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2012 Paper 1 Q3
D: 1516.0
B: 1484.0
integration
curve sketching
inequality
area comparison
trigonometric
exponential
logarithm
geometric proof
- 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*}
