3U Maths Question Thread

[RuiAce]

Hey team, could i please have some help with this inverse function Question.
Q. If the domain is restricted to a monotonic increasing curve, find the inverse function, and state the domain and range of the inverse function.
The original function is y = 1 divided by x^2. Answer is y = -1 divided by root x, domain X>0 range Y<0.

I don't understand how they got their negative in front of the inverse function as well what is the best way to find the domain and range (is graphically a good method)?

Many Thanks!

Do not forget that the \(x\) and \(y\) values essentially swap for the inverse function. This means that the domain and range also swap for the inverse. So since the original domain was \( x < 0\), the range of the inverse is \( y < 0\)

[3.14159265359]

hello!
for this question (2012 q11fii) I was looking at the solutions and I don't understand why I'm not getting the answer.
I find the general term so:
nCr * (2)^(n-r) * (x)^3(n-r) * (-1)^r * (x)^-r
for Constant term
3n-3r-r=0
3n-4r=0
n= 4/3 r

I know that n has to be an integer but I don't get why you don't get an answer doing it this way compared to what the answers had done because according the symmetry of the triangle, both ways should work. can someone please explain why it doesn't? am I doing something wrong?
thank you

[RuiAce]

hello!
for this question (2012 q11fii) I was looking at the solutions and I don't understand why I'm not getting the answer.
I find the general term so:
nCr * (2)^(n-r) * (x)^3(n-r) * (-1)^r * (x)^-r
for Constant term
3n-3r-r=0
3n-4r=0
n= 4/3 r

I know that n has to be an integer but I don't get why you don't get an answer doing it this way compared to what the answers had done because according the symmetry of the triangle, both ways should work. can someone please explain why it doesn't? am I doing something wrong?
thank you

\( n = \frac43 r\) isn't helpful here because this gives you restrictions on \(r\) instead. This tells you that \(r\) must be a multiple of 3, or else \(n\) fails to be an integer.

You actually need \( r = \frac34 n\). This tells you that \(n\) must be a multiple of 4 to ensure that \(r\) is an integer.

[3.14159265359]

\( n = \frac43 r\) isn't helpful here because this gives you restrictions on \(r\) instead. This tells you that \(r\) must be a multiple of 3, or else \(n\) fails to be an integer.

You actually need \( r = \frac34 n\). This tells you that \(n\) must be a multiple of 4 to ensure that \(r\) is an integer.

so technically what I did wasn't wrong? its just that I didn't complete the question, right?

[RuiAce]

so technically what I did wasn't wrong? its just that I didn't complete the question, right?

Yeah

[3.14159265359]

Yeah

thank you!

[3.14159265359]

hello!
I was wondering If something like the hsc 2009 q6b weird binomial question could pop up in this years paper?

[jamonwindeyer]

hello!
I was wondering If something like the hsc 2009 q6b weird binomial question could pop up in this years paper?

It definitely could! It relates directly to stuff you learn in the course that said they have been shying away from those 'weirder' binomial questions lately, they tend to go to the more algebraic proofs

[secretweapon]

Can anyone please check my working out to see if its correct?
if we had the curve, y = -x^2-2 and had to find the area under the curve from -2 to 2, is my working out correct?
top terminal is 2, bottom terminal is -2, so -(top terminal 2, bottom terminal -2)-x^2-2 dx = 40/3
(don't know how to type in latex)
is the bolded - sign needed since the equation is below the x axis?

[jamonwindeyer]

Can anyone please check my working out to see if its correct?
if we had the curve, y = -x^2-2 and had to find the area under the curve from -2 to 2, is my working out correct?
top terminal is 2, bottom terminal is -2, so -(top terminal 2, bottom terminal -2)-x^2-2 dx = 40/3
(don't know how to type in latex)
is the bolded - sign needed since the equation is below the x axis?

That's correct! And yes it is - An area under the x-axis needs a sign change

[secretweapon]

That's correct! And yes it is - An area under the x-axis needs a sign change

Jamon you're a legend 
Also, how would you calculate the area under the curve y = -x^2+1 from 2 to -1 (since it's going from a negative to positive terminal)?

[clovvy]

Jamon you're a legend 
Also, how would you calculate the area under the curve y = -x^2+1 from 2 to -1 (since it's going from a negative to positive terminal)?

I would suggest draw the diagram on paper and shade the area within those curve..... if it was below the x-axis take the absolute value on that thing (usually)

[jamonwindeyer]

Jamon you're a legend 
Also, how would you calculate the area under the curve y = -x^2+1 from 2 to -1 (since it's going from a negative to positive terminal)?

Definitely agree with clovvy! To clarify - You split the area. The bit above the x-axis, keep it the same. The bit under, x=1 to x=2, flip the sign! But you need a diagram to really visualise that properly - Don't skip the diagram, no matter how confident you get

[beeangkah]

Hi, could I please have help with b) and c)

Thanks 

[RuiAce]

Hi, could I please have help with b) and c)

Thanks 

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