VCE Methods Question Thread!

[Shadowxo]

Oh ok thank you! I was typing 5/8 = tan(theta) directly into the calculator.

I always do inverses for those kinds of questions, that should help and it helps that it only gives you one answer that way

[deStudent]

http://m.imgur.com/a/qe6TZ

For Q8h) I'm not exactly sure how to solve this correctly. the correct answer was x=<-2 and x>=1

Thanks

[Sine]

http://m.imgur.com/a/qe6TZ

For Q8h) I'm not exactly sure how to solve this correctly. the correct answer was x=<-2 and x>=1

Thanks

best way to solve any inequality that is not linear is to do it graphically. Graph the function then solve particular points such as function=0 and just read off the set of values that satisfy the inequality

[Shadowxo]

http://m.imgur.com/a/qe6TZ

For Q8h) I'm not exactly sure how to solve this correctly. the correct answer was x=<-2 and x>=1

Thanks

I'd add and subtract 2 from the numerator to make 1 -3/(x+2) >=0
3/(x+2) =<1 and solve from there
If you need extra help just ask

[deStudent]

I'd add and subtract 2 from the numerator to make 1 -3/(x+2) >=0
3/(x+2) =<1 and solve from there
If you need extra help just ask

Thanks. I realised you could also simplify by polynomial long division.

Was initially pretty lost on this question by the way the hyperbola was expressed.

[Shadowxo]

Thanks. I realised you could also simplify by polynomial long division.

Was initially pretty lost on this question by the way the hyperbola was expressed.

No worries,
since x is only to a power of one and is on both the numerator and denominator with the same number on both (single x in this case), changing the way it's expressed so x isn't in the numerator is the best way of starting out most of the time.

[Shinkaze]

https://puu.sh/tI8UN/b93baa2538.png

PLEASEEE and thank you in advance!

[Shadowxo]

https://puu.sh/tI8UN/b93baa2538.png

PLEASEEE and thank you in advance!

f^-1(x) exists where f(x) is a 1:1 function
So I'd find the turning point of it by finding where the derivative = 0 (quotient rule), and that's where it stops being a 1:1 function. In this case, as there are two, since the domain starts at negative infinity, the lower value is the first stationary point
I ended up with -1-√2 and you can see that's where the first turning point on the graph is.

[Shinkaze]

I found that too (just want to make sure my answer is correct) Thanks! But how do I go by finding the equation?

[Shadowxo]

I found that too (just want to make sure my answer is correct) Thanks! But how do I go by finding the equation?

You just swap around the x and y and rearrange to make y the subject
(Alternatively, f(f^-1(x))=x)
so you end up with x = (f^-1(x)+1)/((f^-1(x))²+1)
then (f^-1(x))²x - f^-1(x) +x - 1 = 0
Use the quadratic formula
Find a point on the original equation (I did -3,-1/5) and sub that in to figure out whether the it's positive or negative root ( positive in this case)
My final answer was (1+√(-4x²+4x+1))/2x

[Shinkaze]

so you end up with x = (f^-1(x)+1)/((f^-1(x))²+1)
then (f^-1(x))²x - f^-1(x) +x - 1 = 0
Use the quadratic formula

Thanks This is exactly the part I got stuck with too many different x haha

[samuelbeattie76]

My question is attached, any assistance in helping understand this question further is greatly appreciated. Thank you.

[de]

My question is attached, any assistance in helping understand this question further is greatly appreciated. Thank you.

It might help to do some reading on the Pigeonhole theorem. But anyway, Clearly the least number of hands someone could have shaken is 0 and the most is n-1 (you can't shake your own hand). We want to prove that there exist two people who shook the same number of hands. Suppose, in the hope of a contradiction, that everyone shook a different number of hands. Since there are n possible values for the number of hands a person shakes and there are n people all these values must be used up. Therefore there is someone that shakes 0 hands and someone that shakes n-1 hands-however this person that shook n-1 hands shook everyone's hands but himself INCLUDING the person that shook 0 hands (hand shaking is a mutual thing). So, this is a contradiction, meaning our original assumption is wrong and there exist two people who shook the same number of hands.

[deStudent]

http://imgur.com/361CRwc
http://imgur.com/T9kVfEo
(Question and my working)

I'm not sure what I'm doing wrong? But the correct answer is 2tanxsec^2 (x) after being fully simplified.

Thanks

[sweetcheeks]

http://imgur.com/361CRwc
http://imgur.com/T9kVfEo
(Question and my working)

I'm not sure what I'm doing wrong? But the correct answer is 2tanxsec^2 (x) after being fully simplified.

Thanks

You have made a mistake when solving du/dx. sec²(x) is the same as 1/(cos²(x)). When differentiated it becomes sin(x)/cos³(x). This can now be simplified to tan(x)/cos²(x) and then becomes tan(x)Xsec²(x). Now sub that in as du/dx and you should get the answer.