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January 20, 2017, 03:56:07 am

### AuthorTopic: 3U Maths Question Thread  (Read 40475 times) Tweet Share

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#### shreya_ajoshi

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##### Re: 3U Maths Question Thread
« Reply #1155 on: January 08, 2017, 11:45:46 am »
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How would you do this?
13.Ten points are chosen in a plane, not three of which are collinear.
a) How many lines can be drawn through pairs of the points?
b) How many triangles can be drawn if each of the vertices is at one of the given points
c) How many of the triangles have a particular point as one of the vertices?
d) How many of the triangles have two particular points making up one of the sides?

14.There are ten points in a plane, five of which are collinear. No other sets of three of these points is collinear.
a) How many sets of three points can be selected from those five that are collinear?
b) How many triangles can be formed using the ten points as vertices?

#### RuiAce

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##### Re: 3U Maths Question Thread
« Reply #1156 on: January 08, 2017, 12:00:15 pm »
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How would you do this?
13.Ten points are chosen in a plane, not three of which are collinear.
a) How many lines can be drawn through pairs of the points?
b) How many triangles can be drawn if each of the vertices is at one of the given points
c) How many of the triangles have a particular point as one of the vertices?
d) How many of the triangles have two particular points making up one of the sides?

14.There are ten points in a plane, five of which are collinear. No other sets of three of these points is collinear.
a) How many sets of three points can be selected from those five that are collinear?
b) How many triangles can be formed using the ten points as vertices?
I doubt that these were designed for MX1 students...
$\text{Because no 3 points are collinear, one line can't run through 3 points}\\ \text{Hence, every line must go through 2 and only 2 distinct points.}\\ \text{This is analogous to }\textbf{choosing}\text{ 2 points to draw the line through.}\\ \text{Hence the answer to a) is }\binom{10}2$
$\text{Similarly, for a triangle we must now have 3 distinct points}\\ \text{Hence the answer to b) is }\binom{10}3$
$\text{Part c) is ambiguous. I will have to assume that we are given the particular point}\\ \text{That one point must be included no matter what. Then we have 9 left}\\ \text{to which we must choose 2 of them. Hence the answer is }\binom92$
$\text{And then for d), well, 2 points gone leaves us 8 points left}\\ \text{to which we only choose 1 from. Hence the answer is }\binom81=8$
Hopefully I didn't mess up...
$\text{The answer to part a) is clearly just choosing 3 out of 5, i.e. }\binom53$
$\text{I believe that part a) is a precursor to part b).}\\ \text{We will first consider the number of ways of selecting}\\ \textbf{any 3 points out of the 10}.\\ \text{Which is, of course, }\binom{10}3$
$\text{So in our 10 points for part b), we can choose any 3 in }\binom{10}3\text{ ways.}\\ \text{But there is a twist - 5 of them are collinear!}\\ \text{Because 5 points are collinear, if we choose any 3 of them, they will NOT form a triangle}\\ \text{(instead, they will form a straight line)}$
$\text{So if we want to form TRIANGLES, and not STRAIGHT LINES out of 3 points}\\ \text{We simply must subtract it back out}\\ \text{Hence the answer is }\binom{10}3-\binom53$
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#### shreya_ajoshi

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##### Re: 3U Maths Question Thread
« Reply #1157 on: January 08, 2017, 12:30:51 pm »
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I doubt that these were designed for MX1 students...
$\text{Because no 3 points are collinear, one line can't run through 3 points}\\ \text{Hence, every line must go through 2 and only 2 distinct points.}\\ \text{This is analogous to }\textbf{choosing}\text{ 2 points to draw the line through.}\\ \text{Hence the answer to a) is }\binom{10}2$
$\text{Similarly, for a triangle we must now have 3 distinct points}\\ \text{Hence the answer to b) is }\binom{10}3$
$\text{Part c) is ambiguous. I will have to assume that we are given the particular point}\\ \text{That one point must be included no matter what. Then we have 9 left}\\ \text{to which we must choose 2 of them. Hence the answer is }\binom92$
$\text{And then for d), well, 2 points gone leaves us 8 points left}\\ \text{to which we only choose 1 from. Hence the answer is }\binom81=8$
Hopefully I didn't mess up...
$\text{The answer to part a) is clearly just choosing 3 out of 5, i.e. }\binom53$
$\text{I believe that part a) is a precursor to part b).}\\ \text{We will first consider the number of ways of selecting}\\ \textbf{any 3 points out of the 10}.\\ \text{Which is, of course, }\binom{10}3$
$\text{So in our 10 points for part b), we can choose any 3 in }\binom{10}3\text{ ways.}\\ \text{But there is a twist - 5 of them are collinear!}\\ \text{Because 5 points are collinear, if we choose any 3 of them, they will NOT form a triangle}\\ \text{(instead, they will form a straight line)}$
$\text{So if we want to form TRIANGLES, and not STRAIGHT LINES out of 3 points}\\ \text{We simply must subtract it back out}\\ \text{Hence the answer is }\binom{10}3-\binom53$

Haha thank you!! Would those type of questions pop up in a 3u exam?
I also  need help with probability questions.
5. From a standard pack of 52 cards, three are selected at random. Find that probability that:
a) They are the jack of spades, the two of clubs and the seven of diamonds
b) all 3 are aces
c) they are all diamonds
d) they are all of the same suit
e) they are picture cards
f) two are red and one is black
g) one is a seven, one is an eight and one is a nine
h) two are 7s and one is a 6
i) exactly one is a diamonds
j) at least two of them are diamonds

#### RuiAce

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##### Re: 3U Maths Question Thread
« Reply #1158 on: January 08, 2017, 01:18:28 pm »
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Haha thank you!! Would those type of questions pop up in a 3u exam?
I also  need help with probability questions.
5. From a standard pack of 52 cards, three are selected at random. Find that probability that:
a) They are the jack of spades, the two of clubs and the seven of diamonds
b) all 3 are aces
c) they are all diamonds
d) they are all of the same suit
e) they are picture cards
f) two are red and one is black
g) one is a seven, one is an eight and one is a nine
h) two are 7s and one is a 6
i) exactly one is a diamonds
j) at least two of them are diamonds
Do past papers.
___________________________________
$\text{Assuming that ordering does not matter}\\ \text{Note that for 52 cards, the total possible ways of choosing}\\ \text{any 3 cards is given }\binom{52}3\\ \text{This will always be the denominator from here on because}\\ \text{Probability of an event}=\frac{\text{Favourable outcomes}}{\text{Total outcomes}}\\ \text{The numerator will change for every part}$
$\text{a) Our 3 cards HAVE to be that}\\ \text{So the favourable outcomes is basically just the 1.}$
$\text{b) There are a total of 4 aces in the deck. And we need to choose 3 of them}\\ \text{So the favourable outcomes is }\binom43$
$\text{c) There are 13 possible diamonds to choose 3 from.}\\ \text{So the favourable outcomes is }\binom{13}3$
$\text{d) Pick any random suit, then 13 out of that suit.}\\ \binom{4}{1}\times \binom{13}3$
$\text{e) There are 12 picture cards (J, Q, K of any suit)}\\ \binom{12}3\\ \text{(unless A of spades is counted as a picture card, which would be crazy)}$
$\text{f) There are 26 red cards to choose 2 out of, and 26 to choose 1 black}\\ \binom{26}2\times \binom{26}1$
$\text{g) 4 7's, 4 8's, 4 9's to choose 1}\\ \binom{4}{1}^3$
$\text{h) As above}\\\binom{4}{2}\times \binom{4}{1}$
$\text{i) Choose the diamond out of 13, then rest to not be diamonds from the 39}\\ \binom{13}1\times \binom{39}2$
$\text{At least 2 being diamonds}\\ \text{is the same thing as NOT 0 or 1 are diamonds.}\\ \text{Exactly one diamond is done above in i)}\\ \text{If none are diamonds, well we must choose 3 from all the rest 39 which is }\binom{39}{3}\\ \text{So the answer to j) is }1-\frac{\binom{39}{3}+\binom{13}1\binom{39}2}{\binom{52}3}$
« Last Edit: January 08, 2017, 01:21:04 pm by RuiAce »
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#### Iminschool

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##### Re: 3U Maths Question Thread
« Reply #1159 on: January 09, 2017, 06:19:00 pm »
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For a friend:
2016: Mathematics 92
2017 aims: Physics 96
Chemistry 94
Economics 93
English Advanced 90

ATAR AIM: 98ish

#### RuiAce

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##### Re: 3U Maths Question Thread
« Reply #1160 on: January 09, 2017, 06:29:29 pm »
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For a friend:
$\text{By considering the division transformation}\\ \boxed{P(x)=D(x)Q(x)+R(x)}\\ \text{for some divisor }D(x)\text{, quotient }Q(x)\text{ and remainder }R(x)$
$\text{We have}\\ x^3+ax+b = (x-2)(x+3)Q(x) + 2x+1\\ \text{Note that the quotient is what we don't know.}\\ \text{So since we only want }a\text{ and }b\text{, we should throw out the quotient.}\\ \text{Easiest way of doing this is to just vanish it, i.e. make it equal 0}$
$\text{Note that by substituting }x=2\text{ and }x=-3,\text{ the quotient vanishes}\\ \text{This is because we force the divisor to equal 0.}$$\text{If we let }x=2\text{, we have}\\ 8+2a+b = 4+1\\\ \text{If we let }x=-3\text{, we have}\\ -27 -3a + b = -6+1\\ \text{These are now just simultaneous equations}$
$\text{You can check that the solutions are }a=-5, b=7$
« Last Edit: January 09, 2017, 06:31:06 pm by RuiAce »
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#### Iminschool

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##### Re: 3U Maths Question Thread
« Reply #1161 on: January 09, 2017, 06:40:31 pm »
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$\text{By considering the division transformation}\\ \boxed{P(x)=D(x)Q(x)+R(x)}\\ \text{for some divisor }D(x)\text{, quotient }Q(x)\text{ and remainder }R(x)$
$\text{We have}\\ x^3+ax+b = (x-2)(x+3)Q(x) + 2x+1\\ \text{Note that the quotient is what we don't know.}\\ \text{So since we only want }a\text{ and }b\text{, we should throw out the quotient.}\\ \text{Easiest way of doing this is to just vanish it, i.e. make it equal 0}$
$\text{Note that by substituting }x=2\text{ and }x=-3,\text{ the quotient vanishes}\\ \text{This is because we force the divisor to equal 0.}$$\text{If we let }x=2\text{, we have}\\ 8+2a+b = 4+1\\\ \text{If we let }x=-3\text{, we have}\\ -27 -3a + b = -6+1\\ \text{These are now just simultaneous equations}$
$\text{You can check that the solutions are }a=-5, b=7$

Much appreciated
2016: Mathematics 92
2017 aims: Physics 96
Chemistry 94
Economics 93
English Advanced 90

ATAR AIM: 98ish

#### anotherworld2b

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##### Re: 3U Maths Question Thread
« Reply #1162 on: January 09, 2017, 07:31:18 pm »
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Could i have help with this question? I am not sure how to tackle the multplie pieces of information provided

#### RuiAce

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##### Re: 3U Maths Question Thread
« Reply #1163 on: January 09, 2017, 07:39:35 pm »
+1
Could i have help with this question? I am not sure how to tackle the multplie pieces of information provided
They tell you stuff about the derivatives.

y' = 3ax^2+2bx+c

y" = 6ax + 2b

And then when x=-1:
y=4
y'=8
y"=-24

Just break it down. It's just tedious wording, so think about everything one at a time.

(After subbing, you should get simultaneous equations)
« Last Edit: January 09, 2017, 08:04:43 pm by RuiAce »
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#### anotherworld2b

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##### Re: 3U Maths Question Thread
« Reply #1164 on: January 10, 2017, 12:57:29 pm »
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I attempted to do q44 but i am not sure what to.
I also wanted to ask about differentiating q6. Apparently the answer in the book is 120x +17

They tell you stuff about the derivatives.

y' = 3ax^2+2bx+c

y" = 6ax + 2b

And then when x=-1:
y=4
y'=8
y"=-24

Just break it down. It's just tedious wording, so think about everything one at a time.

(After subbing, you should get simultaneous equations)

#### RuiAce

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##### Re: 3U Maths Question Thread
« Reply #1165 on: January 10, 2017, 01:07:30 pm »
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I attempted to do q44 but i am not sure what to.
I also wanted to ask about differentiating q6. Apparently the answer in the book is 120x +17
Your answer is the correct one for Q6. The book made a bizarre typo there.

With your simultaneous equations, use (1) and (3) to get rid of the c instead of the b
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#### anotherworld2b

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##### Re: 3U Maths Question Thread
« Reply #1166 on: January 10, 2017, 06:12:42 pm »
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I attempted to solve q44 again.
But i got the wrong answers again.
In the answers a= 5 , b=3, c=-1

#### RuiAce

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##### Re: 3U Maths Question Thread
« Reply #1167 on: January 10, 2017, 06:21:58 pm »
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I attempted to solve q44 again.
But i got the wrong answers again.
In the answers a= 5 , b=3, c=-1
Equation 3 is wrong

Should be -a+b-c+5 = 4.
Check the question again.

However putting your wrong equation into WolframAlpha shows that your final answer was correct given the mistake you made earlier on.
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#### anotherworld2b

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##### Re: 3U Maths Question Thread
« Reply #1168 on: January 10, 2017, 07:30:36 pm »
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thank you very much for answering all my questions RuiAce
I finally got it right
Equation 3 is wrong

Should be -a+b-c+5 = 4.
Check the question again.

However putting your wrong equation into WolframAlpha shows that your final answer was correct given the mistake you made earlier on.

#### anotherworld2b

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##### Re: 3U Maths Question Thread
« Reply #1169 on: January 10, 2017, 08:31:26 pm »
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Hi I am back  .
I've been working on this question back i am unable to get the answer for parts c, d and e