Year: 2017
Paper: 1
Question Number: 2
Course: LFM Pure
Section: Integration
The pure questions were again the most popular of the paper with questions 1 and 3 being attempted by almost all candidates. The least popular questions on the paper were questions 10, 11, 12 and 13 and a significant proportion of attempts at these were brief, attracting few or no marks. Candidates generally demonstrated a high level of competence when completing the standard processes and there were many good attempts made when questions required explanations to be given, particularly within the pure questions. A common feature of the stronger responses to questions was the inclusion of diagrams.
Difficulty Rating: 1484.0
Difficulty Comparisons: 1
Banger Rating: 1500.1
Banger Comparisons: 2
\begin{questionparts}
\item The inequality $\dfrac 1 t \le 1$ holds for $t\ge1$.
By integrating both sides of this inequality
over the interval $1\le t \le x$, show that
\[
\ln x \le x-1
\tag{$*$}
\]
for $x \ge 1$. Show similarly
that $(*)$ also holds for $0 < x \le 1$.
\item
Starting from the inequality
$\dfrac{1}{t^2} \le \dfrac1 t $ for $t \ge 1$,
show that
\[
\ln x \ge 1-\frac{1}{x}
\tag{$**$}
\]
for $x > 0$.
\item
Show, by integrating ($*$) and ($**$), that
\[
\frac{2}{ y+1} \le \frac{\ln y}{ y-1} \le \frac{ y+1}{2 y}
\]
for $ y > 0$ and $ y\ne1$.
\end{questionparts}\begin{questionparts}
\item $\,$
\begin{align*}
(x \geq 1): && \int_1^x \frac{1}{t} \d t &\leq \int_1^x 1 \d t \\
\Rightarrow && \ln x - \ln 1 &\leq x - 1 \\
\Rightarrow && \ln x & \leq x - 1 \\
\\
(0 < x \leq 1):&& \int_x^1 1\d t &\leq \int_x^1 \frac{1}{t} \d t \\
\Rightarrow&& 1- x &\leq \ln 1 - \ln x \\
\Rightarrow&& \ln x &\leq x - 1
\end{align*}
\item $\,$
\begin{align*}
(x \geq 1): && \int_1^x \frac{1}{t^2} \d t &\leq \int_1^x \frac{1}{t} \d t \\
\Rightarrow && -\frac1x+1 &\leq \ln x - \ln 1 \\
\Rightarrow && 1 - \frac1x &\leq \ln x \\
\\
(0 < x \leq 1): && \int_x^1 \frac{1}{t} \d t &\leq \int_x^1 \frac{1}{t^2} \d t \\
\Rightarrow && \ln 1 - \ln x & \leq -1 + \frac{1}{x} \\
\Rightarrow && 1 - \frac1x &\leq \ln x \\
\end{align*}
\item $\,$
\begin{align*}
(1 < y): && \int_1^y \left (1 - \frac1{x} \right)\d x &\leq \int_1^y \ln x \d x \\
\Rightarrow && \left [x - \ln x \right]_1^y & \leq \left [ x \ln x - x\right]_1^y \\
\Rightarrow && y - \ln y - 1 &\leq y \ln y - y +1 \\
\Rightarrow && 2y-2 & \leq (y+1) \ln y \\
\Rightarrow && \frac{2}{y+1} & \leq \frac{\ln y}{y-1} \\
(0 < y < 1): && \int_y^1 \left (1 - \frac1{x} \right)\d x &\leq \int_y^1 \ln x \d x \\
\Rightarrow && \left [x - \ln x \right]_y^1 & \leq \left [ x \ln x - x\right]_y^1 \\
\Rightarrow && 1 - (y - \ln y) &\leq -1-(y \ln y-y) \\
\Rightarrow && 2-2y &\leq -(y+1)\ln y \\
\Rightarrow && \frac{2}{y+1} &\leq \frac{-\ln y}{1-y} \tag{$1-y > 0$} \\
\Rightarrow && \frac{2}{y+1} &\leq \frac{\ln y}{y-1}
\\
\\
(1 < y): && \int_1^y \ln x \d x &\leq \int_1^y (x-1) \d x \\
\Rightarrow && \left [x \ln x -x \right]_1^y &\leq \left[ \frac12 x^2 - x \right]_1^y\\
\Rightarrow && y \ln y - y +1 &\leq \frac12y^2 - y+\frac12 \\
\Rightarrow && y \ln y &\leq \frac12 \left (y^2-1 \right) \\
\Rightarrow && \frac{\ln y}{y-1} &\leq \frac{y+1}{2y} \\
\\
(0 < y < 1) && \int_y^1 \ln x \d x &\leq \int_y^1 (x-1) \d x \\
\Rightarrow && \left [x \ln x -x \right]_y^1&\leq \left[ \frac12 x^2 - x \right]_y^1\\
\Rightarrow && -1-(y \ln y - y +1) &\leq-\frac12 - \left ( \frac12y^2 - y\right)\\
\Rightarrow && \frac12 \left (y^2-1 \right) &\leq y \ln y \\
\Rightarrow && \frac{\ln y}{y-1} & \leq \frac{y+1}{2y} \tag{$y-1 < 0$}
\end{align*}
\end{questionparts}
This was the third most popular question on the paper, after questions 1 and 3. Many candidates were able to show the result required in part (i) and to explain why the result also holds for the second range of values of . However, a number of the solutions did not use integration as instructed by the question and so were not able to achieve all of the marks for the question. The second part was also carried out successfully by a large number of candidates and, as there was no specification that integration should be used for this part, a number of alternative solutions were also able to achieve full marks, providing that they started from the given inequality as instructed. In the final part of the question most candidates were able to perform the required integration successfully and, while some were unable to follow through to reach the required final result, many did complete the question successfully and achieve full marks.