3U Maths Question Thread

[hanaacdr]

Hi,
does anyone know where i can find perms and combs questions?
thank you!

[Sukakadonkadonk]

Yep, you're absolutely right! There is a relationship that, for very small theta,



So, if we just let theta equal two alpha, the relationship is also true for



Which is the same as the working out that you posted. Does that make sense?

Ahh, ok
Thank you.

[teapancakes08]

Working my way through past trial papers. Need help on questions involving lnx 

(If the attachment doesn't work...again...reply please...)

[jamonwindeyer]

Hi,
does anyone know where i can find perms and combs questions?
thank you!

Grab em from the Trial papers on THSC! Once you have done everything in your textbook that will be your best bet

[jamonwindeyer]

Working my way through past trial papers. Need help on questions involving lnx 

(If the attachment doesn't work...again...reply please...)

Hey! Good on you for getting practice in so early in the year for Trial papers, I didn't touch them until June, go you

Let me give you the gist of the first one. For (i), you sketch the graph of \(\ln{x}\) as stated (check Wolfram Alpha or Grapher or anything similar to check if you get the shape right, it is something you should remember off by heart), and the other graph you should sketch is \(4-x^2\), an upside down parabola with y-intercept 4 and x-intercepts 2 and -2. Why? Because:



So the roots of the equation are actually just the points of intersection of the two curves you have drawn. Indeed, there is one POI, so only one root.

For (ii), show that the result of the equation is negative for one of 1 and 2, and positive for the other. Therefore, there must be a root between them. For (iii), apply the halving the interval rule - Do a similar sign check on x=1.5 and use that to deduce an interval half as large (you'll get 1.5<x<2 if you do this correctly). Then (iv), just an application of Newton's method! Really, this question doesn't test your knowledge of the logarithm beyond the first bit, it is a test on estimating roots of polynomials! Happy to explain this further if you need

For the second question:



A fairly tricky substitution one there hope it helps!

[hanaacdr]

Grab em from the Trial papers on THSC! Once you have done everything in your textbook that will be your best bet

thank you!

[scienceislife]

"Find the exact area bounded by the curves y=sinx and y=cosx in the domain 0 is less than or equal to x is less than or equal to 2pi."

Find the equation of the normal to the curve y=log₃x at the point where x = 3.

Mod Edit: Posts merged, you can use 'Modify' to add to your previous post if it hasn't been answered yet

[jamonwindeyer]

"Find the exact area bounded by the curves y=sinx and y=cosx in the domain 0 is less than or equal to x is less than or equal to 2pi."

Find the equation of the normal to the curve y=log₃x at the point where x = 3.

Mod Edit: Posts merged, you can use 'Modify' to add to your previous post if it hasn't been answered yet

Hey! Welcome to the forums

For your first one, draw a quick sketch and you'll see that the area we want is in multiple pieces. If you're clever, you'll notice that the areas can be reordered to make it double the area on the left there (not the small chunk, the larger one). How do we find that area? We first need the limits, so solve \(sin{x}=\cos{x}\):



So those become the limits of our integration - We want DOUBLE the area between the curves from \(x=\frac{\pi}{4}\) to \(x=\frac{3\pi}{4}\). In this region, the sine curve is on top, so our area calculation looks like this:



Note that the alternative would be to find each area chunk individually, as is shown on that Wolfram Alpha link above, but this is slower - Try and be clever where you can

For your question on the normal, we need to apply the change of base formula to use the natural logarithm:



So therefore we find the first derivative and use that to find the gradient of the normal at \(x=3\):



Substitute \(x=3\) to find the y-coordinate as well:



Then use the two point formula:



I hope this helps

[bsdfjnlkasn]

Hey!
I was just wondering if using the t-substitutions for integration was common? I've only seen it a few times but don't know if it's actually in the syllabus considering that it isn't addressed in my textbook. If someone could please clarify whether this is an important type of question in trig exams I would really appreciate it! I'm not sure if I should be spending time learning it that's all 

[jamonwindeyer]

Hey!
I was just wondering if using the t-substitutions for integration was common? I've only seen it a few times but don't know if it's actually in the syllabus considering that it isn't addressed in my textbook. If someone could please clarify whether this is an important type of question in trig exams I would really appreciate it! I'm not sure if I should be spending time learning it that's all 

You could have to do it, i don't see why not, but it definitely isn't an important question type - I never got one and I never see it in Trial papers!

[teapancakes08]

Scouring around trial papers for Perms and Combs but also kinda took to the internet...

Found this one: There are 12 people to be seated around two circular tables, the first with seven chairs and second with five chairs. In how many ways can this be done?

If someone could guide me on how to get started that would be a huge help! (I keep getting anything but the answer )

[jamonwindeyer]

Scouring around trial papers for Perms and Combs but also kinda took to the internet...

Found this one: There are 12 people to be seated around two circular tables, the first with seven chairs and second with five chairs. In how many ways can this be done?

If someone could guide me on how to get started that would be a huge help! (I keep getting anything but the answer )

Hey! Can definitely give you a few points in the right direction! With these types of question, try and break it down!

First, pick who goes to which table. Think of it this way -  You are picking 7 people out of 12 to go to a new table, and just leaving the remaining at the 5-seater. How do you pick 7 elements from a group of 12, where order doesn't matter?

Then, arrange the people around the tables. Remember, we can arrange \(n\) things around a circular table in \((n-1)!\) ways, because it doesn't matter where the first person sits.

This might help you get that answer if it still doesn't quite work let me know and I'll show you how I'd do it and (hopefully) show you how to get the answer

[teapancakes08]

Hey! Can definitely give you a few points in the right direction! With these types of question, try and break it down!

First, pick who goes to which table. Think of it this way -  You are picking 7 people out of 12 to go to a new table, and just leaving the remaining at the 5-seater. How do you pick 7 elements from a group of 12, where order doesn't matter?

Then, arrange the people around the tables. Remember, we can arrange \(n\) things around a circular table in \((n-1)!\) ways, because it doesn't matter where the first person sits.

This might help you get that answer if it still doesn't quite work let me know and I'll show you how I'd do it and (hopefully) show you how to get the answer

Hmm...I understand that it's a combination and presumedly that would mean 12C7 and 12C5 would come into play...somewhere (probably wrong though...), and that if we were to arrange each seperate table it would be 6! and 4! respectively. But honestly, I have no clue on where to go next >.<

[RuiAce]

Hmm...I understand that it's a combination and presumedly that would mean 12C7 and 12C5 would come into play...somewhere (probably wrong though...), and that if we were to arrange each seperate table it would be 6! and 4! respectively. But honestly, I have no clue on where to go next >.<

Well firstly note that 12C7 = 12C5 from the symmetry of the binomial coefficient.

We only take one of them. Suppose we consider the table with 7 people first, then we choose 7 out of 12 to deal with. Which means for the second table, we only have 5 people to choose from, and we must choose all 5 people. This can obviously be done in only one way, which makes sense as 5C5 = 1

So the answer is 12C7*6! * 5C5*4!
Or if you choose for the second table first, it becomes 12C5*4! * 7C7*6!
Which gets you the same answer.

[K9810]

Hey!

Can you please help me with these questions?

Thanks