Author Topic: 4U Maths Question Thread (Read 672966 times) Tweet Share

RuiAce · « **Reply #615 on:** October 17, 2016, 10:11:55 pm »

Quote from: massive on October 17, 2016, 10:10:20 pm

how does this line: e^1/(n+1) < 1 + 1/n < e^1/n become this line: (1+1/n)ⁿ < e < (1+1/n)ⁿ⁺¹
Like in the solutions there's no other lines in between, so i was just wondering how they got that second line :S

Hint: Split the inequality up and work on two sides at once. Then recombine them.

massive · « **Reply #616 on:** October 17, 2016, 10:39:37 pm »

Quote from: RuiAce on October 17, 2016, 10:11:55 pm

Hint: Split the inequality up and work on two sides at once. Then recombine them.

genius, thanks i got it now!!

katherine123 · « **Reply #617 on:** October 18, 2016, 05:08:01 pm »

how to do part ii and iv

RuiAce · « **Reply #618 on:** October 18, 2016, 05:54:55 pm »

Quote from: katherine123 on October 18, 2016, 05:08:01 pm

how to do part ii and iv

$\text{For part (iv), note firstly that }p^3\ge 27r\text{ is failed as }8 < 27\\ \text{Hence the equation definitely does not have three distinct (positive) real roots.}$
$\text{Consider the derivative: }\frac{d}{dx}(x^3-2x^2+x-1=3x^2-4x+1)=(3x-1)(x-1)\\ \text{We do so because we have to investigate the stationary points to have a better feel for the graph.}$
$\text{Note that there are two stationary points, namely }\left(\frac{1}{3},-\frac{23}{27}\right)\text{ and }(1,-1)\\ \text{Both of these stationary points satisfy }x> 0\\ \text{This is significant.}$
$\text{Since the turning points of the graph all satisfy }x> 2\text{ we eliminate the possibility of 2 roots}\\ \text{Observe that if two roots were present, we would have one of them being a double root.}\\ \text{This is because our polynomial is of degree 3.}$
$\textit{A cubic either has 3 real, distinct roots, 1 double root and 1 single root, 1 triple root, or 1 real + 2 complex roots.}\\ \text{By eliminating the possibility of the first two cases, there must be 1 root.}$
$\text{Note that the root is definitely positive as }(1,-1)\text{ is below the }x\text{-axis}\\ \text{but }x^3-2x^2+x-1 \to \infty \text{ as }x\to \infty\\ \text{So the cubic does change sign somewhere positive.}$
Note that there are other ways of approaching this question. Please consult past answers wherever possible.

RuiAce · « **Reply #619 on:** October 18, 2016, 06:12:37 pm »

massive · « **Reply #620 on:** October 18, 2016, 06:51:28 pm »

guys how do you do part ii :S

Brenda0708 · « **Reply #621 on:** October 18, 2016, 07:12:17 pm »

Not sure how to do the first 2 questions. Thanks!

RuiAce · « **Reply #622 on:** October 18, 2016, 07:19:32 pm »

Quote from: massive on October 18, 2016, 06:51:28 pm

guys how do you do part ii :S

$\text{Similarly, there exists integers }a_n,b_n\text{ such that}\\ (1-\sqrt{2})^n=a_n-b_n\sqrt{2}$
$\textit{Do you see why?}\\ \text{The proof for part (i) without induction comes from the binomial theorem.}\\ \text{All the minus sign does on the binomial expansion is flip a ton of signs.}\\ \text{Hence the same }a_n\text{ and }b_n\text{ are shared.}$
$\text{If you are unconvinced, try expanding}\\ (1+\sqrt{2})^1, (1-\sqrt{2})^1\\ (1+\sqrt{2})^2, (1-\sqrt{2})^2\\ \text{etc.}$
$\text{Hence multiplying these results:}\\ \begin{align*}(1+\sqrt2)^n(1-\sqrt2)^n&=(a_n+b_n\sqrt2)(a_n-b_n\sqrt2)\\ ((1+\sqrt2)(1-\sqrt2))^n&=a_n^2-2b_n^2\\ a_n^2-2b_n^2&=(-1)^n\end{align*}\\ \text{via difference of two squares}$

RuiAce · « **Reply #623 on:** October 18, 2016, 07:36:27 pm »

Quote from: Brenda0708 on October 18, 2016, 07:12:17 pm

Not sure how to do the first 2 questions. Thanks!

$\text{The total possible outcomes in every case is }31\times28\times20\times25=434000\\ \text{as was correctly stated.}$
$\text{Part b) is considerably annoying. Both methods of analysis require more advanced counting techniques.}$
$\text{By the inclusion-exclusion principle, note that the favourable outcomes are:}\\ \text{No. of ways they have at least same blouse + No. of ways they have at least same skirt}\\ -\text{Number of ways they have both the same}\\ \textit{This is the bane of combinatorics - Double counting. Note that the events are not mutually exclusive.}$
$\text{Case 1: Same blouse. 22 blouses to assign.}\\ \text{Then their skirts don't matter.}\\ \text{So we have }22\times 20 \times 25=11000$
$\text{Case 2: Same skirt. 15 skirts to assign.}\\ \text{Then their blouses don't matter.}\\ \text{So we have }15\times 31\times 28=13020$
$\text{Hence our favourable outcomes equals:}\\ 11000+13020-330=23960$
$\text{Giving us a probability of }\frac{23960}{434000}$
I don't see how you're mistaken for part a) - There are only 22*15 favourable outcomes. Can you please post up the answers in the future? I'm not confident with combinatorics.

massive · « **Reply #624 on:** October 18, 2016, 07:58:30 pm »

guys what do the thetas in conics represent??

RuiAce · « **Reply #625 on:** October 18, 2016, 09:00:58 pm »

Quote from: massive on October 18, 2016, 07:58:30 pm

guys what do the thetas in conics represent??

The angle it makes with the x-axis at the origin, just like with a circle

massive · « **Reply #626 on:** October 18, 2016, 11:50:04 pm »

guys how do you do the attached q

ans: 8!/(4!x4!)

RuiAce · « **Reply #627 on:** October 18, 2016, 11:57:48 pm »

Quote from: massive on October 18, 2016, 11:50:04 pm

guys how do you do the attached q

ans: 8!/(4!x4!)

$\text{Note that to get from }A\text{ to }X\text{ without turning back}\\ \text{means that you can only go up, or right.}\\ \text{Never down or left.}$
$\text{This imposes a restriction.}\\ \text{Because you have to get from one end to the other, you MUST have:}\\ \text{4 ups}\\ \text{4 rights}$
$\text{i.e. a possible combination is }UUUURRRR$
$\text{It turns out that all arrangements have 4 ups and 4 rights.}\\ \text{So we are actually just trying to rearrange the letters of }UUUURRRR\text{ to make 'words'}$
$\text{Which is done in }\frac{8!}{4!4!}\text{ ways}$

massive · « **Reply #628 on:** October 19, 2016, 12:02:11 am »

Quote from: RuiAce on October 18, 2016, 11:57:48 pm

$\text{Note that to get from }A\text{ to }X\text{ without turning back}\\ \text{means that you can only go up, or right.}\\ \text{Never down or left.}$
$\text{This imposes a restriction.}\\ \text{Because you have to get from one end to the other, you MUST have:}\\ \text{4 ups}\\ \text{4 rights}$
$\text{i.e. a possible combination is }UUUURRRR$
$\text{It turns out that all arrangements have 4 ups and 4 rights.}\\ \text{So we are actually just trying to rearrange the letters of }UUUURRRR\text{ to make 'words'}$
$\text{Which is done in }\frac{8!}{4!4!}\text{ ways}$

DAMMNN, thanks fam!

massive · « **Reply #629 on:** October 19, 2016, 01:28:02 pm »

Hey guys whats the general solution to the following qs: (I put more than one coz they're pretty simple.. I think)

1) sin(x-pi/3) = cos2x

2) sin3x + sinx = 0

3) tan2x + cot3x =0

Also is the general solution for cosx = 0 , x= (2n+1)pi OR +/-pi/2 + 2npi ?

Thanks heaps for your help

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