Author Topic: 3U Maths Question Thread (Read 1509869 times) Tweet Share

FallonXay · « **Reply #990 on:** October 25, 2016, 08:06:16 pm »

And I don't understand how to do this question either.

Thanks in advance~

MysteryMarker · « **Reply #991 on:** October 25, 2016, 08:16:49 pm »

Quote from: RuiAce on October 25, 2016, 07:39:57 pm

Let me fix up my typo; a dash was missing there

Oh, could you just go over why it gets times by g'(x)? Sorry, i just find this very confusing.

RuiAce · « **Reply #992 on:** October 25, 2016, 08:21:07 pm »

Quote from: FallonXay on October 25, 2016, 08:06:16 pm

And I don't understand how to do this question either.

Thanks in advance~

Already addressed in post #199

Quote from: MysteryMarker on October 25, 2016, 08:16:49 pm

Oh, could you just go over why it gets times by g'(x)? Sorry, i just find this very confusing.

Chain rule, like I said. Derivative of whole thing times derivative of inside.

massive · « **Reply #993 on:** October 25, 2016, 08:29:14 pm »

hey guys how do you part ii, the solutions say that barbara is infront for exactly half the total arrangements(=8!)
therefore the answer is 8!/2
How do you even know that?

RuiAce · « **Reply #994 on:** October 25, 2016, 08:30:25 pm »

Quote from: massive on October 25, 2016, 08:29:14 pm

hey guys how do you part ii, the solutions say that barbara is infront for exactly half the total arrangements(=8!)
therefore the answer is 8!/2
How do you even know that?

You accidentally posted the image to a 4U question you asked already

FallonXay · « **Reply #995 on:** October 25, 2016, 08:33:01 pm »

Quote from: RuiAce on October 25, 2016, 08:21:07 pm

Already addressed in post #199

How do you do part i/ii tho

massive · « **Reply #996 on:** October 25, 2016, 08:37:16 pm »

Quote from: RuiAce on October 25, 2016, 08:30:25 pm

You accidentally posted the image to a 4U question you asked already

my bad attached the right pic now

RuiAce · « **Reply #997 on:** October 25, 2016, 08:41:40 pm »

Quote from: FallonXay on October 25, 2016, 08:33:01 pm

How do you do part i/ii tho

$\text{Break part (i) down.}\\ \text{A must win on game 7, so that's fixed.}\\ \text{Probability of that happening is }\frac12$
$\text{And then the rest requires a bit of intuition.}\\ \text{Observe that either }A\text{ wins, or }B\text{ wins.}\\ \text{This is binomial.}$
$\text{If a 7th game is to happen, then there must've been 6 games already with no outcome.}\\ \text{Note that therefore }A\text{ must've won 4 games.}\\ \text{Leaving 2 games won by }B\\ \text{So using our binomial probability formula, this is }\binom{6}{4}\left(\frac12\right)^4\left(\frac12\right)^2$
$\text{Multiply it all together to get }\binom64\left(\frac12\right)^7$
I'll let you try ii now. The idea is to consider the cases: Wins on 7th go (done above), 6th go or 5th go

FallonXay · « **Reply #998 on:** October 25, 2016, 08:46:47 pm »

Quote from: RuiAce on October 25, 2016, 08:41:40 pm

$\text{Break part (i) down.}\\ \text{A must win on game 7, so that's fixed.}\\ \text{Probability of that happening is }\frac12$
$\text{And then the rest requires a bit of intuition.}\\ \text{Observe that either }A\text{ wins, or }B\text{ wins.}\\ \text{This is binomial.}$
$\text{If a 7th game is to happen, then there must've been 6 games already with no outcome.}\\ \text{Note that therefore }A\text{ must've won 4 games.}\\ \text{Leaving 2 games won by }B\\ \text{So using our binomial probability formula, this is }\binom{6}{4}\left(\frac12\right)^4\left(\frac12\right)^2$
$\text{Multiply it all together to get }\binom64\left(\frac12\right)^7$
I'll let you try ii now. The idea is to consider the cases: Wins on 7th go (done above), 6th go or 5th go

Why is it a must win on the 7th go?

FallonXay · « **Reply #999 on:** October 25, 2016, 08:47:38 pm »

Quote from: FallonXay on October 25, 2016, 08:46:47 pm

Why is it a must win on the 7th go?

Ohhhhhhh wait, nvm. Was reading the question the wrong way :p

thanks!!!!

:3 · « **Reply #1000 on:** October 25, 2016, 09:11:21 pm »

I am so sorry AN maths gurus, more questions.

How would we go about answering iii) (and these types of binomial questions in general)?

katherine123 · « **Reply #1001 on:** October 25, 2016, 09:11:49 pm »

how to do part iii)

RuiAce · « **Reply #1002 on:** October 25, 2016, 09:14:07 pm »

Quote from: :3 on October 25, 2016, 09:11:21 pm

I am so sorry AN maths gurus, more questions.

How would we go about answering iii) (and these types of binomial questions in general)?
(Image removed from quote.)

That one is an outlier for its spiked difficulty. Already addressed in post #184.

levendibigd · « **Reply #1003 on:** October 25, 2016, 09:18:21 pm »

Could I please have some help with this question. Q.Prove that (2sin^3(x) +2cos^3(x))/ (sinx+cosx) = 2 - sin2x, where sinx+cosx cannot = 0

RuiAce · « **Reply #1004 on:** October 25, 2016, 09:20:25 pm »

Quote from: levendibigd on October 25, 2016, 09:18:21 pm

Could I please have some help with this question. Q.Prove that (2sin^3(x) +2cos^3(x))/ (sinx+cosx) = 2 - sin2x, where sinx+cosx cannot = 0

Hint: Factorise out the 2 in the numerator first, but then use the factorisation of a³+b³

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