HSC 2015 MX1 Marathon (archive) (8 Viewers)

davidgoes4wce · Oct 17, 2015

Re: HSC 2015 3U Marathon

rand_althor said:
$$Evaluate $\int^{3^5}_{2^5}~\frac{1}{x-x^\frac{3}{5}}~\mathrm{d}x$ using the substitution $u=x^{\frac{2}{5}}-1$.$

davidgoes4wce · Oct 17, 2015

Re: HSC 2015 3U Marathon

How do you do that?

InteGrand · Oct 17, 2015

Re: HSC 2015 3U Marathon

davidgoes4wce said:
How do you do that?

$We have$

$$\angle MAD = \angle ABD$ (alternate segment theorem)$

$$=\angle ACD$ (angles at the circumference subtended by same arc $AD$)$

$$=\angle ACB$ (as $CA$ bisects angle $BCD$)$

$$=\angle ADB$ (angles at the circumference subtended by same arc $AB$)$

$\Rightarrow MN \parallel DB$ (alternate angles $MAD$ and $ADB$ are equal). $\blacksquare$

kawaiipotato · Oct 17, 2015

Re: HSC 2015 3U Marathon

$$ By using the fact that $ \int_{0}^{\infty} \frac{\sin x}{x} dx = \frac{\pi}{2} ,$

$$ Evaluate $ \int_{0}^{\infty} \frac{\sin (x^{2015})}{x} dx$

laters · Oct 17, 2015

Re: HSC 2015 3U Marathon

kawaiipotato said:
$$ By using the fact that $ \int_{0}^{\infty} \frac{\sin x}{x} dx = \frac{\pi}{2} ,$

$$ Evaluate $ \int_{0}^{\infty} \frac{\sin (x^{2015})}{x} dx$

Paradoxica · Oct 17, 2015

Re: HSC 2015 3U Marathon

Three tangent lines are constructed on a circle of radius two. Prove, that in the non-degenerate case where the tangents form a triangle, the area of the triangle is equal to the perimeter.

leehuan · Oct 18, 2015

Re: HSC 2015 3U Marathon

Here's a diagram to go with that question.

Crisium · Oct 19, 2015

Re: HSC 2015 3U Marathon

5a) Mathematical Induction

http://www.boardofstudies.nsw.edu.au/hsc_exams/hsc2000exams/hsc00_maths/99MATH34.PDF

My working out is a bit dodgy, I want to see how you guys would set it out

Zlatman · Oct 19, 2015

Re: HSC 2015 3U Marathon

Crisium said:
5a) Mathematical Induction

http://www.boardofstudies.nsw.edu.au/hsc_exams/hsc2000exams/hsc00_maths/99MATH34.PDF

My working out is a bit dodgy, I want to see how you guys would set it out

$I'll skip straight to the assumption, since the base case is pretty straight forward.$

$$Assume true for $ n = k $: $ P_k = 2^k [1 *3 * ... * (2k-1)]$

$$Prove true for $ n = k + 1 $: $ P_{k+1} = 2^{k+1} [1 * 3 * ... * (2k - 1)(2k+1)]$

$P_k = (k+1)(k+2)...(2k-1)(2k)$

$\begin{align*} P_{k+1} &= [(k+1)+1][(k+1)+2]...[2(k+1)-1][2(k+1)] \\ &= (k+2)(k+3)...(2k+1)(2k+2) \\ &= P_k * \frac{(2k+1)(2k+2)}{(k+1)} \\ &= 2^k [1 * 3 * ... * (2k-1)] * 2 * 2k+1 \, \, \, (by \, \, the \, \, assumption) \\ &= 2^{k+1} [1 * 3 * ... * (2k-1) (2k+1)] \, \, \, (as \, \, required) \end{align*}$

$Therefore, by mathematical induction, this statement is true for all positive integer n.$

Crisium · Oct 19, 2015

Re: HSC 2015 3U Marathon

Zlatman said:
$I'll skip straight to the assumption, since the base case is pretty straight forward.$

$$Assume true for $ n = k $: $ P_k = 2^k [1 *3 * ... * (2k-1)]$

$$Prove true for $ n = k + 1 $: $ P_{k+1} = 2^{k+1} [1 * 3 * ... * (2k - 1)(2k+1)]$

$P_k = (k+1)(k+2)...(2k-1)(2k)$

$\begin{align*} P_{k+1} &= [(k+1)+1][(k+1)+2]...[2(k+1)-1][2(k+1)] \\ &= (k+2)(k+3)...(2k+1)(2k+2) \\ &= P_k * \frac{(2k+1)(2k+2)}{(k+1)} \\ &= 2^k [1 * 3 * ... * (2k-1)] * 2 * 2k+1 \, \, \, (by \, \, the \, \, assumption) \\ &= 2^{k+1} [1 * 3 * ... * (2k-1) (2k+1)] \, \, \, (as \, \, required) \end{align*}$

$Therefore, by mathematical induction, this statement is true for all positive integer n.$

That looks a lot better than how I set it out, thanks

Zlatman · Oct 19, 2015

Re: HSC 2015 3U Marathon

$Construct OD, OE and OF.$

$In $ \triangle $OEF, OB is perpendicular to EF, because the tangent is perpendicular to the radius of a circle at its point of contact. Therefore, EF is the base of the triangle, and OB is the height.$

$\begin{align*} A_{\triangle OEF} &= \frac{1}{2} b h \\ &= \frac{1}{2} OB * EF \\ &= EF \, \, \, (OB = r = 2) \end{align*}$

$Similarly, $ A_{\triangle ODF} = DF $ and $ A_{\triangle OED} = ED $.$

$$Therefore, $ A = EF + DF + DE = P$

--------------------------------------------------------------------------------------------------

$NEW QUESTION:$

running31 · Oct 19, 2015

Re: HSC 2015 3U Marathon

this is super exciting I have never written anything on here before

rand_althor · Oct 19, 2015

Re: HSC 2015 3U Marathon

Zlatman said:
$NEW QUESTION:$

(i) 5C3 = 10
(ii) 9C3 - 5C3 = 74

Is that right?

braintic · Oct 19, 2015

Re: HSC 2015 3U Marathon

running31 said:
this is super exciting I have never written anything on here before

It is super exciting to see you posting on here.

Zlatman · Oct 19, 2015

Re: HSC 2015 3U Marathon

rand_althor said:
(i) 5C3 = 10
(ii) 9C3 - 5C3 = 74

Is that right?

I realised after posting the question that I didn't actually have the answers, lol

But yeah, I got that too, so we'll just say it's right

davidgoes4wce · Oct 21, 2015

Re: HSC 2015 3U Marathon

My thinking was

cot (270)= 1/tan (270)

1/tan (270)= cos(270)/sin(270)
=0/-1
=0

InteGrand · Oct 21, 2015

Re: HSC 2015 3U Marathon

davidgoes4wce said:
cot (270)=1/tan(270)

cos(270)/sin(270)
=0/-1
=0

$Best to leave out the first line and go straight to the second, since $\tan(270^\circ)$ is undefined.$

Crisium · Oct 21, 2015

Re: HSC 2015 3U Marathon

Wait I forgot to leave out the part where you have to use the tan(A + B) expansion

davidgoes4wce · Oct 21, 2015

Re: HSC 2015 3U Marathon

Tan is undefined at: -90, 90, 270 .....for every increase of 180 degrees.

davidgoes4wce · Oct 21, 2015

Re: HSC 2015 3U Marathon

Simplify

HSC 2015 MX1 Marathon (archive) (8 Viewers)

Well-Known Member

Well-Known Member

Well-Known Member

Well-Known Member

Member

-insert title here-

Well-Known Member

Pew Pew

Member

Pew Pew

Member

Attachments

New Member

Active Member

Well-Known Member

Member

Well-Known Member

Well-Known Member

Pew Pew

Well-Known Member

Well-Known Member

Users Who Are Viewing This Thread (Users: 0, Guests: 8)