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2017 Paper 1 Q2
D: 1484.0
B: 1500.1
- 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 \(0 < x \le 1\).
- 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 \(x > 0\).
- 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 \( y\ne1\).
Solution:
- \(\,\) \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*}
- \(\,\) \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*}
- \(\,\) \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*}
2008 Paper 3 Q4
D: 1700.0
B: 1516.0
integration
hyperbolic functions
arcosh
inequality
curve sketching
substitution
tanh
proof by integration
- Show, with the aid of a sketch, that \(y> \tanh (y/2)\) for \(y>0\) and deduce that \begin{equation} \textrm{arcosh} x > \dfrac{x-1}{\sqrt{x^2-1}} \text{ for } x>1. \tag{\(*\)} \end{equation}
- By integrating \((*)\), show that $\textrm{arcosh} x > 2 \dfrac{{x-1}}{\sqrt{x^2-1}} \( for \)x>1$.
- Show that $\textrm{arcosh} x >3 \dfrac{\sqrt{x^2-1}}{{x+2}} \( for \)x>1$.
Solution:
- If \(y = \textrm{arcosh} x \), then \(\tanh\textrm{arcosh} x/2 = \sqrt{\frac{\cosh \textrm{arcosh} x-1}{\cosh \textrm{arcosh} x+1}} = \sqrt{\frac{x-1}{x+1}} = \frac{x-1}{\sqrt{x^2-1}}\)
- \begin{align*} \int \textrm{arcosh} x \d x &= \left [x \textrm{arcosh} x \right] - \int \frac{x}{\sqrt{x^2-1}} \d x \\ &= x \textrm{arcosh} x - \sqrt{x^2-1}+C \\ \int \frac{x-1}{\sqrt{x^2-1}} &= \sqrt{x^2-1} - \textrm{arcosh} x +C \end{align*} Therefore \begin{align*} && \int_1^x \textrm{arcosh} t \d t &> \int_1^x \frac{t-1}{\sqrt{t^2-1}} \d t \\ \Rightarrow && x \textrm{arcosh} x - \sqrt{x^2-1} - 0 &> \sqrt{x^2-1} - \textrm{arcosh} x - 0 \\ \Rightarrow && (x+1) \textrm{arcosh} x &> 2\sqrt{x^2-1} \\ \Rightarrow && \textrm{arcosh} x & > 2\frac{\sqrt{x^2-1}}{x+1} \\ &&&= 2 \frac{\sqrt{x-1}}{\sqrt{x+1}} \\ &&&= 2 \frac{x-1}{\sqrt{x^2-1}} \end{align*}
- Integrating both sides again, \begin{align*} && \int_1^x \textrm{arcosh} t \d t &> 2 \int_1^x \frac{t-1}{\sqrt{t^2-1}} \d t \\ \Rightarrow && x \textrm{arcosh} x - \sqrt{x^2-1} &> 2\left (\sqrt{x^2-1} - \textrm{arcosh}x \right) \\ \Rightarrow && (x+2)\textrm{arcosh} x &> 3 \sqrt{x^2-1} \\ \Rightarrow && \textrm{arcosh} x &> 3 \frac{\sqrt{x^2-1}}{x+2} \end{align*}