It is given that \(k(x) = \begin{cases} a \sqrt{1 - \frac{x^2}{4}} & \text{for}\hspace{0.5em} -2

Question

It is given that $k(x) = \begin{cases} a \sqrt{1 - \frac{x^2}{4}} & \text{for}\hspace{0.5em} -2 < x \le 2, \ \frac{a}{2}x - a & \text{for}\hspace{0.5em} 2 < x \le 4, \end{cases}$ And that $k(x) = k(x - 6)$ for all real values of $x$, where $a$ is a real constant. Sketch the graph of $y = k(x)$ for $-3 \le x \le 8$.[3] Using the substitution $x = 2 \sin \theta$, find $\int \sqrt{1 - \frac{x^2}{4}} \,\mathrm{d}x$. [3] Using your answers in (i) and (ii), find the exact value of $\int_{5}^{\sqrt{3} + 6} k(x) \,\mathrm{d}x$ in terms of $a$ and $\pi$. [3]

Accepted Answer

(b)(i) Six-periodic graph shown for $a=1$; scale ordinates by $a$. (b)(ii) $\int\sqrt{1-\frac{x^2}{4}}\,\mathrm{d}x=\frac{x}{4}\sqrt{4-x^2}+\arcsin\frac{x}{2}+C$. (b)(iii) $\frac{a}{2}(\sqrt3+\pi)$.

2022 TGM P1 Q12

2022 TGM P1 Q12

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