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HSC 2013 Maths Marathon (archive) (1 Viewer)

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spanky125

spanky125

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Re: HSC 2013 2U Marathon

The chance of Matt getting laid is equivalent to x in the equation:


Show why Matt has no chance of getting some.

Hint: Discriminant
 
A

andybandy

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Re: HSC 2013 2U Marathon

Youi_ said:
Click to expand...
<a href="http://www.codecogs.com/eqnedit.php?latex=\ln^{2}x&space;-&space;3&space;=&space;2.\ln&space;x" target="_blank"><img src="http://latex.codecogs.com/gif.latex?\ln^{2}x&space;-&space;3&space;=&space;2.\ln&space;x" title="\ln^{2}x - 3 = 2.\ln x" /></a>

<a href="http://www.codecogs.com/eqnedit.php?latex=\textup{let&space;}&space;m&space;=&space;\ln&space;x" target="_blank"><img src="http://latex.codecogs.com/gif.latex?\textup{let&space;}&space;m&space;=&space;\ln&space;x" title="\textup{let } m = \ln x" /></a>

<a href="http://www.codecogs.com/eqnedit.php?latex=m^{2}-2m-3=0" target="_blank"><img src="http://latex.codecogs.com/gif.latex?m^{2}-2m-3=0" title="m^{2}-2m-3=0" /></a>

<a href="http://www.codecogs.com/eqnedit.php?latex=(m-3)(m&plus;1)=&space;0" target="_blank"><img src="http://latex.codecogs.com/gif.latex?(m-3)(m&plus;1)=&space;0" title="(m-3)(m+1)= 0" /></a>

<a href="http://www.codecogs.com/eqnedit.php?latex=m&space;=&space;3,&space;-1" target="_blank"><img src="http://latex.codecogs.com/gif.latex?m&space;=&space;3,&space;-1" title="m = 3, -1" /></a>

<a href="http://www.codecogs.com/eqnedit.php?latex=\ln&space;x&space;=&space;3&space;\textup{&space;}&space;\therefore&space;e^{3}=x" target="_blank"><img src="http://latex.codecogs.com/gif.latex?\ln&space;x&space;=&space;3&space;\textup{&space;}&space;\therefore&space;e^{3}=x" title="\ln x = 3 \textup{ } \therefore e^{3}=x" /></a>

<a href="http://www.codecogs.com/eqnedit.php?latex=\ln&space;x&space;=&space;-1&space;\textup{&space;}&space;\therefore&space;e^{-1}=x" target="_blank"><img src="http://latex.codecogs.com/gif.latex?\ln&space;x&space;=&space;-1&space;\textup{&space;}&space;\therefore&space;e^{-1}=x" title="\ln x = -1 \textup{ } \therefore e^{-1}=x" /></a>

<a href="http://www.codecogs.com/eqnedit.php?latex=x&space;=&space;e^{3},&space;e^{-1}" target="_blank"><img src="http://latex.codecogs.com/gif.latex?x&space;=&space;e^{3},&space;e^{-1}" title="x = e^{3}, e^{-1}" /></a>
 
Last edited:
Menomaths

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Re: HSC 2013 2U Marathon

spanky125 said:
The chance of Matt getting laid is equivalent to x in the equation:


Show why Matt has no chance of getting some.

Hint: Discriminant
Click to expand...
negative discriminant so no real roots? Someone's name's Matt over here btw lol
 
Y

Youi_

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Re: HSC 2013 2U Marathon

spanky125 said:
The chance of Matt getting laid is equivalent to x in the equation:


Show why Matt has no chance of getting some.

Hint: Discriminant
Click to expand...
Matt can only get some if he plans on asking spanky therefore Matt will not get some as he isn't going to ask spanky
 
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leesh95

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Re: HSC 2013 2U Marathon

RealiseNothing said:
It's

Remember that the circle is beneath the x-axis and so it is negative.
Click to expand...
When you get this answer do you leave it as it is? or do you take an absolute value since my teacher told me you can't have a negative area.
 
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andybandy

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Re: HSC 2013 2U Marathon

leesh95 said:
When you get this answer do you leave it as it is? or do you take an absolute value since my teacher told me you can't have a negative area.
Click to expand...
the question didnt specify area, but just an "integral" if it asks to use this integral to find an area you would take absolute value
 
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Youi_

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Re: HSC 2013 2U Marathon

Menomaths said:
negative discriminant so no real roots? Someone's name's Matt over here btw lol
Click to expand...
We regret to inform Matt that he will never get laid. Sorry for any inconvenience.
 
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andybandy

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Re: HSC 2013 2U Marathon

leesh95 said:
Can someone post a solution of this? I can't seem to get it.
Click to expand...
<a href="http://www.codecogs.com/eqnedit.php?latex=\frac{\sec&space;x&space;-&space;\sec&space;x&space;\&space;cos&space;^{4}x}{1&plus;\cos^{2}x}" target="_blank"><img src="http://latex.codecogs.com/gif.latex?\frac{\sec&space;x&space;-&space;\sec&space;x&space;\&space;cos&space;^{4}x}{1&plus;\cos^{2}x}" title="\frac{\sec x - \sec x \ cos ^{4}x}{1+\cos^{2}x}" /></a>

<a href="http://www.codecogs.com/eqnedit.php?latex==&space;\frac{\sec&space;x(1&space;-&space;\cos^{4}x)&space;}{1&plus;\cos^{2}x}" target="_blank"><img src="http://latex.codecogs.com/gif.latex?=&space;\frac{\sec&space;x(1&space;-&space;\cos^{4}x)&space;}{1&plus;\cos^{2}x}" title="= \frac{\sec x(1 - \cos^{4}x) }{1+\cos^{2}x}" /></a>

<a href="http://www.codecogs.com/eqnedit.php?latex==&space;\frac{\sec&space;x(1&space;-&space;\cos^{2}x)(1&plus;\cos^{2}x)&space;}{1&plus;\cos^{2}x}" target="_blank"><img src="http://latex.codecogs.com/gif.latex?=&space;\frac{\sec&space;x(1&space;-&space;\cos^{2}x)(1&plus;\cos^{2}x)&space;}{1&plus;\cos^{2}x}" title="= \frac{\sec x(1 - \cos^{2}x)(1+\cos^{2}x) }{1+\cos^{2}x}" /></a>

<a href="http://www.codecogs.com/eqnedit.php?latex==\sec&space;x(1&space;-&space;\cos^{2}x)" target="_blank"><img src="http://latex.codecogs.com/gif.latex?=\sec&space;x(1&space;-&space;\cos^{2}x)" title="=\sec x(1 - \cos^{2}x)" /></a>

<a href="http://www.codecogs.com/eqnedit.php?latex==\sec&space;x(\sin^{2}x)" target="_blank"><img src="http://latex.codecogs.com/gif.latex?=\sec&space;x(\sin^{2}x)" title="=\sec x(\sin^{2}x)" /></a>

<a href="http://www.codecogs.com/eqnedit.php?latex==&space;\frac{1}{\sec&space;x}.(\sin&space;x.\sin&space;x)" target="_blank"><img src="http://latex.codecogs.com/gif.latex?=&space;\frac{1}{\sec&space;x}.(\sin&space;x.\sin&space;x)" title="= \frac{1}{\sec x}.(\sin x.\sin x)" /></a>

<a href="http://www.codecogs.com/eqnedit.php?latex==\frac{\sin&space;x}{\cos&space;x}.\sin&space;x" target="_blank"><img src="http://latex.codecogs.com/gif.latex?=\frac{\sin&space;x}{\cos&space;x}.\sin&space;x" title="=\frac{\sin x}{\cos x}.\sin x" /></a>

<a href="http://www.codecogs.com/eqnedit.php?latex==\sin&space;x\tan&space;x" target="_blank"><img src="http://latex.codecogs.com/gif.latex?=\sin&space;x\tan&space;x" title="=\sin x\tan x" /></a>
 
Last edited:
Y

Youi_

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Re: HSC 2013 2U Marathon

andybandy said:
<a href="http://www.codecogs.com/eqnedit.php?latex=\frac{\sec&space;x&space;-&space;\sec&space;x&space;\&space;cos&space;^{4}x}{1+\cos^{2}x}" target="_blank"><img src="http://latex.codecogs.com/gif.latex?\frac{\sec&space;x&space;-&space;\sec&space;x&space;\&space;cos&space;^{4}x}{1+\cos^{2}x}" title="\frac{\sec x - \sec x \ cos ^{4}x}{1+\cos^{2}x}" /></a>

<a href="http://www.codecogs.com/eqnedit.php?latex==&space;\frac{\sec&space;x(1&space;-&space;\cos^{4}x)&space;}{1+\cos^{2}x}" target="_blank"><img src="http://latex.codecogs.com/gif.latex?=&space;\frac{\sec&space;x(1&space;-&space;\cos^{4}x)&space;}{1+\cos^{2}x}" title="= \frac{\sec x(1 - \cos^{4}x) }{1+\cos^{2}x}" /></a>

<a href="http://www.codecogs.com/eqnedit.php?latex==&space;\frac{\sec&space;x(1&space;-&space;\cos^{2}x)(1+\cos^{2}x)&space;}{1+\cos^{2}x}" target="_blank"><img src="http://latex.codecogs.com/gif.latex?=&space;\frac{\sec&space;x(1&space;-&space;\cos^{2}x)(1+\cos^{2}x)&space;}{1+\cos^{2}x}" title="= \frac{\sec x(1 - \cos^{2}x)(1+\cos^{2}x) }{1+\cos^{2}x}" /></a>

<a href="http://www.codecogs.com/eqnedit.php?latex==\sec&space;x(1&space;-&space;\cos^{2}x)" target="_blank"><img src="http://latex.codecogs.com/gif.latex?=\sec&space;x(1&space;-&space;\cos^{2}x)" title="=\sec x(1 - \cos^{2}x)" /></a>

<a href="http://www.codecogs.com/eqnedit.php?latex==\sec&space;x(\sin^{2}x)" target="_blank"><img src="http://latex.codecogs.com/gif.latex?=\sec&space;x(\sin^{2}x)" title="=\sec x(\sin^{2}x)" /></a>

<a href="http://www.codecogs.com/eqnedit.php?latex==\frac{\sin&space;x}{\cos&space;x}.\sin&space;x" target="_blank"><img src="http://latex.codecogs.com/gif.latex?=\frac{\sin&space;x}{\cos&space;x}.\sin&space;x" title="=\frac{\sin x}{\cos x}.\sin x" /></a>

<a href="http://www.codecogs.com/eqnedit.php?latex==\sin&space;x\tan&space;x" target="_blank"><img src="http://latex.codecogs.com/gif.latex?=\sin&space;x\tan&space;x" title="=\sin x\tan x" /></a>
Click to expand...
Your version of latex looks very confusing :S
 
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hanif

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Re: HSC 2013 2U Marathon

andybandy said:
<a href="http://www.codecogs.com/eqnedit.php?latex=\frac{\sec&space;x&space;-&space;\sec&space;x&space;\&space;cos&space;^{4}x}{1+\cos^{2}x}" target="_blank"><img src="http://latex.codecogs.com/gif.latex?\frac{\sec&space;x&space;-&space;\sec&space;x&space;\&space;cos&space;^{4}x}{1+\cos^{2}x}" title="\frac{\sec x - \sec x \ cos ^{4}x}{1+\cos^{2}x}" /></a>

<a href="http://www.codecogs.com/eqnedit.php?latex==&space;\frac{\sec&space;x(1&space;-&space;\cos^{4}x)&space;}{1+\cos^{2}x}" target="_blank"><img src="http://latex.codecogs.com/gif.latex?=&space;\frac{\sec&space;x(1&space;-&space;\cos^{4}x)&space;}{1+\cos^{2}x}" title="= \frac{\sec x(1 - \cos^{4}x) }{1+\cos^{2}x}" /></a>

<a href="http://www.codecogs.com/eqnedit.php?latex==&space;\frac{\sec&space;x(1&space;-&space;\cos^{2}x)(1+\cos^{2}x)&space;}{1+\cos^{2}x}" target="_blank"><img src="http://latex.codecogs.com/gif.latex?=&space;\frac{\sec&space;x(1&space;-&space;\cos^{2}x)(1+\cos^{2}x)&space;}{1+\cos^{2}x}" title="= \frac{\sec x(1 - \cos^{2}x)(1+\cos^{2}x) }{1+\cos^{2}x}" /></a>

<a href="http://www.codecogs.com/eqnedit.php?latex==\sec&space;x(1&space;-&space;\cos^{2}x)" target="_blank"><img src="http://latex.codecogs.com/gif.latex?=\sec&space;x(1&space;-&space;\cos^{2}x)" title="=\sec x(1 - \cos^{2}x)" /></a>

<a href="http://www.codecogs.com/eqnedit.php?latex==\sec&space;x(\sin^{2}x)" target="_blank"><img src="http://latex.codecogs.com/gif.latex?=\sec&space;x(\sin^{2}x)" title="=\sec x(\sin^{2}x)" /></a>

<a href="http://www.codecogs.com/eqnedit.php?latex==&space;\frac{1}{\sec&space;x}.(\sin&space;x.\sin&space;x)" target="_blank"><img src="http://latex.codecogs.com/gif.latex?=&space;\frac{1}{\sec&space;x}.(\sin&space;x.\sin&space;x)" title="= \frac{1}{\sec x}.(\sin x.\sin x)" /></a>

<a href="http://www.codecogs.com/eqnedit.php?latex==\frac{\sin&space;x}{\cos&space;x}.\sin&space;x" target="_blank"><img src="http://latex.codecogs.com/gif.latex?=\frac{\sin&space;x}{\cos&space;x}.\sin&space;x" title="=\frac{\sin x}{\cos x}.\sin x" /></a>

<a href="http://www.codecogs.com/eqnedit.php?latex==\sin&space;x\tan&space;x" target="_blank"><img src="http://latex.codecogs.com/gif.latex?=\sin&space;x\tan&space;x" title="=\sin x\tan x" /></a>
Click to expand...
How did you go from the first line to the second?
 
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andybandy

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Re: HSC 2013 2U Marathon

hanif said:
How did you go from the first line to the second?
Click to expand...
i took a common factor of secx

if your talking about the change from cos to the power of 4 to them two brackets, you use this rule

<a href="http://www.codecogs.com/eqnedit.php?latex=(x^{2}-1)&space;=&space;(x&space;-&space;1)(x&plus;1)" target="_blank"><img src="http://latex.codecogs.com/gif.latex?(x^{2}-1)&space;=&space;(x&space;-&space;1)(x&plus;1)" title="(x^{2}-1) = (x - 1)(x+1)" /></a>
 
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