2011 A-level H2 Mathematics (9740) Paper 2 Question 4 Suggested Solutions

All solutions here are SUGGESTED. Mr. Teng will hold no liability for any errors. Comments are entirely personal opinions.

(a)
(i)
\int_0^n x^2 e^{-2x} dx

= [x^2 (\frac{e^{-2x}}{-2})]_0^n - \int_0^n (\frac{e^{-2x}}{-2})2x dx

= -\frac{n^2e^{-2n}}{2} + \int_0^n xe^{-2x} dx

= -\frac{n^2e^{-2n}}{2} + [x(\frac{e^{-2x}}{-2})]_0^n-\int_0^n \frac{e^{-2x}}{-2} dx

= -\frac{n^2e^{-2n}}{2} - \frac{ne^{-2n}}{2} + \frac{1}{2} [\frac{e^{-2x}}{-2}]_0^n dx

= \frac{1}{4} - (\frac{n^2}{2} + \frac{n}{2} + \frac{1}{4})e^{-2n}

(ii)
\int_0^{\infty} x^2 e^{-2x} dx

\lim_{n \rightarrow \infty} \frac{1}{4} - (\frac{n^2}{2} + \frac{n}{2} + \frac{1}{4})e^{-2n}

= \frac{1}{4}

(b)
Let x = \mathrm{tan}\theta \Rightarrow \frac{dx}{d\theta} = \mathrm{sec}^2 \theta

When x=0, \theta = 0

When x=1, \theta = \frac{\pi}{4}

Required Volume
= \pi \int_0^1 (\frac{4x}{x^2+1})^2 dx

= 16 \pi \int_0^{\frac{\pi}{4}} \frac{\mathrm{tan}^2 \theta}{(\mathrm{tan}^2 \theta + 1)^2} \mathrm{sec}^2 \theta d\theta

= 16 \pi \int_0^{\frac{\pi}{4}} \mathrm{sin}^2 \theta d\theta

= 8 \pi \int_0^{\frac{\pi}{4}} 1 - \mathrm{cos}^2 2\theta d\theta

= 8 \pi [\theta - \frac{\mathrm{sin}2\theta}{2}]_0^{\frac{\pi}{4}}

= 2\pi^2 -4\pi

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