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March 29, 2024, 01:05:58 am

Author Topic: 4U Maths Question Thread  (Read 659826 times)  Share 

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RuiAce

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Re: 4U Maths Question Thread
« Reply #2295 on: June 02, 2019, 03:54:08 pm »
+2
Hey, I'm a bit stuck over this integration question.

Prove the integral of 0 to 1 (x^m)(1-x)^n dx = (m!n!)/(m+n+1)! Hint: think of binomial theorem.

I tried using integration by parts but got stuck.  I'm not sure how to use binomial theorem since we haven't really covered that in 3U yet, even though it looks like the binomial theorem formula from my textbook.
What textbook is this? This is usually proven via a double-recurrence formula which is outside of the scope of the MX2 course.

martinstran

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Re: 4U Maths Question Thread
« Reply #2296 on: June 03, 2019, 08:04:56 pm »
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Hey everyone, I'm quite stuck on this Harder 3U inequalities question:
aČ+bČ≥ab+a+b-1
I know that aČ+bČ≥2ab from the AM≥GM inequality, so I've attempted to prove that  2ab≥ab+a+b-1 hence, ab-a-b+1≥0, but I can't prove that either.
There are no restrictions on the values of a and b. Any help is much appreciated :)
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RuiAce

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Re: 4U Maths Question Thread
« Reply #2297 on: June 03, 2019, 08:53:58 pm »
+2
Hey everyone, I'm quite stuck on this Harder 3U inequalities question:
aČ+bČ≥ab+a+b-1
I know that aČ+bČ≥2ab from the AM≥GM inequality, so I've attempted to prove that  2ab≥ab+a+b-1 hence, ab-a-b+1≥0, but I can't prove that either.
There are no restrictions on the values of a and b. Any help is much appreciated :)
For this question I took on a bit of an unorthodox method. I'm not too sure at this stage how I'd approach it in a sense similar to AM-GM. For now I'll post what I've seen, but it may not be the intended approach.
\[ \text{In my approach, I first note that the statement is equivalent to}\\ a^2+b^2-ab-a-b+1 \geq 0.\\ \text{Which is equivalent to}\\ a^2-a(b+1) + (b^2-b+1) \geq 0. \]
\[ \text{Therefore I consider the quadratic expression}\\ a^2-(b+1)a+(b^2-b+1)\\ \text{as a quadratic in terms of }a. \]
Note that the quadratic could've been re-expressed in terms of \(b\), and the following proof should still work upon adapting.
\[ \text{The leading coefficient is 1, which is positive.}\\ \text{The discriminant of this quadratic is}\\ \begin{align*}\Delta &= [-(b+1)]^2 - 4(b^2-b+1)\\ &= b^2+2b+1 - 4b^2+4b-4\\ &= -3b^2+6b-3\\ &= -3(b-1)^2\\ &\leq 0\end{align*} \\ \text{for }\textbf{all}\text{ real }b.\]
\[ \text{Therefore the expression will never have two distinct real roots}\\ \text{for }a,\text{ in terms of }b.\\ \text{So since the leading coefficient is positive,}\\ \text{it follows that }a^2-(b+1)a+(b^2-b+1)\geq 0\text{ for all }a,b. \]
Which rearranges to produce the desired result.

david.wang28

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Re: 4U Maths Question Thread
« Reply #2298 on: June 11, 2019, 11:59:55 am »
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Hello,
I am stuck on Q 23 in the link below. I got the first part and attempted the second part, but I don't know where to go. Can anyone please help me out? Thanks :)
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david.wang28

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Re: 4U Maths Question Thread
« Reply #2299 on: June 11, 2019, 04:52:52 pm »
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Hello,
I have another question on 4u, this time on integration. The question is evaluate the integral ∫(cos6x*cos2x)dx. My working out is shown, but I can't seem to get the answer using substitution. Can anyone please help me out? Thanks :)
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RuiAce

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Re: 4U Maths Question Thread
« Reply #2300 on: June 11, 2019, 05:23:54 pm »
+2
Hello,
I have another question on 4u, this time on integration. The question is evaluate the integral ∫(cos6x*cos2x)dx. My working out is shown, but I can't seem to get the answer using substitution. Can anyone please help me out? Thanks :)
It's way easier to use the product-to-sum identities (which should be derived) for integrals of that form. Here, we derive:
\begin{align*}
\frac12 [\cos (A-B) + \cos(A+B)] &= \frac12[(\cos A \cos B + \sin A \sin B)-(\cos A \cos B -  \sin A \sin B)]\\
&= \cos A \cos B
\end{align*}
\begin{align*}
\therefore \int \cos 6x \cos 2x\,dx &= \frac12 \int \cos 4x + \cos 8x\,dx\\
&= \frac{\sin 4x}{8} + \frac{\sin 8x}{16}+C
\end{align*}
The method is outlined in my April lecture slides.

fun_jirachi

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Re: 4U Maths Question Thread
« Reply #2301 on: June 11, 2019, 05:25:10 pm »
+2
There's also a way without sums-to-products, albeit longer:



Hope this helps :)
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RuiAce

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Re: 4U Maths Question Thread
« Reply #2302 on: June 11, 2019, 10:40:01 pm »
+1
Hello,
I am stuck on Q 23 in the link below. I got the first part and attempted the second part, but I don't know where to go. Can anyone please help me out? Thanks :)
Just taking a look at this now:

With b) (i), it's basically relating back to where banked tracks originate from. The idea is that the velocity you're travelling at is faster than the velocity that gives you the optimal banking angle. The train therefore has a tendency to skid up the track.

To ensure this does not happen, the lateral force must bring the train somehow downwards. The lateral force must therefore be directed inwards to the centre of the circle of motion.

This can only be done by the rail that is further outwards. The outer rail effectively 'pushes' the train back inwards.

With b) (iii), if you draw your force diagram correctly you should now arrive at \(N\cos \theta -F\sin \theta = mg\) and \(N\sin\theta + F\cos\theta = \frac{mv^2}{500} \). Hint: Multiply both sides of the first equation by \(\sin\theta\), and both sides of the second by \(\cos\theta\)...

david.wang28

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Re: 4U Maths Question Thread
« Reply #2303 on: June 11, 2019, 10:45:05 pm »
0
Just taking a look at this now:

With b) (i), it's basically relating back to where banked tracks originate from. The idea is that the velocity you're travelling at is faster than the velocity that gives you the optimal banking angle. The train therefore has a tendency to skid up the track.

To ensure this does not happen, the lateral force must bring the train somehow downwards. The lateral force must therefore be directed inwards to the centre of the circle of motion.

This can only be done by the rail that is further outwards. The outer rail effectively 'pushes' the train back inwards.

With b) (iii), if you draw your force diagram correctly you should now arrive at \(N\cos \theta -F\sin \theta = mg\) and \(N\sin\theta + F\cos\theta = \frac{mv^2}{500} \). Hint: Multiply both sides of the first equation by \(\sin\theta\), and both sides of the second by \(\cos\theta\)...
Thank you Rui! :)
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Shaynell01

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Re: 4U Maths Question Thread
« Reply #2304 on: June 12, 2019, 03:25:18 pm »
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Can someone please help me rearrange to form an integratable expression.
cosinverse(pi*x) * sin(pi*x)

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Re: 4U Maths Question Thread
« Reply #2305 on: June 14, 2019, 03:16:32 pm »
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What textbook is this? This is usually proven via a double-recurrence formula which is outside of the scope of the MX2 course.

This was from tutoring but I asked both my tutor and my class teacher and they both say it's outside the scope, so I shouldn't worry about it!
I have some other questions though (from past assessments from my school).

1.  Explain why multiplying a complex number z by cistheta rotates z anticlockwise about the origin, through an angle of theta.
    I get the theory behind this, but I'm confused on how to set this out. Would it be best to let z= rcis(delta) and then use the multiplication of polar numbers:
    so you get that: z*cistheta= rcis(delta+theta)? But I'm still a bit confused on how to prove it shifts anticlockwise

2. Assume w is a nth root of infinity. Using geometric sums, prove that w+w^2+w^3+...w^n=0
    When I tried doing this, I got this: w(1-w^n)/(1-w)= w(1-w^n)/w(1/w-1)                           
                                                                   = (1-w^n)/(1/w-1)
             was I meant to use smth to do with the property 1+w+w^2=0??

3. If the mod of z=1, prove |a+bz|= |az+b|. You may assume a and b are elements of the Real numbers,.
What I did was:
     square both sides: |a+bz|^2= |az+b|^2 . [[I'm just going to let the conjugates = capitalised version of the letter to make it easier to read]
   (a+bz)(conjugate(a)+conjugate(bz))= (az+b)conjugate(az)+conjugate(b))
(a+bz)(A+BZ)= (az+b)(AZ+B)
aA+Abz+aBZ+BbZz=aAzZ+AZb+aBz+Bb (but Zz=1)
LHS= |a| +Abz+ aBZ+|b|
RHS= |a|+AZb+aBz +|b|

but these aren't the same? if Zz=1, isn't Z=1/z? So I'm not sure how to get the answer.

Thank you!!

RuiAce

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Re: 4U Maths Question Thread
« Reply #2306 on: June 14, 2019, 03:32:52 pm »
+2
This was from tutoring but I asked both my tutor and my class teacher and they both say it's outside the scope, so I shouldn't worry about it!
I have some other questions though (from past assessments from my school).

1.  Explain why multiplying a complex number z by cistheta rotates z anticlockwise about the origin, through an angle of theta.
    I get the theory behind this, but I'm confused on how to set this out. Would it be best to let z= rcis(delta) and then use the multiplication of polar numbers:
    so you get that: z*cistheta= rcis(delta+theta)? But I'm still a bit confused on how to prove it shifts anticlockwise

2. Assume w is a nth root of infinity. Using geometric sums, prove that w+w^2+w^3+...w^n=0
    When I tried doing this, I got this: w(1-w^n)/(1-w)= w(1-w^n)/w(1/w-1)                           
                                                                   = (1-w^n)/(1/w-1)
             was I meant to use smth to do with the property 1+w+w^2=0??

3. If the mod of z=1, prove |a+bz|= |az+b|. You may assume a and b are elements of the Real numbers,.
What I did was:
     square both sides: |a+bz|^2= |az+b|^2 . [[I'm just going to let the conjugates = capitalised version of the letter to make it easier to read]
   (a+bz)(conjugate(a)+conjugate(bz))= (az+b)conjugate(az)+conjugate(b))
(a+bz)(A+BZ)= (az+b)(AZ+B)
aA+Abz+aBZ+BbZz=aAzZ+AZb+aBz+Bb (but Zz=1)
LHS= |a| +Abz+ aBZ+|b|
RHS= |a|+AZb+aBz +|b|

but these aren't the same? if Zz=1, isn't Z=1/z? So I'm not sure how to get the answer.

Thank you!!

If they have an answer I'm genuinely curious to see tbh, because I can't see how the coefficients work nicely when the binomial theorem is used.
__________________________________________________

1.That is because the angle is taken with respect to the positive real axis (i.e. positive \(x\)-axis).

As the argument increases, the more we go counterclockwise, because it's always up and outwards from the positive \(x\)-axis. Or else you cannot, in some sense, make the 'angle' 'bigger'.

Clockwise shifts occur when the argument is decreased.

2. Because \(\omega\) is an nth root of unity (watch out for the typo), by definition it means that \( \boxed{\omega^n = 1} \). Which once you sub that in, the expression becomes 0.

3. Firstly, never assume what you're trying to prove, or else you will receive 0 marks. You've started from the final result and worked backwards from it, but that can only be done when you know the final result is true, not when you are trying to prove it.

To work around this issue, you should only work on one side of the equation at a time. Not both sides simultaneously.
\begin{align*} LHS&=|a+bz|^2 \\&= (a+bz)\overline{(a+bz)}\\ &= (a+bz)(a+b\overline{z}) \tag{since a and b are real}\\ &= a^2 + abz + ab\overline{z} + b^2z\overline{z}\\ &= a^2+ab(z+\overline{z})+b^2 \tag{given |z|=1} \end{align*}
\begin{align*} RHS&=|az+b|^2\\ &= (az+b)\overline{(az+b)}\\ &= (az+b)(a\overline{z}+b) \tag{again, a and b are real}\\ &= az\overline{z} + ab\overline{z}+abz+b^2\\ &= a|z|^2 +abz+ ab\overline{z} + b^2\\ &= a^2+ ab(z+\overline{z})+b^2\\ & =LHS \end{align*}
However it is worthwhile to note that yes, in general \(z^{-1} = \frac{\overline{z}}{|z|^2}\). But when \(|z|=1\), the denominator basically collapses, and we arrive at \( z^{-1}=\overline{z}\)

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Re: 4U Maths Question Thread
« Reply #2307 on: June 14, 2019, 08:58:26 pm »
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Took me ages bc Word crashed but here's my "latexed" version of their solution


RuiAce

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Re: 4U Maths Question Thread
« Reply #2308 on: June 14, 2019, 09:14:15 pm »
+1
Took me ages bc Word crashed but here's my "latexed" version of their solution


That's basically the double recurrence relation method actually. It's not the binomial theorem.

(Double recurrences are outside the scope of MX2 but as seen in that example, they're handled the same way as usual recurrence formulas. i.e. with integration by parts. But I feel as though the question is misleading to mention the binomial theorem in that case.)

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Re: 4U Maths Question Thread
« Reply #2309 on: June 15, 2019, 02:05:36 pm »
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I'm struggling with a volumes question help would be great thanks.

A hemisphere has radius r. By considering cross-sections parallel to the base of the hemisphere, show the volume is given by V=2/3(pi)r^3.