2008 NJC P1 Q3

2008 NJC P1 Q3

5 marks
Prelims

The graph of \(y={{2}^{x}}\) for \(0\le x\le 1\), is shown in the diagram below. Rectangles, each of width \(\frac{1}{n}\), where \(n\) is an integer are drawn under the curve.

Show that the total area of all the \({n}\) rectangles is \({\frac{1}{n\left( 2^{\frac{1}{n}} - 1 \right)}}\).[2]

By considering the area of the region bounded by the curve \(y={{2}^{x}}\), \(x=1\) and the axes, briefly explain why \(\ln 2<n\left( {{2}^{\frac{1}{n}}}-1 \right)\).[3]

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Answer:\(\displaystyle \frac1{n(2^{1/n}-1)}<\frac1{\ln2}\), hence \(n(2^{1/n}-1)>\ln2\)

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