not required here as is the upper limit to choose) and therefore arrived at far more complicated expressions than were needed. Often this led to a doomed attempt to handle the algebra that was left. For those who correctly chose the limits of the integration the part of the inequality whose solution involved still proved a challenge for some.