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VCE Stuff => VCE Mathematics => VCE Mathematics/Science/Technology => VCE Subjects + Help => VCE Mathematical Methods CAS => Topic started by: srxmos on July 10, 2016, 01:57:10 pm

Title: Worked solution/explanation please?
Post by: srxmos on July 10, 2016, 01:57:10 pm: (http://uploads.tapatalk-cdn.com/20160709/74cb83396540c6b702d7e5aa091782bf.jpg)
Calculator allowed too
Title: Re: Worked solution/explanation please?
Post by: RuiAce on July 10, 2016, 02:06:34 pm: $\text{For Q12, begin by considering the first derivative }\frac{dy}{dx}=12x^2+54x-30=6(2x-1)(x+5)$
$\text{Recall that the gradient is negative when }\frac{dy}{dx} < 0$
$\therefore \text{We need the solution to }6(2x-1)(x+5) < 0$
$\text{Sketch the graph of the curve }\frac{dy}{dx}=6(2x-1)(x+5). \text{The intercepts are at }x=\frac{1}{2}, x=5$
$\text{From the graph, we can deduce that the solutions are }-\frac{1}{2} < x < 5\\ \text{In interval form, this is equivalent to }x \in (0.5,5)$
Title: Re: Worked solution/explanation please?
Post by: RuiAce on July 10, 2016, 02:09:37 pm: Quote from: srxmos on July 10, 2016, 01:57:10 pm
(http://uploads.tapatalk-cdn.com/20160709/74cb83396540c6b702d7e5aa091782bf.jpg)
Calculator allowed too
$\text{For Q13, we wish to calculate the areas of both rectangles and add them up.}\\ \text{Carefully study the diagram. The width of both rectangles is just }\frac{\pi}{6}$
$\text{The height requires us to actually find some function values.}$
$\text{When }x=\frac{\pi}{6}, \, y=\sin \frac{\pi}{6}=\frac{1}{2}\\ \text{Hence the area of the left rectangle is }A_1=\frac{1}{2}\times \frac{\pi}{6} = \frac{\pi}{12}$
$\text{When }x=\frac{\pi}{3}, \, y=\sin \frac{\pi}{3}=\frac{\sqrt{3}}{2}\\ \text{Hence the area of the right rectangle is }A_2=\frac{\sqrt{3}}{2}\times \frac{\pi}{6}=\frac{\sqrt{3}\pi}{12}$
$\text{Sum them up:}\\ A=\frac{\pi}{12}+\frac{\sqrt{3}\pi}{12}=\frac{(\sqrt{3}+1)\pi}{12}$