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March 29, 2024, 10:20:44 am

Author Topic: 3U Maths Question Thread  (Read 1230427 times)  Share 

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Opengangs

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Re: 3U Maths Question Thread
« Reply #4080 on: May 27, 2019, 11:55:06 pm »
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Hey, thank you so much for your help but I am still confused as to why we are using x=1 when the point is (4,1) so shouldn't X be 4. I'm sorry


Thanks again for your help
The point is (4, 1) when we refer to the function
\[ y = f^{-1}(x). \]

If we were talking about \( y = f(x) \), then we use (1, 4). To see why this matters, take \( x = 1 \). Then \( f(1) = 1^3 + 3 = 4 \).

So IF we were talking about \( y = f(x) \), then we use the ordered pair (1, 4). Otherwise, you’d be absolutely right in saying that \( x = 4 \).

Minivasili

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Re: 3U Maths Question Thread
« Reply #4081 on: May 28, 2019, 12:35:28 am »
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Thank you so much for your help, you are a life saver

spnmox

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Re: 3U Maths Question Thread
« Reply #4082 on: June 01, 2019, 12:10:16 pm »
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Need help with this please!

If AC is a direct common tangent, prove that angle ABC = 90 degrees.

That extra tangent line I drew is supposed to be a hint I think.

RuiAce

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Re: 3U Maths Question Thread
« Reply #4083 on: June 01, 2019, 12:16:15 pm »
+2
Need help with this please!

If AC is a direct common tangent, prove that angle ABC = 90 degrees.

That extra tangent line I drew is supposed to be a hint I think.
Hint: Label \(\angle AMB = \theta\). Note that \(AM = BM\) and \( BM = CM\) since tangents to an external point are equal. You then have two isosceles triangles and you also know that \(\angle AMC = 180^\circ\)...

Shiv3n

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Re: 3U Maths Question Thread
« Reply #4084 on: June 03, 2019, 02:56:22 pm »
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Heyya! This is my first time posting so fingers crossed this is in the right place  :)
I've been struggling with two questions, one is a logarithm question and the other is a binomial theorem question.
"Solve the equation: log2(x^(2)-6x)=3+log2(1-x)"
"The nth term of (a + bx + cx^(2) + dx^(3) + ex^(4))(1-x)^(-5)    is    n^(4)x^(n-1). Determine the values of 'a' 'b' 'c' 'd' and 'e'.


Any help would be appreciated!!!
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Shiv3n

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Re: 3U Maths Question Thread
« Reply #4085 on: June 03, 2019, 04:35:06 pm »
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Heyya! This is my first time posting so fingers crossed this is in the right place  :)
I've been struggling with two questions, one is a logarithm question and the other is a binomial theorem question.
"Solve the equation: log2(x^(2)-6x)=3+log2(1-x)"
"The nth term of (a + bx + cx^(2) + dx^(3) + ex^(4))(1-x)^(-5)    is    n^(4)x^(n-1). Determine the values of 'a' 'b' 'c' 'd' and 'e'.


Any help would be appreciated!!!

Okay, I think I have solved the logarithm question (attached)  ;D  but I'm still unable to work something out for the binomial theorem question (also attached)  :'( .
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RuiAce

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Re: 3U Maths Question Thread
« Reply #4086 on: June 03, 2019, 06:23:00 pm »
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Heyya! This is my first time posting so fingers crossed this is in the right place  :)
I've been struggling with two questions, one is a logarithm question and the other is a binomial theorem question.
"Solve the equation: log2(x^(2)-6x)=3+log2(1-x)"
"The nth term of (a + bx + cx^(2) + dx^(3) + ex^(4))(1-x)^(-5)    is    n^(4)x^(n-1). Determine the values of 'a' 'b' 'c' 'd' and 'e'.


Any help would be appreciated!!!
Assuming that you meant  \( (a+bx+cx^2+dx^3+ex^4)(1-x)^{-5} \), i.e. \( \frac{a+bx+cx^2+dx^3+ex^4}{(1-x)^5} \) I will have to request a sanity check. No matter what polynomial long division is done, there will always be a \( \frac{C}{1-x} \) term present in the final result, which cannot be expressed in the form \(n^4 x^{n-1}\). (Unless we have some extra assumption that \( |x| < 1\) and thus can use geometric series.)

What is the source of this question?

Shiv3n

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Re: 3U Maths Question Thread
« Reply #4087 on: June 03, 2019, 08:53:06 pm »
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Assuming that you meant  \( (a+bx+cx^2+dx^3+ex^4)(1-x)^{-5} \), i.e. \( \frac{a+bx+cx^2+dx^3+ex^4}{(1-x)^5} \) I will have to request a sanity check. No matter what polynomial long division is done, there will always be a \( \frac{C}{1-x} \) term present in the final result, which cannot be expressed in the form \(n^4 x^{n-1}\). (Unless we have some extra assumption that \( |x| < 1\) and thus can use geometric series.)

What is the source of this question?

That is the correct question. This question is actually from a handout given to us in 3-unit, the teacher hasn't referenced the sources from which the questions are taken from. I do believe, we can go off that assumption and use a geometric series using n^(4)x^(n-1).
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RuiAce

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Re: 3U Maths Question Thread
« Reply #4088 on: June 03, 2019, 08:58:41 pm »
+1
That is the correct question. This question is actually from a handout given to us in 3-unit, the teacher hasn't referenced the sources from which the questions are taken from. I do believe, we can go off that assumption and use a geometric series using n^(4)x^(n-1).
If geometric series is along the right line of thinking, I'll have to assume that \(|x|<1\) and hence a limiting sum is defined. Are you willing to accept this assumption?

(Due to the numerator being restricted to a quartic polynomial, I'm not really convinced that a finite geometric sum will work.)

Edit: I'm now convinced that assuming \(|x| < 1\) must be necessary. If we let \(x\to \infty\), then the expression \( \frac{a+bx+cx^2+dx^3+ex^4}{(1-x)^5} \) goes all the way down to \(0\), but \(1+x+x^2+\dots + x^n\) will also tend to \(\infty\) for any positive integer \(n\). The outline in the method below will therefore incorporate the assumption \(|x|<1\) by default.
« Last Edit: June 03, 2019, 09:17:24 pm by RuiAce »

RuiAce

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Re: 3U Maths Question Thread
« Reply #4089 on: June 03, 2019, 09:18:20 pm »
+4
\[ \text{If we are prepared to take the assumption that }|x| < 1\\ \text{then we may resort to differentiation to obtain the desired answer.} \]
\[ \text{Note that differentiation of the infinite geometric series formula}\\ \text{(or any infinite series for that matter) is not well justified in the HSC}\\ \text{since there is no coverage on Taylor series at the HSC level.} \]
\[ \text{Nevertheless, the following computations are a classic way of forcing out}\\ \text{the end result. Note that due to the large number of derivatives}\\ \text{that would need to be typed up, I use computations from WolframAlpha.} \]
In practice, the quotient rule computations start getting a bit catastrophic after a while. Careful factorisation and use of chain rule is needed to avoid expanding nastily long binomial terms.

Note that whilst ridiculous for MX1, this is actually a fairly common technique in first year/second year university. It just so happens that we need to differentiate a lot.
_______________________________________________________________________
\[ \text{Begin by writing}\\ \frac{1}{1-x} = 1 + \sum_{n=1}^\infty x^n. \]
Note that the reason why the \(x^0\) term was pulled out is because it vanishes after the first derivative. Yet the following computations are rigged to ensure no other term vanishes.
\[ \text{Differentiating both sides gives}\\ \frac{1}{(1-x)^2} = \sum_{n=1}^\infty nx^{n-1}.\\ \text{But multiplying both sides now by }x\text{ gives}\\ \frac{x}{(1-x)^2} = \sum_{n=1}^\infty nx^n. \]
\[ \text{Differentiating both sides again now gives}\\ \frac{x+1}{(1-x)^3} = \sum_{n=1}^\infty n^2 x^{n-1}.\\ \text{Again multiplying both sides by }x\text{ gives}\\ \frac{x^2+x}{(1-x)^3} = \sum_{n=1}^\infty n^2 x^n. \]
Essentially the idea is to rinse and repeat.
\[ \text{Once again, differentiating,}\\ \frac{x^2+4x+1}{(1-x)^4} = \sum_{n=1}^\infty n^3 x^{n-1}.\\ \text{Multiplying }x\text{ to both sides yields}\\ \frac{x^3+4x^2+x}{(1-x)^4} = \sum_{n=1}^\infty n^3 x^n \]
\[ \text{Differentiating one last time gives}\\ \frac{x^3+11x^2+11x+1}{(1-x)^5} = \sum_{n=1}^\infty n^4 x^{n-1}. \]
\[ \text{Observe that the }RHS\text{ satisfies the criteria we require, i.e.}\\ \text{the }n\text{-th term of the sum, when expanded out,}\\ \text{will be }n^4 x^{n-1}. \]
\[ \text{Therefore equating coefficients,}\\ \begin{align*}a&=1\\ b&=11\\ c&=11\\ d&=1\\ e&=0 \end{align*}\]

Shiv3n

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Re: 3U Maths Question Thread
« Reply #4090 on: June 03, 2019, 09:54:02 pm »
0
\[ \text{If we are prepared to take the assumption that }|x| < 1\\ \text{then we may resort to differentiation to obtain the desired answer.} \]
\[ \text{Note that differentiation of the infinite geometric series formula}\\ \text{(or any infinite series for that matter) is not well justified in the HSC}\\ \text{since there is no coverage on Taylor series at the HSC level.} \]
\[ \text{Nevertheless, the following computations are a classic way of forcing out}\\ \text{the end result. Note that due to the large number of derivatives}\\ \text{that would need to be typed up, I use computations from WolframAlpha.} \]
In practice, the quotient rule computations start getting a bit catastrophic after a while. Careful factorisation and use of chain rule is needed to avoid expanding nastily long binomial terms.

Note that whilst ridiculous for MX1, this is actually a fairly common technique in first year/second year university. It just so happens that we need to differentiate a lot.
........
Wowzers! Thanks Rui!!!
I ended up doing it like this (working attached) before checking the answers against yours but I'm going to have a go at understanding how you went about solving the question because the way I did it seems a bit dodgy. Thanks again!  ;D
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RuiAce

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Re: 3U Maths Question Thread
« Reply #4091 on: June 03, 2019, 10:35:40 pm »
+1
Wowzers! Thanks Rui!!!
I ended up doing it like this (working attached) before checking the answers against yours but I'm going to have a go at understanding how you went about solving the question because the way I did it seems a bit dodgy. Thanks again!  ;D
I had a look and thinking about it your method should be fine (and perhaps easier :P). For some reason I discarded that idea when it first went through my head, but presumably that was prior to suspecting the need for geometric series. Just have to be careful when equating coefficients after all, because each time we go further down, we bump into more terms upon equating.

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Re: 3U Maths Question Thread
« Reply #4092 on: June 06, 2019, 08:24:33 pm »
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Hi all!

I need a little help on the circled questions attached.
For 13(b), I subbed in the value of 1 in the x of the velocity formula I got in part (a) but the answer was square root 45.

For the second question (sorry, I didn't circle it but the question I'm stuck on is 19(c) ), I can't seem to get the right answer.

Thank you!
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Re: 3U Maths Question Thread
« Reply #4093 on: June 07, 2019, 10:16:03 am »
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Can someone please help me find the volume of the first function (red) on the list when it is rotated around the x-axis from x=1/6 to x=1/3.

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Re: 3U Maths Question Thread
« Reply #4094 on: June 07, 2019, 04:35:07 pm »
+2
Hi all!

I need a little help on the circled questions attached.
For 13(b), I subbed in the value of 1 in the x of the velocity formula I got in part (a) but the answer was square root 45.

For the second question (sorry, I didn't circle it but the question I'm stuck on is 19(c) ), I can't seem to get the right answer.

Thank you!

Hey there,

You haven't attached question 13! But as for question 19C there are a few approaches. The easiest way I can think of is to try expressing the displacement in terms of a single trigonometric function, and here it's going to be sine or cosine. Then, differentiate the function with respect to t, and the coefficient will be your maximum velocity since the range of sine or cosine is -1<=f(x)<=1. Give it a go! If you don't understand or haven't used the method before, ask again :)

Hope this helps :)
 
Answer
On inspection, the max velocity should be (but tell me if it's wrong!! might be wrong :( )


Can someone please help me find the volume of the first function (red) on the list when it is rotated around the x-axis from x=1/6 to x=1/3.

Remember that the volume of a solid rotated around the x-axis is expressed by


Try evaluating this either using a u substitution (Hint: u=cos x) or the reverse chain rule (which is a load quicker if you can recognise it).

Hope this helps :)
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