4U Maths Question Thread

[RuiAce]

I'm not quite sure, my substitute teacher gave us a bunch of maths questions to do, saying her son said was appropriate for us (I think some of it is uni maths since there were some she marked wasn't in our syllabus). I thought it was just vector addition like in physics, but I'm not sure how to go about it

I'm not convinced that they understand the current 4U syllabus. It's actually doable within the boundaries of the new MX1 syllabus but not with any of the current syllabuses. Although if you're really interested in a solution, you can post it in the first year uni maths question thread and I'm happy to address it there.

It requires slightly more than the tip-to-tail stuff you learn in physics.

[david.wang28]

Hello,
I have partially done the working out, but I don't know what to do after. Can anyone please help me with this question? Thanks

[RuiAce]

Hello,
I have partially done the working out, but I don't know what to do after. Can anyone please help me with this question? Thanks

Follow the usual partial fractions approach. Sub \(x=1\), then \(x=2\) and then \(x=3\) to find your coefficients \(A\), \(B\) and \(C\).

[david.wang28]

Follow the usual partial fractions approach. Sub \(x=1\), then \(x=2\) and then \(x=3\) to find your coefficients \(A\), \(B\) and \(C\).

Thank you!

[veronicaaat]

Does anyone know how to attempt the answer to q19 and 20?
Thank you so much !

[RuiAce]

Does anyone know how to attempt the answer to q19 and 20?
Thank you so much !

Considering you've posted two potentially long questions, what attempts have you made so far? Please provide any understanding relevant.

[goodluck]

Hi, I have an induction question that I'm a bit stuck on

Let A1=1 and for n>1, let A(n+1) (the n+1 is a subscript sorry if it's not clear) = sqrt(1+A(n)). let G=(1+sqrt(5))/2

a) Prove that if A(n) < G, then A(n+1) <G
b) Prove that for all n, A(n+1) > A(n)

I'm a bit confused on whether I should use induction for the first part or for the second part, or whether there's an easier way of doing the question?

Thanks for your help!

[RuiAce]

Hi, I have an induction question that I'm a bit stuck on

Let A1=1 and for n>1, let A(n+1) (the n+1 is a subscript sorry if it's not clear) = sqrt(1+A(n)). let G=(1+sqrt(5))/2

a) Prove that if A(n) < G, then A(n+1) <G
b) Prove that for all n, A(n+1) > A(n)

I'm a bit confused on whether I should use induction for the first part or for the second part, or whether there's an easier way of doing the question?

Thanks for your help!

You can actually use induction for both. However, the key thing to remember for both parts is that if \(n \geq 1\), then by definition, \(\boxed{A_{n+1} = \sqrt{1 + A_n}} \). Note - You wrote \(n >1 \), however because of how you defined it I'm actually convinced that you should've written \(n \geq 1\). Because otherwise we cannot compute what \(A_2\) is.
\[ \text{For the second one, we can check that when }n=1,\\ A_2 = \sqrt{1+A_1} = \sqrt{1+1} = \sqrt{2} > 1 = A_1\\ \text{as required.} \]
\[ \text{Now assume that for any integer }k\geq 1,\\ \boxed{A_k > A_{k-1}} \]
\[ \text{Then we have}\\ \begin{align*} A_{k+1} &= \sqrt{1+A_k} \tag{by definition}\\ &>\sqrt{1+A_{k-1}} \tag{inductive assumption}\\ &= A_k \tag{by definition} \end{align*}\\ \text{and hence the statement holds when }n=k+1\text{ as well.}\]
Note that we had used the definition twice.
________________________________________________________
\[ \text{For the first, we can check that when }n=1,\\ A_1 = 1 = \frac{1+\sqrt1}{2} < \frac{1+\sqrt5}{2} = G\\ \text{as required.}\]
\[ \text{Now assume that for any integer }k\geq 1,\\ \boxed{A_k < G}\]
\[ \text{Then we have}\\ \begin{align*} A_{k+1} &= \sqrt{1+A_k}\tag{by definition}\\ &< \sqrt{1+G}\tag{inductive assumption}\\ &= \sqrt{1+\frac{1+\sqrt5}{2}} \\ &= \sqrt{\frac{3+\sqrt5}{2}}\\ &= \sqrt{\frac{6+2\sqrt5}{4}}\\ &= \sqrt{\frac{1+2\sqrt5+5}{4}} \tag{completing the square}\\ &= \sqrt{\frac{(1+\sqrt5)^2}4} \\ &= \frac{1+\sqrt5}{2}\\ &= G\end{align*}\\ \text{as required, having taken positive root}\\ \text{noting that }\frac{1+\sqrt5}{2}\text{ is positive.}\]

[goodluck]

You can actually use induction for both. However, the key thing to remember for both parts is that if \(n \geq 1\), then by definition, \(\boxed{A_{n+1} = \sqrt{1 + A_n}} \). Note - You wrote \(n >1 \), however because of how you defined it I'm actually convinced that you should've written \(n \geq 1\). Because otherwise we cannot compute what \(A_2\) is.
\[ \text{For the second one, we can check that when }n=1,\\ A_2 = \sqrt{1+A_1} = \sqrt{1+1} = \sqrt{2} > 1 = A_1\\ \text{as required.} \]
\[ \text{Now assume that for any integer }k\geq 1,\\ \boxed{A_k > A_{k-1}} \]
\[ \text{Then we have}\\ \begin{align*} A_{k+1} &= \sqrt{1+A_k} \tag{by definition}\\ &>\sqrt{1+A_{k-1}} \tag{inductive assumption}\\ &= A_k \tag{by definition} \end{align*}\\ \text{and hence the statement holds when }n=k+1\text{ as well.}\]
Note that we had used the definition twice.
________________________________________________________
\[ \text{For the first, we can check that when }n=1,\\ A_1 = 1 = \frac{1+\sqrt1}{2} < \frac{1+\sqrt5}{2} = G\\ \text{as required.}\]
\[ \text{Now assume that for any integer }k\geq 1,\\ \boxed{A_k < G}\]
\[ \text{Then we have}\\ \begin{align*} A_{k+1} &= \sqrt{1+A_k}\tag{by definition}\\ &< \sqrt{1+G}\tag{inductive assumption}\\ &= \sqrt{1+\frac{1+\sqrt5}{2}} \\ &= \sqrt{\frac{3+\sqrt5}{2}}\\ &= \sqrt{\frac{6+2\sqrt5}{4}}\\ &= \sqrt{\frac{1+2\sqrt5+5}{4}} \tag{completing the square}\\ &= \sqrt{\frac{(1+\sqrt5)^2}4} \\ &= \frac{1+\sqrt5}{2}\\ &= G\end{align*}\\ \text{as required, having taken positive root}\\ \text{noting that }\frac{1+\sqrt5}{2}\text{ is positive.}\]

I'll ask the teacher about the greater than and equal to sign, since the greater than sign was what was written on our worksheet. But if it's any comfort, I completely agree

I have another from a tutoring centre and I'm not sure whether it's just a bit advanced for the course or where it fits in exactly into harder 3u (i think it's applications of inequalities)

Considering 1/x (take x>0), explain why 1/n+1 < the integral from (n+1) to n of dx/x < 1/n
As the integral can be simplified to ln(1 +1/x), do we just graph the intervals to prove that it works (I checked on desmos and it fits nicely)

b) hence or otherwise, deduce (1+1/n)^n < e < (1 + 1/n)^(n+1)
         Do we separate the two sides from part a) and then raise it to e (e.g. from 1+1/k to e^ (1/k+1)) and multiply by the two sides and somehow manipulate the result? I'm not sure on how to get the lone e

c) this one is an extension question but it goes like this:
((1+n)^n)/(e^n) <n! < ((1+n)^(1+n))/(e^n)

Would that be relevant for this course or not?

[RuiAce]

I'll ask the teacher about the greater than and equal to sign, since the greater than sign was what was written on our worksheet. But if it's any comfort, I completely agree

I have another from a tutoring centre and I'm not sure whether it's just a bit advanced for the course or where it fits in exactly into harder 3u (i think it's applications of inequalities)

Considering 1/x (take x>0), explain why 1/n+1 < the integral from (n+1) to n of dx/x < 1/n
As the integral can be simplified to ln(1 +1/x), do we just graph the intervals to prove that it works (I checked on desmos and it fits nicely)

b) hence or otherwise, deduce (1+1/n)^n < e < (1 + 1/n)^(n+1)
         Do we separate the two sides from part a) and then raise it to e (e.g. from 1+1/k to e^ (1/k+1)) and multiply by the two sides and somehow manipulate the result? I'm not sure on how to get the lone e

c) this one is an extension question but it goes like this:
((1+n)^n)/(e^n) <n! < ((1+n)^(1+n))/(e^n)

Would that be relevant for this course or not?

Like yeah it's just harder 3U. Since you posted it on the 4U thread that's fine. I'm gonna swap out all the \(n\)'s for \(k\)'s, except for in the last part.

Yeah so for part i), just sketch the right branch of \(y= \frac1x\) and plot the points \( \left(k, \frac1k\right) \) and \(\left( k+1, \frac{1}{k+1} \right)\) on it. Note that the area under \(y = \frac{1}{x}\) from \(k\) to \(k+1\) has area given by \( \int_k^{k+1} \frac{1}{x}\,dx \). Note that a rectangle with length \(1\) and breadth \( \frac{1}{k+1} \) can be drawn right below it, so comparing areas we have \( \frac{1}{k+1} < \int_k^{k+1} \frac{1}{x}\,dx\). The other part is done similarly.

And yep. Upon evaluating, \( \int_k^{k+1} \frac{1}{x}\,dx = \ln \left( 1 + \frac1{k}\right)\), so you can rearrange by chopping the three-sided inequality into 2 two-sided inequalities first.
\[ \text{With your third part, note that rewriting the inequality in part ii) we have}\\ \boxed{ \left(\frac{k+1}{k}\right)^k < e < \left(\frac{k+1}{k}\right)^{k+1} } \]
\[ \text{Now observe that by subbing in }k=1,2,3,\dots, n\text{ we obtain}\\ \left(\frac21\right)^1 < e < \left(\frac21\right)^2\\ \left(\frac32\right)^2 < e < \left(\frac32\right)^2\\ \left( \frac43\right)^2 < e < \left(\frac43\right)^3\\ \vdots\\ \left( \frac{n+1}{n}\right)^n < e < \left(\frac{n+1}{n}\right)^{n+1}\]
\[ \text{Upon multiplying all of that out, we obtain}\\ \boxed{\left(\frac21\right)^1 \left(\frac32\right)^2 \left( \frac43\right)^3 \dots \left( \frac{n+1}{n}\right)^n < e^n < \left( \frac21\right)^2\left(\frac32\right)^3\left(\frac43\right)^4\dots\left(\frac{n+1}n\right)^{n+1}}\]
\[ \text{On the LHS, observe that}\\ 2^1\text{ in the top cancels out with }2^2\text{ in the bottom to yield }\frac12.\\ 3^2\text{ in the top cancels out with }3^3\text{ in the bottom to yield }\frac13\\ \text{and this cancellation process continues. Ultimately}\\ \text{we're just left with }\frac{(n+1)^n}{n!}\text{ after doing all the cancellations.} \]
Note that the \( (n+1)^n\) in the top has nothing in the bottom to be cancelled out with.
\[ \text{An essentially similar thing occurs on the RHS and leaves us }\frac{(n+1)^{n+1}}{n!}\\ \text{So the whole thing boils down to}\\ \boxed{\frac{(n+1)^n}{n!} < e^n < \frac{(n+1)^{n+1}}{n!}}.\]
This immediately rearranges to produce the required result.

[Fatemah.S]

Hey!
The question I have attached is finding volumes using cylindrical/shell method. I am able to get the diagram and what I think is the right volume integration. However, I am struggling to progress with it.  (6.2 Question 7 Cambridge)

[RuiAce]

Hey!
The question I have attached is finding volumes using cylindrical/shell method. I am able to get the diagram and what I think is the right volume integration. However, I am struggling to progress with it.  (6.2 Question 7 Cambridge)

Your integral should look something like \(V = 2\pi \int_0^2 2x\sqrt{1-(x-1)^2}\,dx\). For this integral, first sub \(u=x-1\). Then, split the integral up, and use the area of a semi-circle and properties of odd functions to help out.

[david.wang28]

Hello,
I am stuck on one integration by parts question in the link below (my working out is below as well). Can anyone please help me with this question? Thanks

[fun_jirachi]

Skipped a few steps, but happy to explain further Hope this helps

[david.wang28]

Skipped a few steps, but happy to explain further Hope this helps

I don't understand how you got from the second step to the third step.