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March 29, 2024, 09:43:20 pm

Author Topic: 3U Maths Question Thread  (Read 1230631 times)  Share 

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RuiAce

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Re: 3U Maths Question Thread
« Reply #3600 on: August 18, 2018, 02:09:05 pm »
+1
is it the only way of doing this question?
since it is a mcq, could the process of elimination be used instead?
Yeah pretty much the above.

The only "other way" is to use the same method you'd use for something ugly like \( (2x-5)^{12} (x-4)^8\) where you keep extracting terms and find all the ways of building up terms involving \(x^2\). Which is honestly just overkill when the powers are this small.

Not every multiple choice question is designed so that there HAS to be a way to do it by process of elimination.

arii

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Re: 3U Maths Question Thread
« Reply #3601 on: August 18, 2018, 11:15:56 pm »
0
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arii

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Re: 3U Maths Question Thread
« Reply #3602 on: August 19, 2018, 02:40:09 am »
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Another question (Not sure how to do it):

Brine containing 2g of salt/L flows into a container initially filled with 50L of water containing 10g of salt. The brine enters the tank at 5 L/min, the concentration is kept uniform by stirring and the mixture flows out at the same rate. P(t) is the amount of salt (in grams) in the tank after t minutes. Find the amount of salt in the tank after 10 minutes.

Answer: 66.9g (1 d.p.)
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dermite

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Re: 3U Maths Question Thread
« Reply #3603 on: August 19, 2018, 11:42:40 am »
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hi there, i dont understand the bit in the blue box. if anyone can explain it to me, i'd greatly appreciate it.
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jamonwindeyer

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Re: 3U Maths Question Thread
« Reply #3604 on: August 19, 2018, 12:13:27 pm »
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.

Hey! So for this one you are using the chain rule. The result you want can be found as:



The first of those derivatives, \(\frac{dM}{dv}\), you can calculate by differentiating the expression you were given. The second derivative, \(\frac{dv}{dt}\), you are given! Then you can just substitute in \(0.6c\) for \(v\) into the result once you've multiplied them together! ;D

Another question (Not sure how to do it):

Brine containing 2g of salt/L flows into a container initially filled with 50L of water containing 10g of salt. The brine enters the tank at 5 L/min, the concentration is kept uniform by stirring and the mixture flows out at the same rate. P(t) is the amount of salt (in grams) in the tank after t minutes. Find the amount of salt in the tank after 10 minutes.

Answer: 66.9g (1 d.p.)

This one you can also approach with the change rule, but I prefer to think of it intuitively. So we know that 5L of water flows into the tank per minute, and that each litre contains 2g of salt. So, the salt going into the tank is 10g/min. At this stage:



However, we also have some amount going out of the tank. Let's call this \(\frac{dA}{dt}\), just for reference. At any given time, we know that the amount of water leaving the tank is 5L per minute. The concentration of salt in the tank is always \(\frac{p(t)}{50}\) litres, however much salt there is divided by 50. So the amount of salt leaving the tank is:



Put that together with what we have at the top, and we have the actual expression we want:




If we're clever, we notice that this matches the expression for exponential growth and/or decay. We can form this equation using our knowledge of that topic (let me know if you need this explained):



Now that we have it, just a matter of substituting \(t=10\):


jamonwindeyer

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Re: 3U Maths Question Thread
« Reply #3605 on: August 19, 2018, 12:17:36 pm »
+1
hi there, i dont understand the bit in the blue box. if anyone can explain it to me, i'd greatly appreciate it.

Hey! All that Is saying is if \(k\ge3\), then all of those inequalities will be true. \(4^{k-1}>k^2\) for sure, since \(4^{3-1}>3^2\) and exponentials grow faster than quadratics. \(k^2>1\) is clearly true for anything above \(k=1\), the same is true for \(2k^2>2k\). Therefore, if \(k\ge3\), then they are all true ;D

dermite

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Re: 3U Maths Question Thread
« Reply #3606 on: August 19, 2018, 03:44:45 pm »
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can someone help me with this question? thanks
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RuiAce

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Re: 3U Maths Question Thread
« Reply #3607 on: August 19, 2018, 04:08:09 pm »
+3
can someone help me with this question? thanks

\begin{align*} x^{\log_2 x} &= 8x^2\\ \log_2 \left(x^{\log_2 x} \right) &= \log_2 (8x^2)\\ (\log_2 x)(\log_2 x) &= 3 + 2\log_2 x\\ (\log_2 x)^2 - 2 \log_2 x - 3 &= 0\\ \left( \log_2 x - 3\right)\left(\log_2 x + 1\right)&=0\\ \log_2 x &= -1, 3\\ x&= \frac12, 8 \end{align*}

dermite

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Re: 3U Maths Question Thread
« Reply #3608 on: August 19, 2018, 04:18:44 pm »
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\begin{align*} x^{\log_2 x} &= 8x^2\\ \log_2 \left(x^{\log_2 x} \right) &= \log_2 (8x^2)\\ (\log_2 x)(\log_2 x) &= 3 + 2\log_2 x\\ (\log_2 x)^2 - 2 \log_2 x - 3 &= 0\\ \left( \log_2 x - 3\right)\left(\log_2 x + 1\right)&=0\\ \log_2 x &= -1, 3\\ x&= \frac12, 8 \end{align*}

how did you get from the 2nd line to the 3rd?
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RuiAce

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Re: 3U Maths Question Thread
« Reply #3609 on: August 19, 2018, 04:21:09 pm »
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how did you get from the 2nd line to the 3rd?
Log laws.
\begin{align*}\log a^m &= m \log a\\ \log (ab) &= \log a + \log b \end{align*}

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Re: 3U Maths Question Thread
« Reply #3610 on: August 20, 2018, 12:14:08 am »
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Hi! Some motion questions:

a) A particle is moving in SHM according to the equation: x = 2 - 4cos^2 (2t+ pi/3). In which interval does this particle oscillate?
(I thought it would be -2 to 6 but the answers say otherwise)

b) A particle in SHM satisfies the velocity-displacement equation: v^2= 12 + 4x- x^2. What is a possible displacement-time equation of the particle?
   answers include: x = 2 +- 4sin(t), x= 2+- 4cos(t), x = 2 +- 4cos(t + pi/3), and all of above

Thanks :)


dermite

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Re: 3U Maths Question Thread
« Reply #3611 on: August 20, 2018, 07:24:30 am »
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hi, i need some help with this projectile motion question:

A gun is aimed at a target on the ground 150m away. If the initial velocity is 125ms-1, find the angles at which the gun could be fired to reach the target (use g = 10ms-2).
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RuiAce

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Re: 3U Maths Question Thread
« Reply #3612 on: August 20, 2018, 09:00:23 am »
+1
Hi! Some motion questions:

a) A particle is moving in SHM according to the equation: x = 2 - 4cos^2 (2t+ pi/3). In which interval does this particle oscillate?
(I thought it would be -2 to 6 but the answers say otherwise)

b) A particle in SHM satisfies the velocity-displacement equation: v^2= 12 + 4x- x^2. What is a possible displacement-time equation of the particle?
   answers include: x = 2 +- 4sin(t), x= 2+- 4cos(t), x = 2 +- 4cos(t + pi/3), and all of above

Thanks :)


In a), you need to note the significance of the cos-squared. Decompose via the formula \( \cos^2 \theta = \frac12(1 + \cos 2\theta) \)
\begin{align*}x&= 2 - 4\cos^2 \left( 2t + \frac\pi3 \right)\\ &= 2 - 2\left[ 1+ \cos\left( 4t + \frac{2\pi}{3} \right)\right] \\ &= -2\cos\left( 4t + \frac\pi3 \right)\end{align*}
So -2 to 2.
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RuiAce

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Re: 3U Maths Question Thread
« Reply #3613 on: August 20, 2018, 09:11:05 am »
+3
hi, i need some help with this projectile motion question:

A gun is aimed at a target on the ground 150m away. If the initial velocity is 125ms-1, find the angles at which the gun could be fired to reach the target (use g = 10ms-2).



and then divide by 2.

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Re: 3U Maths Question Thread
« Reply #3614 on: August 21, 2018, 06:00:10 pm »
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Hi, I'm fairly slow at extension maths and struggle with time management in the exam. For trials there were quite a few questions I couldn't complete simply because of time (I think almost 10 marks worth). I usually get most things that I manage to complete correct, but I think that's because I take more time to do the questions. Especially for multiple choice, I find they take a long time to work out for just one mark. Does anyone else struggle with this, or have any tips to improve?
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