Author Topic: 4U Maths Question Thread (Read 665237 times) Tweet Share

RuiAce · « **Reply #540 on:** October 05, 2016, 11:47:18 pm »

Quote from: nay103 on October 05, 2016, 11:33:26 pm

Hi! I'm a bit confused about this 2011 HSC question:
8(b) A bag contains seven balls numbered from 1 to 7. A ball is chosen at random
and its number is noted. The ball is then returned to the bag. This is done a total
of seven times.
(i) What is the probability that each ball is selected exactly once?
(ii) What is the probability that at least one ball is not selected?
(iii) What is the probability that exactly one of the balls is not selected?

I am fine with parts i and ii, but iii had me a bit confused, and I'm not really sure I quite understand the given answer. My answer was
7/7x6/7x5/7x4/7x3/7x2/7x6/7 -> i.e. probability of six balls being selected exactly once, then one of those six balls being selected again. I don't see what is incorrect with this, if you could explain that would be great!

$\text{Your approach just considers things step by step. It is a bit too neglecting of special factors.}\\ \text{Here are some explanations:}$
$\text{a) You don't explicitly state which one we discard. There are 7 numbers which we discard any one from.}\\ \text{b) The fact that one ball is repeated means that the order they're drawn can be switched around.}$
$\text{A correct approach would be achieved through just the formula}\\ \text{Probability}=\frac{\text{Favourable outcomes}}{\text{Total outcomes}}\\ \text{For part a), the favourable outcomes are }7!\text{ and total are }7^7$
$\text{For part c), we choose one ball to eliminate straight away. This is done in }\binom{7}{1}=7\text{ ways}\\ \text{Then, of the remaining 6 balls, we choose one to be selected again: }\binom{6}{1}=6\\ \text{Then, this is now just an arrangements problem. 7 objects, 2 repeated: }\frac{7!}{2!}\\ \text{So the total outcomes is }42\times \frac{7!}{2!}$

nay103 · « **Reply #541 on:** October 05, 2016, 11:53:00 pm »

Quote from: RuiAce on October 05, 2016, 11:47:18 pm

$\text{Your approach just considers things step by step. It is a bit too closed-minded.}\\ \text{The ball that gets repeated does not have to be chosen last specifically.}\\ \text{They can be chosen at the very start if we wanted that to happen.}$
$\text{Some possible other explanations are that a) the ordering of the balls is unimportant.}\\ \text{b) The fact that one ball is repeated means that the order they're drawn can be switched around.}$
$\text{A correct approach would be achieved through just the formula}\\ \text{Probability}=\frac{\text{Favourable outcomes}}{\text{Total outcomes}}\\ \text{For part a), the favourable outcomes are }7!\text{ and total are }7^7$
$\text{For part c), we choose one ball to eliminate straight away. This is done in }\binom{7}{1}=7\text{ ways}\\ \text{Then, of the remaining 6 balls, we choose one to be selected again: }\binom{6}{1}=6\\ \text{Then, this is now just an arrangements problem. 7 objects, 2 repeated: }\frac{7!}{2!}\\ \text{So the total outcomes is }42\times \frac{7!}{2!}$

Ah, this makes a lot more sense now. Thank you!

jamonwindeyer · « **Reply #542 on:** October 06, 2016, 12:08:43 am »

Quote from: RuiAce on October 05, 2016, 11:47:18 pm

$\text{Your approach just considers things step by step. It is a bit too neglecting of special factors.}\\ \text{Here are some explanations:}$

Love this question (and the amount of the questions that I chose appearing makes me think that I really smashed you guys in that MX1 Revision Lecture)

katherine123 · « **Reply #543 on:** October 06, 2016, 01:48:28 am »

how to do part iv)

RuiAce · « **Reply #544 on:** October 06, 2016, 08:22:18 am »

Quote from: katherine123 on October 06, 2016, 01:48:28 am

how to do part iv)

Evaluating:
P(0)=-8
P(1)=2
Hence P(x) must have some root that is real, between 0 and 1.

But there's more to it. Remember that a complex root comes with its complex conjugate. This is because the coefficients on P(x) are all real

Since there are 4 roots, we must have either:
No complex roots - Part iii says that this is impossible
2 complex roots
4 complex roots

So since we've just deduced that at least 1 of the roots is real, it can't have 4 complex roots either.

Hence there are exactly two complex roots.

massive · « **Reply #545 on:** October 06, 2016, 04:23:27 pm »

if 1<x<y , is x-y>0 ??

jakesilove · « **Reply #546 on:** October 06, 2016, 04:51:47 pm »

Quote from: massive on October 06, 2016, 04:23:27 pm

if 1<x<y , is x-y>0 ??

Clearly, y is greater than x, so x-y is going to be negative!

RuiAce · « **Reply #547 on:** October 06, 2016, 05:25:42 pm »

Quote from: massive on October 06, 2016, 04:23:27 pm

if 1<x<y , is x-y>0 ??

1 < x < y

So x < y
So x - y < 0

Yeah, like Jake said, it's not positive; it's negative.

massive · « **Reply #548 on:** October 06, 2016, 06:08:31 pm »

thanks guys, also how do you do the q attached.

the question asks to solve for x

RuiAce · « **Reply #549 on:** October 06, 2016, 06:19:08 pm »

Quote from: massive on October 06, 2016, 06:08:31 pm

thanks guys, also how do you do the q attached.

the question asks to solve for x

$\textbf{That is really nasty.}\\ \text{Note that }x\neq -5,2\\ \text{Remember to use the square of the denominator trick, unless you want 4 different cases.}$
$\text{Case 1: }x\ge -1\\ \begin{align*}\frac{|x+1|-1}{(5+x)(2-x)}&\ge 0\\ \frac{x}{(5+x)(2-x)} &\ge 0\\ x(5+x)(2-x)&\ge 0\\ x(x+5)(x-2)&\le 0\end{align*}$
$\text{From a free-hand sketch of }y=x(x+5)(x-2)\\ \text{We see that the solutions are }x < -5, 0 \le x < 2\\ \text{So since }x\ge -1\text{, combining the restriction we have }0 \le x < 2$
$\text{Case 2: }x < -1\\ \begin{align*}\frac{|x+1|-1}{(5+x)(2-x)}&\ge 0\\ \frac{-x-2}{(5+x)(2-x)} &\ge 0\\ (-x-2)(5+x)(2-x)&\ge 0\\ (x+2)(x+5)(x-2)&\ge 0\end{align*}$
$\text{From a free-hand sketch of }y=(x+2)(x+5)(x-2)\\ \text{We see that the solutions are }-5 < x \le -2, x \ge 2\\ \text{So since }x < -1\text{, combining the restriction we have }-5 < x \le -2$
$\text{So recollecting solutions, we have}\\ \boxed{-5 < x \le -2, 0 \le x < 2}$

massive · « **Reply #550 on:** October 06, 2016, 07:55:46 pm »

Quote from: RuiAce on October 06, 2016, 06:19:08 pm

$\textbf{That is really nasty.}\\ \text{Note that }x\neq -5,2\\ \text{Remember to use the square of the denominator trick, unless you want 4 different cases.}$
$\text{Case 1: }x\ge -1\\ \begin{align*}\frac{|x+1|-1}{(5+x)(2-x)}&\ge 0\\ \frac{x}{(5+x)(2-x)} &\ge 0\\ x(5+x)(2-x)&\ge 0\\ x(x+5)(x-2)&\le 0\end{align*}$
$\text{From a free-hand sketch of }y=x(x+5)(x-2)\\ \text{We see that the solutions are }x < -5, 0 \le x < 2\\ \text{So since }x\ge -1\text{, combining the restriction we have }0 \le x < 2$
$\text{Case 2: }x < -1\\ \begin{align*}\frac{|x+1|-1}{(5+x)(2-x)}&\ge 0\\ \frac{-x-2}{(5+x)(2-x)} &\ge 0\\ (-x-2)(5+x)(2-x)&\ge 0\\ (x+2)(x+5)(x-2)&\ge 0\end{align*}$
$\text{From a free-hand sketch of }y=(x+2)(x+5)(x-2)\\ \text{We see that the solutions are }-5 < x \le -2, x \ge 2\\ \text{So since }x < -1\text{, combining the restriction we have }-5 < x \le -2$
$\text{So recollecting solutions, we have}\\ \boxed{-5 < x \le -2, 0 \le x < 2}$

wow that was real neat. Thanks Rui. So whenever we have absolute values and we need to solve do we always use cases like you did?

RuiAce · « **Reply #551 on:** October 06, 2016, 09:01:46 pm »

Quote from: massive on October 06, 2016, 07:55:46 pm

wow that was real neat. Thanks Rui. So whenever we have absolute values and we need to solve do we always use cases like you did?

Yeah. Whenever a course of action isn't obvious, split the absolute values into cases.

nay103 · « **Reply #552 on:** October 06, 2016, 09:23:51 pm »

Back with another question!
I'm super confused about the last part. Doesn't it essentially mean having to prove the existence of the circle in the very question??
Thanks

RuiAce · « **Reply #553 on:** October 06, 2016, 10:02:09 pm »

$\textbf{1996 HSC 4U Additional}$

Quote from: nay103 on October 06, 2016, 09:23:51 pm

Back with another question!
I'm super confused about the last part. Doesn't it essentially mean having to prove the existence of the circle in the very question??
Thanks

$\text{Essentially, we are trying to prove the }\textit{converse}\text{ of (i)}\\ \text{This is a proof of uniqueness - The circle must be tangential to all 4 sides}\\ \text{Not just three sides and the fourth ignored.}$
i.e. If that circle is tangential to 3 sides, it must be tangential to the fourth as well. That's why the implication is reversed in comparison to part (i).
_________________________________________
$\text{We start by using part (ii), because it's the same quadrilateral.}\\ \text{According to part (ii), there will be a circle such that }\\ AB, BC\text{ and }CD\text{ are tangents of the circle.}\\ RTP:\, AD\text{ is also a tangent to this circle.}$
$\text{Somehow we should use part (b). Here's where the warning comes in.}\\ \textbf{Questions in the current HSC will not require this level of depth.}\\ \textbf{The ways of thought are now a bit more exceeding of an average student.}$

$\text{Suppose, by way of contradiction, }AD\text{ is not tangent.}\\ \text{Instead, suppose }AD^\prime\text{ is tangent, where }D^\prime\text{ either}\\ \text{lies on }CD\text{, or }CD\text{ produced.}$
$\text{From part b), if...}\\ D^\prime\text{ lies between C, D, then }DA-DC> D^\prime A - D^\prime C\\ D^\prime\text{ lies on it produced, then }D^\prime A - D^\prime C > DA - DC$
$\text{We are given}\\ AB+CD=AD+BC\tag{1}$
$\text{From part (i), however, because we SUPPOSED that }AD^\prime\text{ is tangent}\\ AB+CD^\prime = AD^\prime + BC\tag{2}$
$\text{Subtracting (1) and (2) we get }D^\prime A - D^\prime C = DA - DC\\ \text{This clearly contradicts with the above scenarios!}$
$\text{Hence, }D\text{ and }D^\prime\text{ must be the same point, effectively forcing }AD\text{ to be tangential.}\\ \text{Q.E.D.}$
Answers

katherine123 · « **Reply #554 on:** October 07, 2016, 01:34:23 am »

how to do part iii)

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