HSC 2013 Maths Marathon (archive) (5 Viewers)

[leesh95]

Re: HSC 2013 2U Marathon

x=e^3?

 

[spanky125]

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

 

[andybandy]

Re: HSC 2013 2U Marathon

<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>

 

[Menomaths]

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

negative discriminant so no real roots? Someone's name's Matt over here btw lol

 

[spanky125]

Re: HSC 2013 2U Marathon

negative discriminant so no real roots? Someone's name's Matt over here btw lol

Correct haha maybe it's an omen?

 

[Badsiii]

Re: HSC 2013 2U Marathon

I don't think so

x=e^3?

Yeh that's correct

 

[Menomaths]

Re: HSC 2013 2U Marathon

Yeh that's correct

Why you quoting me for?

 

[Youi_]

Re: HSC 2013 2U Marathon

x=e^3?

 

[Youi_]

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

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

 

[leesh95]

Re: HSC 2013 2U Marathon

It's

Remember that the circle is beneath the x-axis and so it is negative.

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.

 

[andybandy]

Re: HSC 2013 2U Marathon

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.

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

 

[Youi_]

Re: HSC 2013 2U Marathon

negative discriminant so no real roots? Someone's name's Matt over here btw lol

We regret to inform Matt that he will never get laid. Sorry for any inconvenience.

 

[leesh95]

Re: HSC 2013 2U Marathon

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

Okay thanks

 

[leesh95]

Re: HSC 2013 2U Marathon

Show that
sec(x) - sec(x)cos^4(x) = sin(x)tan(x)
1 + cos^2(x)

Can someone post a solution of this? I can't seem to get it.

 

[Youi_]

Re: HSC 2013 2U Marathon

Can someone post a solution of this? I can't seem to get it.

Numerator: take out sec(x) as a factor and use the fact that



You should get it from there

 

[andybandy]

Re: HSC 2013 2U Marathon

Can someone post a solution of this? I can't seem to get it.

<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>

 

[Youi_]

Re: HSC 2013 2U Marathon

<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>

Your version of latex looks very confusing :S

 

[andybandy]

Re: HSC 2013 2U Marathon

Your version of latex looks very confusing :S

Haha, what do you mean? though the second last line does seem confusing

 

[hanif]

Re: HSC 2013 2U Marathon

<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>

How did you go from the first line to the second?

 

[andybandy]

Re: HSC 2013 2U Marathon

How did you go from the first line to the second?

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>