<p>It is given that</p>
<p style="text-align: center">$\mathrm{f}\left( x \right)=\left\{ \begin{matrix}
&amp; \sqrt{1-\frac{{{x}^{2}}}{{{a}^{2}}}}\,\,\,\,\mathrm{for}\,\,-a\le x\le a,\\&amp;0\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\mathrm{for}\,\,a&lt;x\le2a, \\\end{matrix} \right.$</p>
<p>and that $\mathrm{f}\left( x+3a \right)=\mathrm{f}\left( x \right)$ for all real values of $x$, where $a$ is a real constant.</p>
<ol class="i-list"><li><div class="list-item-content"><span class="list-item-text">Sketch the graph of $y=\mathrm{f}\left( x \right)$ for $-4a\le x\le 6a$.</span></div><p style="text-align: right;">[3]</p></li><li><div class="list-item-content"><span class="list-item-text">Use the substitution $x=a\sin \theta $ to find the exact area of $\displaystyle\int\nolimits_{\frac{1}{2}a}^{\frac{\sqrt{3}}{2}a}{\mathrm{f}\left( x \right)\,\mathrm{d}x}$ in terms of $a$ and $\pi $.</span></div><p style="text-align: right;">[5]</p></li></ol>

$\mathrm{f}\left( x \right)=\left\{ \begin{matrix}
&amp; \sqrt{1-\frac{{{x}^{2}}}{{{a}^{2}}}}\,\,\,\,\mathrm{for}\,\,-a\le x\le a,\\&amp;0\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\mathrm{for}\,\,a&lt;x\le2a, \\\end{matrix} \right.$

and that $\mathrm{f}\left( x+3a \right)=\mathrm{f}\left( x \right)$ for all real values of $x$, where $a$ is a real constant.

Sketch the graph of $y=\mathrm{f}\left( x \right)$ for $-4a\le x\le 6a$.

Use the substitution $x=a\sin \theta $ to find the exact area of $\displaystyle\displaystyle \displaystyle\int\nolimits_{\frac{1}{2}a}^{\frac{\sqrt{3}}{2}a}{\mathrm{f}\left( x \right)\,\mathrm{d}x}$ in terms of $a$ and $\pi $.

Modulus Functions

Functions

International Baccalaureate Higher Level Mathematics

IB HL Math

N2013 P1 Q5

$y = \sqrt{1 - \frac{x^{2}}{a^{2}}}$ is the upper half of the ellipse $\frac{x^{2}}{a^{2}} + y^{2} = 1$ (semicircle of radius $a$ scaled).

$\mathrm{f}(x+3a) = \mathrm{f}(x)$: period $3a$. Sketch shows semicircular bumps (max height 1) centered at $x = 0, \pm 3a, \pm 6a, \ldots$ with zero regions between.

$\displaystyle\int_{a/2}^{\sqrt{3}a/2} \sqrt{1 - \dfrac{x^{2}}{a^{2}}}\;\mathrm{d}x$. Let $x = a\sin\theta$, $\mathrm{d}x = a\cos\theta\;\mathrm{d}\theta$.

When $x = \frac{a}{2}$: $\theta = \frac{\pi}{6}$. When $x = \frac{\sqrt{3}a}{2}$: $\theta = \frac{\pi}{3}$.

$= a\displaystyle\int_{\pi/6}^{\pi/3} \cos^{2}\theta\;\mathrm{d}\theta = \dfrac{a}{2}\displaystyle\int_{\pi/6}^{\pi/3} (1 + \cos 2\theta)\;\mathrm{d}\theta$

$= \dfrac{a}{2}\left[\theta + \dfrac{\sin 2\theta}{2}\right]_{\pi/6}^{\pi/3}$

$= \dfrac{a}{2}\left[\left(\dfrac{\pi}{3} + \dfrac{\sqrt{3}}{4}\right) - \left(\dfrac{\pi}{6} + \dfrac{\sqrt{3}}{4}\right)\right] = \dfrac{a}{2} \times \dfrac{\pi}{6} = \dfrac{a\pi}{12}$

N2013 P1 Q5

N2013 P1 Q5

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