HSC 2016 MX1 Marathon (archive) (1 Viewer)

davidgoes4wce · Nov 17, 2015

Re: HSC 2016 3U Marathon

From the centre point, divide 360 degrees by 12= 30 degrees. From the centre we can treat each triangle as an isosceles triangle so the interior angles are 75 degrees . The sum of an angle of a triangle is = 75+75+30=180 degrees.

The exterior angle is equal to : 180-150= 30 degrees

davidgoes4wce · Nov 17, 2015

Re: HSC 2016 3U Marathon

I had a bit of trouble visualising this question.

From Fitzpatrick

P and Q are points on the parabola x^2=4ay. Tangents TP and TQ are drawn from an external point T to cut the x-axis at A and B. Show that the joining the focus to the midpoint of AB is perpendicular to the chord of contact PQ.

davidgoes4wce · Nov 17, 2015

Re: HSC 2016 3U Marathon

This was my attempt at the diagram:

leehuan · Nov 17, 2015

Re: HSC 2016 3U Marathon

Call the midpoint of AB something like M, but I don't think F, M and T have to be collinear do they? In other words I reckon T doesn't have to lie on FM.

I think:

a) Find the gradient of PQ
b) Find the equation of the tangents
c) Sub y=0 to find x-intercepts
d) Find midpoint
e) Focus is obviously (0,a) so use gradient formula with focus and midpoint.
f) Prove perpendicularity
-------------------
P.S. If you want to know why I put the easy question
http://community.boredofstudies.org/10/maths/344794/hard-vce-maths-question.html

leehuan · Nov 18, 2015

Re: HSC 2016 3U Marathon

NEXT QUESTION
$\\ \int { \cot { x } \, dx } \\ $using the substitution $u=\sin x$

Drsoccerball · Nov 18, 2015

Re: HSC 2016 3U Marathon

leehuan said:
NEXT QUESTION
$\\ \int { \cot { x } \, dx } \\ $using the substitution $u=\sin x$

Reverse chain rule would be one step.

InteGrand · Nov 18, 2015

Re: HSC 2016 3U Marathon

Drsoccerball said:
Reverse chain rule would be one step.

Reverse chain rule is substitution done in the head.

davidgoes4wce · Nov 18, 2015

Re: HSC 2016 3U Marathon

leehuan said:
NEXT QUESTION
$\\ \int { \cot { x } \, dx } \\ $using the substitution $u=\sin x$

davidgoes4wce · Nov 18, 2015

Re: HSC 2016 3U Marathon

Apologies for writing a double absolute value sign. Its 11:25pm and I had a long day of work

leehuan · Nov 19, 2015

Re: HSC 2016 3U Marathon

Just a tip, I would write du = cos(x) dx instead. But keep everything else except for your little accident the same. This is because you maintain the variables in the integral the same if you just directly substituted.

davidgoes4wce said:
Apologies for writing a double absolute value sign. Its 11:25pm and I had a long day of work

Np

----------------------------------------------------------------
NEXT QUESTION
$\\ $Solve $\arccos { x } -\arcsin { x } =\frac { \pi }{ 3 }$

Note: arcsin just means sine inverse, aka sin^-1

integral95 · Nov 19, 2015

Re: HSC 2016 3U Marathon

leehuan said:
Just a tip, I would write du = cos(x) dx instead. But keep everything else except for your little accident the same. This is because you maintain the variables in the integral the same if you just directly substituted.

Np
----------------------------------------------------------------
NEXT QUESTION
$\\ $Solve $\arccos { x } -\arcsin { x } =\frac { \pi }{ 3 }$

Note: arcsin just means sine inverse, aka sin^-1

Using the identity $\\ \arccos { x } + \arcsin { x } =\frac { \pi }{ 2 } $ for$ -1 \leq x \leq 1$

We add those 2 expressions together to get

$2\arccos { x } = \frac { 5 \pi }{ 6 } \Rightarrow \\ \\ \\ \ x = cos(\frac { 5 \pi }{ 12 })$

Then you can use the double angle formulas to get that

leehuan · Nov 19, 2015

Re: HSC 2016 3U Marathon

integral95 said:
Using the identity $\\ \arccos { x } + \arcsin { x } =\frac { \pi }{ 2 } $ for$ -1 \leq x \leq 1$

We add those 2 expressions together to get

$2\arccos { x } = \frac { 5 \pi }{ 6 } \Rightarrow \\ \\ \\ \ x = cos(\frac { 5 \pi }{ 12 })$

Then you can use the double angle formulas to get that

*Compound
----------------------------------------------
NEXT QUESTION

$Write down the probability that Drsoccerball will continuously gain attention mocking InteGrand in troll questions like this one.$

Drsoccerball · Nov 19, 2015

Re: HSC 2016 3U Marathon

leehuan said:
*Compound
----------------------------------------------
NEXT QUESTION

$Write down the probability that Drsoccerball will continuously gain attention mocking InteGrand in troll questions like this one.$

They're legit questions :L...

leehuan · Nov 19, 2015

Re: HSC 2016 3U Marathon

I believe the maths behind them is actually legit tbh

you just pick him a lot
--------------------------
NEXT QUESTION (for realsies)

$\\\int { \sin ^{ 3 }{ x } \, dx } \\ $by reversing the substitution $du=\sin { x } \, dx$

davidgoes4wce · Nov 19, 2015

Re: HSC 2016 3U Marathon

leehuan said:
I believe the maths behind them is actually legit tbh you just pick him a lot
--------------------------
NEXT QUESTION (for realsies)

$\\\int { \sin ^{ 3 }{ x } \, dx } \\ $by reversing the substitution $du=\sin { x } \, dx$

Drsoccerball · Nov 19, 2015

Re: HSC 2016 3U Marathon

I think we should give the 2016'ers a chance...

dan964 · Nov 20, 2015

Re: HSC 2016 3U Marathon

davidgoes4wce said:
From the centre point, divide 360 degrees by 12= 30 degrees. From the centre we can treat each triangle as an isosceles triangle so the interior angles are 75 degrees . The sum of an angle of a triangle is = 75+75+30=180 degrees.

The exterior angle is equal to : 180-150= 30 degrees

There happens to be a simpler method, remembering that the exterior angle sum of any regular polygon is 360.
Divide that by 12, to get 30 degrees which is the answer.

leehuan · Nov 20, 2015

Re: HSC 2016 3U Marathon

Trying to keep questions at an easy level to promote activity from the 2016ers tbh...
------------------------------------
NEXT QUESTION
$\\ $By using the definitions of the trigonometric ratios, prove the ratio identity $ \tan x = \frac {\sin x} {\cos x} $ for acute angles$.$

InteGrand · Nov 20, 2015

Re: HSC 2016 3U Marathon

leehuan said:
Trying to keep questions at an easy level to promote activity from the 2016ers tbh...
------------------------------------
NEXT QUESTION
$\\ $By using the definitions of the trigonometric ratios, prove the ratio identity $ \tan x = \frac {\sin x} {\cos x} $ for acute angles$.$

This is the definition of tan (or the most common one anyway).

leehuan · Nov 20, 2015

Re: HSC 2016 3U Marathon

True (I think) but when the trigonometric ratios are introduced to the Yr 9 student they're introduced with aid of SOH-CAH-TOA and only for the right-angled triangle.

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