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Here is Q16(a): y = 5x - 7 \ \fbox{1} y' = 4x - 3 \ \fbox{2} $From$ \ \fbox{1} \ m = 5 $From$ \ \fbox{2} \ y' = 4x - 3 = 5 \therefore \ x = 2 \ $and$ \ y = 3 \int \ 4x - 3 \ dx = 2x^2 - 3x + C $Using$ \ (2, \ 3), \ C = 1 \therefore \ y = 2x^2 - 3x +1

\tan \frac{\pi}{6} = \frac{4}{BC} \therefore \ BC = 4\sqrt{3} $By Pythagoras,$ \ 13^2 - (4\sqrt{3})^2 = AC^2 AC = 11, \ $meaning,$ \ AD = 11 - 4 = 7 \angle BDC = 180 - 90 - 30 = 60 \therefore \ \angle ADB = 180 - 60 = 120 = \frac{2\pi}{3} \frac{13}{\sin \frac{2\pi}{3}} = \frac{7}{\sin x}...

I got a big fraction as well.

It was 83..

Yes I heard this was true.

You are meant to make a series, find the common ratio and then calculate the limiting sum.

I think it did..

Definitely lower. I'm thinking about 76-80.

Considering last years cut off was 83 and the exam was considerably easier, I highly doubt the cut off for this exam will be 82.

Much harder than last year. I was stuck on the trigonometry question asking to find the expression of r^2. Used the cosine rule but just couldn't get it.

Re: HSC 2013 2U Marathon Hmm makes sense..

Re: HSC 2013 2U Marathon There are actually 3 if you also consider AB to be a diameter.

Re: HSC 2013 2U Marathon How can you tell which pair of points is the diagonal and which is the side of the parallelogram?

Re: HSC 2013 2U Marathon It is the cosine formula.. \cos C = \frac{a^2 + b^2 - c^2}{2ab}

Re: HSC 2013 2U Marathon Here's the continuation of the question: $Hence, given that$ \ \cos\alpha = \frac{x^2 - 32}{4x}, \ $show that the$ $value of$ \ x \ $is a solution of$ \ x^4 - 56x^2 + 640 = 0

Its true haha.

Re: HSC 2013 2U Marathon Not sure why you aren't getting x^2 -16 then.

Re: HSC 2013 2U Marathon Oh I realise what you've done. When squaring 2\sqrt{5} you must also square the 2 in the front. So you would get 4 x 5 = 20.

Re: HSC 2013 2U Marathon That's correct. Are you sure you're applying the cosine rule correctly?

Re: HSC 2013 2U Marathon $By considering triangle PBA, show that$ \ \sin\alpha = \frac{x^2 - 16}{4x}