2004 HSC Mathematics Extension 2 Paper (1 Viewer)

frog0101 · Oct 10, 2018

Hi,
For Q7(a)(iii), how did students know that they needed to sub the left hand side of the inequality into the inequality proved in the part prior, what thought process lead them to that?

Thanks

2004 HSC Mathematics Extension 2:
http://www.k6.boardofstudies.nsw.edu.au/wps/wcm/connect/c03be093-eab2-405b-9de4-b2f9986e3f96/maths-ext2-hsc-exam-2004.pdf?MOD=AJPERES&CACHEID=ROOTWORKSPACE-c03be093-eab2-405b-9de4-b2f9986e3f96-lGd7Z3z

Jonomyster · Oct 10, 2018

There are several useful hints that you can look for:
1.) These part (iii) style of questions nearly always use the previous part, so it is useful to think about how you can incorporate part (ii) into part (iii)
2.) On part (iii), there are 3 terms on LHS, but 3^2 = 9 and there is a 9 on RHS (think of n^2 from part(ii)). This is another hint to substitute
3.) In (iii), note that there is 1/sin^2(x), 1/cos^2(x) and 1/tan^2(x), this is similar to the 1/a1, 1/a2, 1/a3 from part (ii), a clue you should substitute them

Using these clues you can rearrange part (ii) into:

$\dfrac{1}{a_1} + \dfrac{1}{a_2} + \dfrac{1}{a_3} \geq \dfrac{9}{a_1 + a_2 + a_3} \text{ Then try the substitution:}$
$\text{cosec}^2(x) + \text{sec}^2(x) + \text{cot}^2(x) \geq \dfrac{9}{\sin^2(x) + \cos^2(x) + \tan^2(x)}$

Then we solve from here:

$\dfrac{9}{\sin^2(x) + \cos^2(x) + \tan^2(x)}$
$= \dfrac{9\cos^2(x)}{\sin^2(x)\cos^2(x) + \cos^4(x) + \sin^2(x)}$
$= \dfrac{9\cos^2(x)}{\cos^2(x) (\cos^2(x) + \sin^2(x)) + \sin^2(x)}$
$= \dfrac{9\cos^2(x)}{\cos^2(x) + \sin^2(x)}$
$= 9\cos^2(x)$

2004 HSC Mathematics Extension 2 Paper (1 Viewer)

frog0101

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Jonomyster

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