##### DSE_Math Notes

• It is given that $$2 + \frac{1}{{2 + {\textstyle{1 \over {2\, + \,{\textstyle{1 \over {2\; + \; \cdots }}}}}}}}$$ is a fixed value and let x be the value.
(a) Prove that $${x^2} + 2x - 1 = 0$$.
(b) Hence find the value of $$2 + \frac{1}{{2 + {\textstyle{1 \over {2\, + \,{\textstyle{1 \over {2\; + \; \cdots }}}}}}}}$$.
• $$\begin{array}{c}(a)x = 2 + \frac{1}{{2 + {\textstyle{1 \over {2\, + \,{\textstyle{1 \over {2\; + \; \cdots }}}}}}}}\\ = 2 + \frac{1}{x}\\{x^2} - 2x - 1 = {0^{}}\end{array}$$

$$\begin{array}{1}(b){x^2} - 2x - 1 = 0\\x = \frac{{ - ( - 2) \pm \sqrt {{{( - 2)}^2} - 4(1)( - 1)} }}{{2(1)}}\\ = \frac{{2 \pm \sqrt 8 }}{2}\\ = \frac{{2 \pm 2\sqrt 2 }}{2}\\ = 1 \pm \sqrt 2 \end{array}$$
∵ $$2 + \frac{1}{{2 + {\textstyle{1 \over {2\, + \,{\textstyle{1 \over {2\; + \; \cdots }}}}}}}}$$ is a positive number.
∴ $$2 + \frac{1}{{2 + {\textstyle{1 \over {2\, + \,{\textstyle{1 \over {2\; + \; \cdots }}}}}}}} = \underline{\underline {1 + \sqrt 2 }}$$

• Consider $$\frac{{\cos \theta }}{{{{\sin }^2}\theta }} = \frac{1}{3}$$.
(a) Rewrite the equation in the form of $$a{\cos ^2}\theta + b\cos \theta + c = 0$$, where a, b and c are integers.
(b) Hence solve $$\frac{{\cos \theta }}{{{{\sin }^2}\theta }} = \frac{1}{3}$$, where $$0^\circ \le \theta \le 360^\circ$$.
• $$\begin{array}{1}(a) \frac{{\cos \theta }}{{{{\sin }^2}\theta }} = \frac{1}{3}\\3\cos \theta = {\sin ^2}\theta \\3\cos \theta = 1 - {\cos ^2}\theta \\{\cos ^2}\theta + 3\cos \theta - 1 = 0\end{array}$$

(b) $${\cos ^2}\theta + 3\cos \theta - 1 = 0$$
$$\begin{array}{1}\cos \theta = \frac{{ - {\rm{ }}3 \pm \sqrt {{3^2} - 4(1)( - 1)} }}{{2(1)}}\\ = \frac{{ - {\rm{ }}3 \pm \sqrt {13} }}{2}\end{array}$$
$$\cos \theta = \frac{{ - {\rm{ }}3 + \sqrt {13} }}{2}$$ or $$\cos \theta = \frac{{ - {\rm{ }}3 - \sqrt {13} }}{2}$$ (rejected)
$$\theta = 72.4^\circ$$ or $$360^\circ - 72.4^\circ$$ (corr. to 1 d.p.)
∴ $$\theta = \underline{\underline {72.4^\circ \;\;{\rm{o}}{\kern 1pt} {\rm{r}}\;\;287.6^\circ }}$$

• A group of students have to share the expense of $2 640 equally for a Christmas party. Now 3 of them cannot join the party and the rest of the students need to pay$8 more each.
(a) If the final number of students joining the party is n, express the original amount payable for each student in terms of n.
(b) How many students will join the party finally?
• (a) ∵ $$(n + 3)$$ students will join the party originally.
∴ Original amount payable by each student$$= \underline{\underline {\ \frac{{2{\rm{ }}640}}{{n + 3}}}}$$

(b) Final amount payable by each student$$= \ \frac{{2{\rm{ }}640}}{n}$$
$$\begin{array}{1}\frac{{2{\rm{ }}640}}{n} - \frac{{2{\rm{ }}640}}{{n + 3}} = 8\\2{\rm{ }}640(n + 3) - 2{\rm{ }}640n = 8n{(n + 3)^{^{}}}\\2{\rm{ }}640n + 7{\rm{ }}920 - 2{\rm{ }}640n = 8{n^2} + 24{n^{^{}}}\\{n^2} + 3n - 990 = 0\\(n - 30)(n + 33) = {0^{}}\end{array}$$
$$n = 30$$ or $$n = - 33$$ (rejected)
∴ 30 students will join the party finally.