2015 A-level H2 Mathematics (9740) Paper 1 Question 10 Suggested Solutions

By KS Teng

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

(i)
A_1 = \int_0^{\frac{\pi}{4}} \mathrm{sin}x~ dx + \int_{\frac{\pi}{4}}^{\frac{\pi}{2}} \mathrm{cos}x~ dx

= 2 - \sqrt{2}

A_2 = \int_0^{\frac{\pi}{4}} \mathrm{cos}x - \mathrm{sin}x~ dx

= \sqrt{2} - 1

\frac{A_1}{A_2} = \frac{2-\sqrt{2}}{\sqrt{2}-1} = \frac{\sqrt{2}(\sqrt{2}-1)}{\sqrt{2}-1} = \sqrt{2}

(ii)
y = \mathrm{sin} x \Rightarrow x = \mathrm{sin^{-1}}y

Vol = \pi \int_0^{\sqrt{2}/2} x^2~ dy

= \pi \int_0^{\sqrt{2}/2} (\mathrm{sin^{-1}}y)^2~ dy

(iii)
When y = \frac{\sqrt{2}}{2}, u = \frac{\pi}{4}

When y = 0, x = 0

y = \mathrm{sin}u

\mathrm{sin^{-1}}y = u \Rightarrow \frac{dy}{dx}=\mathrm{cos}u

Vol = \pi \int_0^{\sqrt{2}/2} (\mathrm{sin^{-1}}y)^2~ dy

= \pi \int_0^{\pi / 4} u^2 \mathrm{cos}u~du

\therefore, a = 0, b = \frac{\pi}{4}

\pi \int_0^{\pi / 4} u^2 \mathrm{cos}u~du

= \pi \{u^2 \mathrm{sin}u\bigl|_0^{\pi /4} - \int_0^{\pi /4} 2u \mathrm{sin} u ~ du\}

= \pi \{\frac{\sqrt{2}\pi ^2}{32} + 2u \mathrm{cos}u \bigl|_0^{\pi / 4} - \int_0^{\pi /4} 2\mathrm{cos}u~du\}

= \pi \{\frac{\sqrt{2}\pi ^2}{32} + \frac{\sqrt{2} \pi}{4} - 2 \mathrm{sin}u \bigl|_0^{\pi /4} \}

= \pi \{\frac{\sqrt{2}\pi ^2}{32} + \frac{\sqrt{2} \pi}{4} - \sqrt{2}\}

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