It is given that \({\mathrm{f}(x) = \begin{cases} 2 + \sqrt{9 - (x-3)^2} & \text{for}\hspace{0.5em}

Question

It is given that ${\mathrm{f}(x) = \begin{cases} 2 + \sqrt{9 - (x-3)^2} & \text{for}\hspace{0.5em} 0 < x < 6, \ \qquad\qquad 2 & \text{for}\hspace{0.5em} 6 \leq x \leq 12. \end{cases}}$ and that $\mathrm{f}\left( x \right)=\mathrm{f}\left( x+12 \right)$ for all real values of $x$. Sketch the graph of $y=\mathrm{f}\left( x \right)$ for $-11\le x\le 12$.[3] Find the exact value of $\displaystyle\displaystyle \displaystyle\int_{-9}^{10}{\mathrm{f}\left( x \right)\,\mathrm{d}x}$.[2] The curve $C$ has equation $y=\frac{2{{x}^{2}}+3}{{{x}^{2}}-6}$. Using an algebraic method, find the set of value of $y$ that $C$ can take.[3]

Accepted Answer

(a)(ii) $38+\frac{27\pi}{4}$ (b) $y\leq-\frac12$ or $y>2$

2019 ASRJC Promo Q6

2019 ASRJC Promo Q6

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