HSC 2016 Maths Marathon (archive) (1 Viewer)

davidgoes4wce · Feb 12, 2016

Re: HSC 2016 2U Marathon

$I found out that the period is given in the question, using the formula$ $T=\frac{2 \pi}{n}, n= 4. Amplitude =4.$ $Could the answer be both 'a' \and 'd'?$

InteGrand · Feb 12, 2016

Re: HSC 2016 2U Marathon

davidgoes4wce said:
$I found out that the period is given in the question, using the formula$ $T=\frac{2 \pi}{n}, n= 4. Amplitude =4.$ $Could the answer be both 'a' \and 'd'?$

Yes

davidgoes4wce · Feb 12, 2016

Re: HSC 2016 2U Marathon

Thanks integrand.

davidgoes4wce · Feb 12, 2016

Re: HSC 2016 2U Marathon

$$My thinking for this question was the period has to be doubled, so I multiplied it by 2. The period of the function 'f' is $ \frac{\pi}{2} , $ so for function 'g' I doubled it and it gave me a value of $ \pi .$

InteGrand · Feb 12, 2016

Re: HSC 2016 2U Marathon

davidgoes4wce said:
$$My thinking for this question was the period has to be doubled, so I multiplied it by 2. The period of the function 'f' is $ \frac{\pi}{2} , $ so for function 'g' I doubled it and it gave me a value of $ \pi .$

The period is halved though, they said. So the answer must be (A).

davidgoes4wce · Feb 12, 2016

Re: HSC 2016 2U Marathon

Yep I misread that one. My bad.

davidgoes4wce · Feb 12, 2016

Re: HSC 2016 2U Marathon

Is there a mistake in this question as well?

According to the formula book

$a\ cos \theta - b \ sin \theta= A \ cos (\theta +\alpha) , where \ tan \alpha=\frac{b}{a} \ and \ A=\sqrt{a^2+b^2}$

Bestintheworld · Feb 16, 2016

Re: HSC 2016 2U Marathon

Can anyone solve this for me please?

A rectangle is cut from a circular disc of radius 6 cm. Find the area of the largest rectangle that can be produced.

davidgoes4wce · Feb 16, 2016

Re: HSC 2016 2U Marathon

Bestintheworld said:
Can anyone solve this for me please?

A rectangle is cut from a circular disc of radius 6 cm. Find the area of the largest rectangle that can be produced.

Do you have an answer for it?

Bestintheworld · Feb 16, 2016

Re: HSC 2016 2U Marathon

^yes

leehuan · Feb 16, 2016

Re: HSC 2016 2U Marathon

Bestintheworld said:
Can anyone solve this for me please?

A rectangle is cut from a circular disc of radius 6 cm. Find the area of the largest rectangle that can be produced.

This may be of use...

Because of all shapes a rectangle was sandwiched in a circle, the intersection of the diagonals of the rectangle is also the centre of the circle. Hence, the diagonal of the rectangle is actually the diameter of the circle.

(Proof: According to Ext 1 circle geometry, the angle in a semicircle is a right angle, and all angles on a rectangle...are right angles)

So if you label one of the sides 'x' and then according to Pythagoras' theorem the other side is √(36-x^2)... A = x√(36-x^2)
-------> A^2 = x^2(36-x^2)
A^2 = 36x^2 - x^4

d(A^2)/dx = 72x - 4x^3 [4u students can rewrite d(A^2)/dx as 2A dA/dx using implicit diff if they want]

Then let d(A^2)/dx = 0

Trust me, letting d(A^2)/dx = 0 has the same effect as dA/dx = 0

Bestintheworld · Feb 16, 2016

Re: HSC 2016 2U Marathon

can anyone attempt to do it? I wonder how long it will take you guys lol coz it took me so long to figure it out, seriously.

leehuan · Feb 16, 2016

Re: HSC 2016 2U Marathon

Bestintheworld said:
can anyone attempt to do it? I wonder how long it will take you guys lol coz it took me so long to figure it out, seriously.

Well the full method only hit me a few seconds ago so read my above edited post.

Bestintheworld · Feb 16, 2016

Re: HSC 2016 2U Marathon

leehuan said:
Well the full method only hit me a few seconds ago so read my above edited post.

Great, it took me a while to realise that I could use pythagoras theorem. Nevertheless, this is what I did:

leehuan · Feb 16, 2016

Re: HSC 2016 2U Marathon

Lol I knew it. I mixed up r=6 and 2r=12 again. But now, observe the SHORTCUT methods I suggested.

Direct use of product and chain rule is unnecessary brute force

InteGrand · Feb 16, 2016

Re: HSC 2016 2U Marathon

Bestintheworld said:
Great, it took me a while to realise that I could use pythagoras theorem. Nevertheless, this is what I did:

$\noindent As leehuan said, work with $A^2$ rather than $A$, this really simplifies the differentiation required, and $A$ is optimised precisely when $A^2$ is, since $A \geq 0$ and the function $f(x) = x^2$ is increasing on $\left[0, \infty\right)$.$

davidgoes4wce · Feb 17, 2016

Re: HSC 2016 2U Marathon

The point P(x,y) moves so that it is equidistant from L and the line y=0. Show that the locus of P is a parabola, and find its vertex.

Answer:

$(x+1)^2=6(y-\frac{2}{3}), vertex (-1\frac{3}{2})$

davidgoes4wce · Feb 17, 2016

Re: HSC 2016 2U Marathon

^^^Disregard the last question, just realized it was a continuation from Q10a in Cambridge.

davidgoes4wce · Feb 19, 2016

Re: HSC 2016 2U Marathon

This is from the Pender et al book Year 12 2 Unit Enhanced text:

$With regards to the 2nd part of the question, why does the book write the repayment along the lines of$

$A_{520}= M \times 0.0015 + M \times 0.0015^2 + ..... + M \times 0.0015^{20}$

$$ I was thinking more along the lines of $ A_{520} = M \times 1.0015 + M \times 1.0015^2 + ......+ M \times 1.0015^{520}$

davidgoes4wce · Feb 19, 2016

Re: HSC 2016 2U Marathon

$My thinking is based on the previous example from Pender et al, they should have worded it : $'' $The 1st instalment is invested for 520 weeks and so amounts to M$ $ \times 1.0015^{520} ''$

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