3U Maths Question Thread

[jamonwindeyer]

Heyy whats locus for

Distance of P from line x=-3 is 2/5 of its distance from y=-1

Hey! So we form the condition:



Now \(d_1\), the distance to the line \(x=-3\), is just the difference in the x-coordinates. Similarly, \(d_2\) is the difference in the y-coordinates. Meaning:



We use absolute value signs because we don't care about sign when considering distance (note what I've done here is a shortcut from the perpendicular distance to a line formula, which also has absolute values, you could use that instead if you prefer).

To finish, square both sides so we don't have to deal with the absolute value:



And I'll leave it to you to get it into an appropriate form this locus is actually a pair of intersecting lines. You could handle each separately - This approach handles them together

[anotherworld2b]

Can I have help with this question ? I'm not sure how to even begin

[bsdfjnlkasn]

Hey there,

Was wondering how you would approach the following multiple choice - the answer's D

Thank you!

[jamonwindeyer]

Can I have help with this question ? I'm not sure how to even begin

Hey! The first one is just substitution, \(f_1=20\) and \(f_2=50\)

The second one, \(x=3\), and you are rearranging to make \(f_2=\text{Something}\). The start would be:



Reckon you could take the reins from there? Let us know

[jamonwindeyer]

Hey there,

Was wondering how you would approach the following multiple choice - the answer's D

Thank you!

Hey! We get that angle with the formula:



Those are the velocities in the vertical and horizontal directions. Neglecting air resistance, \(\dot{x}=v\), since it is launched horizontally. The vertical is tougher, but there is no initial vertical velocity, so we can get it fairly simply from the usual formula:

\(\dot{y}=-gt\)

Now we could get this in terms of \(v\) and \(h\) as well, but let's get intuitive. The taller the cliff, the longer it will fall. Therefore, the taller the cliff, the larger \(\dot{y}\) will be!

To maximise \(\theta\), we need to maximise \(\tan{\theta}\). This means a big vertical velocity and a small horizontal velocity. So, make \(v\) small and make \(h\) large - That matches D as our answer

Note you could have done the whole thing intuitively, or the whole thing mathematically - I did a mix

[bsdfjnlkasn]

Hey! We get that angle with the formula:



Those are the velocities in the vertical and horizontal directions. Neglecting air resistance, \(\dot{x}=v\), since it is launched horizontally. The vertical is tougher, but there is no initial vertical velocity, so we can get it fairly simply from the usual formula:

\(\dot{y}=-gt\)

Now we could get this in terms of \(v\) and \(h\) as well, but let's get intuitive. The taller the cliff, the longer it will fall. Therefore, the taller the cliff, the larger \(\dot{y}\) will be!

To maximise \(\theta\), we need to maximise \(\tan{\theta}\). This means a big vertical velocity and a small horizontal velocity. So, make \(v\) small and make \(h\) large - That matches D as our answer

Note you could have done the whole thing intuitively, or the whole thing mathematically - I did a mix

Awesome, makes sense now!

I was wondering if you could help explain the logic for the following polynomials question (solution attached). I haven't seen this approach before and don't really understand what they've set up in the beginning of method 2. Any guidance would be super appreciated. I understand Method 1, but was wondering if the sum in pairs squared business was a result we had to know in 3U. Finally, how did they get the order of the coefficients correct? How can we deduce them from the different root equations (the sum, sum in pairs and product)?

Thank you!!

[jamonwindeyer]

Awesome, makes sense now!

I was wondering if you could help explain the logic for the following polynomials question (solution attached). I haven't seen this approach before and don't really understand what they've set up in the beginning of method 2. Any guidance would be super appreciated. I understand Method 1, but was wondering if the sum in pairs squared business was a result we had to know in 3U. Finally, how did they get the order of the coefficients correct? How can we deduce them from the different root equations (the sum, sum in pairs and product)?

Thank you!!

Sure! So Method 2, essentially what they've done is they've said okay, if the original roots are \(x\) values, let's setup a new variable corresponding to the squares of those roots, as required by the question. \(y=x^2\). By rearranging, we can then substitute into the original, and we end up with the new polynomial straight away. This isn't the typical method for this question, Method 1 is definitely what would be expected imo.

That sum of pairs stuff in Method 1 isn't necessarily something you should memorise, but it is absolutely something to be aware of and be ready to use (even if you have to do some scribbling to get it back into your mind) for questions. A HSC question might even lead you through the process of obtaining the result yourself. But could it show up in an exam in some way? Absolutely

[RuiAce]

Sure! So Method 2, essentially what they've done is they've said okay, if the original roots are \(x\) values, let's setup a new variable corresponding to the squares of those roots, as required by the question. \(y=x^2\). By rearranging, we can then substitute into the original, and we end up with the new polynomial straight away. This isn't the typical method for this question, Method 1 is definitely what would be expected imo.

Remark: This is as far as 3U goes. Method 2 is far more common in 4U.

[legorgo18]

Hello! Needs some help with this probability question: A pack of 36 cards includes 20 numbered cards from 6-10 inclusive, 4 aces and 12 picture cards. If a hand of 5 cards is selected at random find the probability of receiving at least 2 aces.
Thank you

[RuiAce]

Hello! Needs some help with this probability question: A pack of 36 cards includes 20 numbered cards from 6-10 inclusive, 4 aces and 12 picture cards. If a hand of 5 cards is selected at random find the probability of receiving at least 2 aces.
Thank you

Hint: The complement may be easier to consider here. There's less work to do if you're dealing with just one ace or no aces.

That being said, please post up some relevant ideas that you tried to employ, or any appropriate train of thought.

[legorgo18]

Rui, btw what is this question 3u or 4u?

The region R, bounded by 0 ≤ x ≤ 2, 0 ≤ y ≤ 4x2 – x4, is rotated about the y-axis. The solid so formed is sliced by planes perpendicular to the y-axis. Express the areas of the cross-sections so formed as a function of y, the distance of the plane from the origin. Use this result to calculate the volume of the solid.

For that probability q, i just got confused with how to do picture and number cards cases whilst also having the ace cases.

[RuiAce]

Rui, btw what is this question 3u or 4u?

The region R, bounded by 0 ≤ x ≤ 2, 0 ≤ y ≤ 4x2 – x4, is rotated about the y-axis. The solid so formed is sliced by planes perpendicular to the y-axis. Express the areas of the cross-sections so formed as a function of y, the distance of the plane from the origin. Use this result to calculate the volume of the solid.

For that probability q, i just got confused with how to do picture and number cards cases whilst also having the ace cases.

I feel as though that volumes question is 4U. It's doable within the scope of 2U but the amount of work you have to put into it is honestly bizarre for even 3U.
____________________

[legorgo18]

oh i get it now you just add the pictures and numbers together as 1 thing as there is no limit on it... silly me

[RuiAce]

Basically yeah. It's important that you know what's not restricted because most of the time the unrestricted objects can be treated all at once.

[anotherworld2b]

I'm not sure what I did wrong