HSC 2012 MX1 Marathon #1 (archive) (2 Viewers)

pokka

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Re: 2012 HSC MX1 Marathon

 
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Nooblet94

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Re: 2012 HSC MX1 Marathon

You've made an error in the expression for Tr+1 of the second binomial, although coincidentally one of your answers is the correct one.
 

pokka

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Re: 2012 HSC MX1 Marathon

Oh woops I just realised haha my bad...fixed it anyway :D
 

deswa1

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Re: 2012 HSC MX1 Marathon

This question was fun to do. Enjoy :)

Find the value of k for which the line y=kx bisects the area enclosed by the curve 4y=4x-x^2 and the x-axis.
 

Carrotsticks

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Re: 2012 HSC MX1 Marathon

This question was fun to do. Enjoy :)

Find the value of k for which the line y=kx bisects the area enclosed by the curve 4y=4x-x^2 and the x-axis.
I love questions like these. Don't know why.
 

zeebobDD

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Re: 2012 HSC MX1 Marathon

Use induction to prove that; ∑_(r=1)^n▒1/r^2 ≤2-1/n for n=1,2,3….

bascially 1/n^2≤ 2-(1/n)
 

bleakarcher

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Re: 2012 HSC MX1 Marathon

I first found the points of intersection between the parabola and the line clearly the first one being O(0,0) and the second being (4(1-k),f(4(1-k))) where f(x)=x-(1/4)x^2.
After this I found the area bounded by the line and curve in terms of k and set this equal to the half the area bounded by f(x)=x-(1/4)x^2 and the x-axis.
 

Aesytic

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Re: 2012 HSC MX1 Marathon

prove true for n=1,
lhs = 1/1^2 = 1
rhs = 2-1/1 = 1
1≤1
.'. true for n=1

assume true for n=k
1/k^2 ≤ 2 - 1/k

RTP 1/(k+1)^2 ≤ 2 - 1/(k+1)
k>0, .'. k+1 > 1
.'. (k+1)^2 > k^2
.'. 1/(k+1)^2 < 1/k^2
using the assumption,
1/(k+1)^2 < 1/k^2 ≤ 2 - 1/k
.'. 1/(k+1)^2 ≤ 2 - 1/k
since k +1 > k,
.'. 1/(k+1)< 1/k
2-1/(k+1)> 2-1/k [multiply both sides by -1 and add 2]
.'. 2-1/(k+1) > 2-1/k > 1/(k+1)^2
.'. 1/(k+1)^2 ≤ 2 - 1/(k+1)
.'. true for n=k+1 if true for n=k
.'. by the process of mathematical induction is true for n=1+1=2, and 2+1=3, etc. and hence true for all real integers n
 

deswa1

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Re: 2012 HSC MX1 Marathon

Maybe you guys can help me out on this problem. I tried it and what I'm doing should work but it isn't (have no idea where I made my mistake). Here's the question:
The tangent to the curve y=x^3 has equation y=12x-16 and meets the curve at (2,8) and (-4,-64). Find the area enclosed between the curve and the tangent.

Thanks :)
 

kingkong123

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Re: 2012 HSC MX1 Marathon

Maybe you guys can help me out on this problem. I tried it and what I'm doing should work but it isn't (have no idea where I made my mistake). Here's the question:
The tangent to the curve y=x^3 has equation y=12x-16 and meets the curve at (2,8) and (-4,-64). Find the area enclosed between the curve and the tangent.

Thanks :)
is the answer 108 u^2?
 

Carrotsticks

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Re: 2012 HSC MX1 Marathon

.'. by the process of mathematical induction is true for n=1+1=2, and 2+1=3, etc. and hence true for all real integers n
Alternatively, you could draw a little square.
 

deswa1

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Re: 2012 HSC MX1 Marathon

is the answer 108 u^2?
Yep. How did you do it? If you can't type it up, can you just summarise the method and then I'll be able to see where I stuffed up.
 

Nooblet94

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Re: 2012 HSC MX1 Marathon

Yep. How did you do it? If you can't type it up, can you just summarise the method and then I'll be able to see where I stuffed up.
<a href="http://www.codecogs.com/eqnedit.php?latex=\\ \begin{align*} A &=\left|\int^{2}_{-4} (x^3-12x@plus;16)dx\right|\\ &=\left|\left[ \frac{x^4}{4}-6x^2@plus;16x\right]^{2}_{-4} \right|\\ &=\left|\left( \frac{2^4}{4}-6\cdot 2^2@plus;16\cdot 2\right)- \left( \frac{(-4)^4}{4}-6\cdot (-4)^2@plus;16\cdot (-4)\right)\right|\\ &=\left |12@plus;96 \right|\\ &=\left |108 \right|\\ &=108~\textrm{units}^2 \end{align*}" target="_blank"><img src="http://latex.codecogs.com/gif.latex?\\ \begin{align*} A &=\left|\int^{2}_{-4} (x^3-12x+16)dx\right|\\ &=\left|\left[ \frac{x^4}{4}-6x^2+16x\right]^{2}_{-4} \right|\\ &=\left|\left( \frac{2^4}{4}-6\cdot 2^2+16\cdot 2\right)- \left( \frac{(-4)^4}{4}-6\cdot (-4)^2+16\cdot (-4)\right)\right|\\ &=\left |12+96 \right|\\ &=\left |108 \right|\\ &=108~\textrm{units}^2 \end{align*}" title="\\ \begin{align*} A &=\left|\int^{2}_{-4} (x^3-12x+16)dx\right|\\ &=\left|\left[ \frac{x^4}{4}-6x^2+16x\right]^{2}_{-4} \right|\\ &=\left|\left( \frac{2^4}{4}-6\cdot 2^2+16\cdot 2\right)- \left( \frac{(-4)^4}{4}-6\cdot (-4)^2+16\cdot (-4)\right)\right|\\ &=\left |12+96 \right|\\ &=\left |108 \right|\\ &=108~\textrm{units}^2 \end{align*}" /></a>
 
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deswa1

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Re: 2012 HSC MX1 Marathon

Just a quick Q. If a question in a three unit paper says solve this integral using the substitution... or otherwise, can you use integration by parts or are you limited to 3U methods?
 

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