Author Topic: 4U Maths Question Thread (Read 669404 times) Tweet Share

RuiAce · « **Reply #645 on:** October 19, 2016, 11:57:23 pm »

Quote from: massive on October 19, 2016, 11:54:46 pm

Howdu figure that out :O

Since there are 6 intervals, consider the difference in the boundaries and divide by 6

(π-0)/6 = π/6

Hence, start at 0 and increase by π/6, six times. It's actually a standard trick that 2U kids need to know, although Simpson's rule is, of course, extremely rare in 4U

massive · « **Reply #646 on:** October 20, 2016, 12:41:53 am »

Quote from: RuiAce on October 19, 2016, 11:57:23 pm

Since there are 6 intervals, consider the difference in the boundaries and divide by 6

(π-0)/6 = π/6

Hence, start at 0 and increase by π/6, six times. It's actually a standard trick that 2U kids need to know, although Simpson's rule is, of course, extremely rare in 4U

that's so cool, thanks for sharing!

massive · « **Reply #647 on:** October 20, 2016, 12:43:36 am »

hey quickly, is the q attached wrong? because cos and sin aren't for hyperbolas, or can this q still be done?

amandali · « **Reply #648 on:** October 20, 2016, 02:13:25 am »

for answer to part c (below) is it correct to have dotted line for the outer line of circle and the vertical line

how to do part b

RuiAce · « **Reply #649 on:** October 20, 2016, 08:13:29 am »

Quote from: massive on October 20, 2016, 12:43:36 am

hey quickly, is the q attached wrong? because cos and sin aren't for hyperbolas, or can this q still be done?

It can still be done. It's just extensively unconventional.

RuiAce · « **Reply #650 on:** October 20, 2016, 08:15:18 am »

Quote from: amandali on October 20, 2016, 02:13:25 am

(Image removed from quote.)
for answer to part c (below) is it correct to have dotted line for the outer line of circle and the vertical line

(Image removed from quote.)

how to do part b

(Image removed from quote.)

Technically yes, the dotted line is correct for c). But I never drew it like that in the HSC. I just kept it blocked because of the ≤.

It's only technically correct because the outer regions are useless. But I believe the HSC will be more lenient on this one if you just draw it blocked and not dotted.

RuiAce · « **Reply #651 on:** October 20, 2016, 08:24:48 am »

Quote from: amandali on October 20, 2016, 02:13:25 am

(Image removed from quote.)
for answer to part c (below) is it correct to have dotted line for the outer line of circle and the vertical line

(Image removed from quote.)

how to do part b

(Image removed from quote.)

b):
$\text{For (i), it becomes a bit too tedious to do a case by case analysis}\\ \text{because there are too many cases.}\\ \text{When case by case analysis seems too exhaustive, attempt the complement.}$
$\text{The complement, which is what we }\textit{don't}\text{ want, is the}\\ \text{number of arrangements such that one of the rooms will be empty.}\\ \text{Note that given two rooms, you can only have one empty or no empty.}$
$\text{For the total outcomes, we just note that the }n\text{ people can go in any of the 2 rooms.}\\ \text{This is achieved in }2^n \text{ ways.}$
$\text{For the unfavourable outcomes, note that there's actually only 2 ways one room can be empty.}\\ \text{It means that either everyone was in room A, or everyone was in room B}.$
$\text{So the favourable outcomes is }2^n-2$
__________________
$\text{Once again, the analysis would be exhaustive so we turn to the complement.}\\ \text{The total outcomes, in a similar way, is just }3^5$
$\text{The unfavourable outcomes are a bit more tricky now.}\\ \text{What's unfavourable is the number of ways 1 and/or 2 rooms are empty.}$
$\text{If two rooms are empty, that's simple enough. There are only 3 ways that can happen.}$
$\text{It's harder when there's only one room.}\\\text{We need to choose two rooms to not be empty: }\binom{3}{2}$
$\text{Once we have 2 (of 3) rooms, this is now applying part a)}\\ \text{Reason being that both rooms must be empty or we have overlap.}\\ \text{Hence we have }2^5-2\text{ to multiply to }\binom{3}{2}$
$\text{So the answer is }3^5-\left[3+\binom{3}{2}(2^5-2)\right]$

That's the MANSW book. That already has all the answers you need. If it's explaining an answer that doesn't make sense you're more than welcome to post that up as well but that should be all you need.

Brenda0708 · « **Reply #652 on:** October 20, 2016, 12:18:24 pm »

Hi, is there a way to do ii) without finding the equation of the ellipse? Is the locus saying the distance of z from the origin to (1+3i) + distance of z from the origin to (9+3i) = 10? I.e. PS + PS' = 10? For so then wouldn't the origin be a point on the ellipse?

RuiAce · « **Reply #653 on:** October 20, 2016, 12:20:04 pm »

Quote from: Brenda0708 on October 20, 2016, 12:18:24 pm

(Image removed from quote.)(Image removed from quote.)

Hi, is there a way to do ii) without finding the equation of the ellipse? Is the locus saying the distance of z from the origin to (1+3i) + distance of z from the origin to (9+3i) = 10? I.e. PS + PS' = 10? For so then wouldn't the origin be a point on the ellipse?

(1+3i) and (9+3i) are the foci.

S and S' do NOT lie on the x-axis. So the origin doesn't have to play a role here.

And of course, PS+PS' = 10 implies that the length of the major axis is 10 (hence the length of the semi-major axis is 5)

So your interpretation is correct. But your interpretation does not mean that the origin has to lie on the ellipse.
_______________________
$\text{To check: When }P\text{ is the origin }O(0,0)\\ |(0+0i)-1-3i| + |(0+0i)-9-3i| \neq 10$

massive · « **Reply #654 on:** October 20, 2016, 12:40:16 pm »

guys is the equation for asymptotes y=+/-b/a x right?, but how come the answer to this q is +/4/3 x not +/-3/4 x :S. In conics does 'a' ALWAYS represent the number under x, i always thought a was the bigger number and b was the smaller no.

RuiAce · « **Reply #655 on:** October 20, 2016, 12:41:49 pm »

Quote from: massive on October 20, 2016, 12:40:16 pm

guys is the equation for asymptotes y=+/-b/a x right?, but how come the answer to this q is +/4/3 x not +/-3/4 x :S. In conics does 'a' ALWAYS represent the number under x, i always thought a was the bigger number and b was the smaller no.

$\text{The RHS has to be }1.\\ \frac{x^2}{9}-\frac{y^2}{16}=1$
$\text{So since the hyperbola is shaped like }> \, <\\ \text{The asymptotes are }y=\pm \frac{4}{3}x$

massive · « **Reply #656 on:** October 20, 2016, 01:15:27 pm »

Quote from: RuiAce on October 20, 2016, 12:41:49 pm

$\text{The RHS has to be }1.\\ \frac{x^2}{9}-\frac{y^2}{16}=1$
$\text{So since the hyperbola is shaped like }> \, <\\ \text{The asymptotes are }y=\pm \frac{4}{3}x$

wait so is 'a' always the one under x and 'b' the one under y? like when you use b^2=a^2(e^21), B^2 = 16 and a^2=9 right?

RuiAce · « **Reply #657 on:** October 20, 2016, 01:19:46 pm »

Quote from: massive on October 20, 2016, 01:15:27 pm

wait so is 'a' always the one under x and 'b' the one under y? like when you use b^2=a^2(e^21), B^2 = 16 and a^2=9 right?

For x^2/a^2-y^2/b^2=1 yeah

massive · « **Reply #658 on:** October 20, 2016, 01:44:44 pm »

Quote from: RuiAce on October 20, 2016, 01:19:46 pm

For x^2/a^2-y^2/b^2=1 yeah

is it the same for ellipses as well?

jakesilove · « **Reply #659 on:** October 20, 2016, 02:00:18 pm »

Quote from: massive on October 20, 2016, 01:44:44 pm

is it the same for ellipses as well?

Hey! Looks like you're mixing up your formulas a bit; I would take a look at this seriously comprehensive document HERE and come back if you have any more questions!

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Author Topic: 4U Maths Question Thread (Read 669404 times) Tweet Share

RuiAce

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