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

[Nooblet94]

Re: 2012 HSC MX1 Marathon

<a href="http://www.codecogs.com/eqnedit.php?latex=\\ f(x)=\sin x -x@plus; \frac{x^3}{6}\\ f'(x)=\cos x -1@plus;\frac{x^2}{2}\\ f''(x)=-\sin x @plus;x\\ f'''(x)=-\cos x @plus;1\\ ~\\ f(0)=\sin 0 -0@plus; \frac{0^3}{6}=0\\ f'(0)=\cos 0 -1@plus;\frac{0^2}{2}=0\\ f''(0)=-\sin 0 @plus;0=0\\ ~\\ \textrm{Now,}~f'''(x)\geq 0~\textrm{for}~x>0~\textrm{(Since the maximum value of}~\cos x~\textrm{is 1)}\\ ~\\ \textrm{Hence, by the given theorem}\\ f''(x)\geq 0~\textrm{for}~x>0\\ \Rightarrow f'(x)\geq 0~\textrm{for}~x>0\\ \Rightarrow f(x)\geq 0~\textrm{for}~x>0\\" target="_blank"><img src="http://latex.codecogs.com/gif.latex?\\ f(x)=\sin x -x+ \frac{x^3}{6}\\ f'(x)=\cos x -1+\frac{x^2}{2}\\ f''(x)=-\sin x +x\\ f'''(x)=-\cos x +1\\ ~\\ f(0)=\sin 0 -0+ \frac{0^3}{6}=0\\ f'(0)=\cos 0 -1+\frac{0^2}{2}=0\\ f''(0)=-\sin 0 +0=0\\ ~\\ \textrm{Now,}~f'''(x)\geq 0~\textrm{for}~x>0~\textrm{(Since the maximum value of}~\cos x~\textrm{is 1)}\\ ~\\ \textrm{Hence, by the given theorem}\\ f''(x)\geq 0~\textrm{for}~x>0\\ \Rightarrow f'(x)\geq 0~\textrm{for}~x>0\\ \Rightarrow f(x)\geq 0~\textrm{for}~x>0\\" title="\\ f(x)=\sin x -x+ \frac{x^3}{6}\\ f'(x)=\cos x -1+\frac{x^2}{2}\\ f''(x)=-\sin x +x\\ f'''(x)=-\cos x +1\\ ~\\ f(0)=\sin 0 -0+ \frac{0^3}{6}=0\\ f'(0)=\cos 0 -1+\frac{0^2}{2}=0\\ f''(0)=-\sin 0 +0=0\\ ~\\ \textrm{Now,}~f'''(x)\geq 0~\textrm{for}~x>0~\textrm{(Since the maximum value of}~\cos x~\textrm{is 1)}\\ ~\\ \textrm{Hence, by the given theorem}\\ f''(x)\geq 0~\textrm{for}~x>0\\ \Rightarrow f'(x)\geq 0~\textrm{for}~x>0\\ \Rightarrow f(x)\geq 0~\textrm{for}~x>0\\" /></a>


Here's one from my MX1 assessment task last year that hardly anyone got:
<a href="http://www.codecogs.com/eqnedit.php?latex=\\ \textrm{The constant term in each of the binomials} \left(x@plus;\frac{4}{x}\right)^{2n}~\textrm{and}~\left(\frac{x}{a}@plus;\frac{b}{x} \right )^{2n}~\textrm{are equal.}\\ ~\\ \textrm{If}~a@plus;b=-5~\textrm{find the values of}~a~\textrm{and}~b" target="_blank"><img src="http://latex.codecogs.com/gif.latex?\\ \textrm{The constant term in each of the binomials} \left(x+\frac{4}{x}\right)^{2n}~\textrm{and}~\left(\frac{x}{a}+\frac{b}{x} \right )^{2n}~\textrm{are equal.}\\ ~\\ \textrm{If}~a+b=-5~\textrm{find the values of}~a~\textrm{and}~b" title="\\ \textrm{The constant term in each of the binomials} \left(x+\frac{4}{x}\right)^{2n}~\textrm{and}~\left(\frac{x}{a}+\frac{b}{x} \right )^{2n}~\textrm{are equal.}\\ ~\\ \textrm{If}~a+b=-5~\textrm{find the values of}~a~\textrm{and}~b" /></a>

 

[pokka]

Re: 2012 HSC MX1 Marathon

 

[Nooblet94]

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]

Re: 2012 HSC MX1 Marathon

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

 

[deswa1]

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]

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.

 

[bleakarcher]

Re: 2012 HSC MX1 Marathon

k=1-cuberoot(0.5)?

 

[zeebobDD]

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)

 

[deswa1]

Re: 2012 HSC MX1 Marathon

k=1-cuberoot(0.5)?

Yep. How did you approach it (the way I did it was a bit messy...)

 

[bleakarcher]

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]

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]

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]

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]

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.

 

[SpiralFlex]

Re: 2012 HSC MX1 Marathon

Lol ^ On sugar Carrotsticks?

 

[Aesytic]

Re: 2012 HSC MX1 Marathon

Alternatively, you could draw a little square.

what do you mean by a little square?

 

[deswa1]

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.

 

[cutemouse]

Re: 2012 HSC MX1 Marathon

what do you mean by a little square?

A little square symbol means "End of proof". Or you could use "QED".

 

[Nooblet94]

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>

 

[deterministic]

Re: 2012 HSC MX1 Marathon

what do you mean by a little square?

http://en.wikipedia.org/wiki/Tombstone_(typography)