HSC 2012 MX1 Marathon #1 (archive) (1 Viewer)

AAEldar

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Re: 2012 HSC MX1 Marathon

So how do you do addition/subtraction/multiplication/division of trigonometry without calcs or tables?

For example:

Tan^260/sin45-cos30

OR

Sin^2 15 + sin^2 30 + cos^2 15 - cos^2 45
For the first one, use your exact values. You know them for 30, 45 and 60 and so just by subbing them in you'll reach the answer.

For the second one, same thing. But you don't know the exact value of 15 however you do know that so group the terms and use that.
 
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seanieg89

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Re: 2012 HSC MX1 Marathon

Wouldn't necessarily call any of the school courses "fun" but at least 4U had the potential for some of the questions to be interesting. Higher mathematics is a lot more "fun" because there is room for creativity and originality.
 

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Re: 2012 HSC MX1 Marathon

Though what do you do with tan260?
 

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Re: 2012 HSC MX1 Marathon

Hmm, I seem to be having a lot of trouble with these. How did you guys learn how to do addition, subtraction, multiplication and division of trigonometry?
 

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Re: 2012 HSC MX1 Marathon

Find the values of m for which the equation 4x^2-mx+9=0 has real roots.

^Need help
 

deswa1

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Re: 2012 HSC MX1 Marathon

You have real roots when the disriminant is greater than or equal to zero...

<a href="http://www.codecogs.com/eqnedit.php?latex=\Delta =m^2-4(4)(9) \\ \Delta \geq 0\\ \therefore m^2-144\geq 0" target="_blank"><img src="http://latex.codecogs.com/gif.latex?\Delta =m^2-4(4)(9) \\ \Delta \geq 0\\ \therefore m^2-144\geq 0" title="\Delta =m^2-4(4)(9) \\ \Delta \geq 0\\ \therefore m^2-144\geq 0" /></a>

You can solve the rest :)
 

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Re: 2012 HSC MX1 Marathon

x^2+(k+6)x-2k=0 Find the discriminant.

This is my working out

(k+6)^2-4(1)(-2k)

k^2+12k+36-4+8k

k^2+20k+32=0 <-Is this the discriminant?
 

Carrotsticks

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Re: 2012 HSC MX1 Marathon

x^2+(k+6)x-2k=0 Find the discriminant.

This is my working out

(k+6)^2-4(1)(-2k)

k^2+12k+36-4+8k

k^2+20k+32=0 <-Is this the discriminant?
Check your working out in this line.
 

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Re: 2012 HSC MX1 Marathon

Find the values of k for which the quad eqn x^2-(k+3)x+(k+6)=0

My working
(K+3)^2-4(1)(K+6)
k^2+6k+9-4k-24
K^2+2k-15
(k+5)(k-3)

Answers are -5,3

Right/wrong?
 
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Carrotsticks

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Re: 2012 HSC MX1 Marathon

Find the values of k for which the quad eqn x^2-(k+3)x+(k+6)=0

My working
(K+3)^2-4(1)(K=6)
k^2+6k+9-4k-24
K^2+2k-15
(k+5)(k-3)

Answers are -5,3

Right/wrong?
Not quite, you have to use the quadratic equation to solve for x.

Then make the inside of the square root (from the quadratic equation) > or equal to 0.

You just found the values of K such that the quadratic eqn has 1 real root.
 

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Re: 2012 HSC MX1 Marathon

Not quite, you have to use the quadratic equation to solve for x.

Then make the inside of the square root (from the quadratic equation) > or equal to 0.

You just found the values of K such that the quadratic eqn has 1 real root.
I'm not sure if I follow :/
 

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