# Edexcel A2 C3 Mathematics 12th June 2015Watch

This discussion is closed.
3 years ago
#561
I don't get the final question on C3 june 2014, I thought the minimum value on a sin graph is -1 which occurs at 3/2pi? Surely if the minimum value is -1, and you are stretching it by 100 then adding 4 the minimum becomes -96?!
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3 years ago
#562
(Original post by Dylann)
I don't get the final question on C3 june 2014, I thought the minimum value on a sin graph is -1 which occurs at 3/2pi? Surely if the minimum value is -1, and you are stretching it by 100 then adding 4 the minimum becomes -96?!
The sine function inside the brackets has been squared in the question, which means H(ϑ) cannot be negative. Therefore, the smallest value it could be would be 0, since all the negative values would become positives.

Then you equate H(ϑ) = 4 +5(0) = 4, and solve normally.
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3 years ago
#563
Whoops my question was too hard for edexcel students never mind.

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3 years ago
#564
(Original post by Supergirlxxxxxx)
Whoops my question was too hard for edexcel students never mind.

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Cause it's a c4 question this is a c3 thread

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3 years ago
#565
(Original post by Supergirlxxxxxx)
With the binomial shown in my notes, which validity would I choose? Because you work out the validity of both expansions but overall have to choose one, how do I work out which one it is? Attachment 406349

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regardless of the fact that this is a C3 thread; for that validity question, you're meant to chose the value that satisfies both of the equations/statements/validity binomial equations.

In short, your answer would be: (1+4x) ^ -1.

this is because, the modulus 'x' value in that equation is 0.25; which is less than the other modulus 'x' value (which is: 0.3) and as such, it satisfies both equations.

if you're still confused:
the (1+4x) ^ -1 equation has: *'modulus x modulus' = 1/4*. this means that it has a numerical x value (on an x-axis in a graph) that starts from: -0.25 to 0.25.

the other one, the (1+3x) ^ -1 equation has: *'modulus x modulus' = 1/3*. this means that it has a numerical x value (on an x-axis in a graph) that goes from: -0.33 to 0.33.

therefore i say go for the (1+4x) ^ -1, as that value goes from -0.25 to 0.25. that is, these are x values that would appear in both x-axis ranges. you wouldn't choose the other equation as,

is that clear or have i made a complete fool of myself either with the working out or with the explanation? :P
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3 years ago
#566
(Original post by kprime2)
The sine function inside the brackets has been squared in the question, which means H(ϑ) cannot be negative. Therefore, the smallest value it could be would be 0, since all the negative values would become positives.

Then you equate H(ϑ) = 0, and solve normally.
Bloody marvelous! Makes sense, cheers
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3 years ago
#567
(Original post by theDanIdentity)
regardless of the fact that this is a C3 thread; for that validity question, you're meant to chose the value that satisfies both of the equations/statements/validity binomial equations.

In short, your answer would be: (1+4x) ^ -1.

this is because, the modulus 'x' value in that equation is 0.25; which is less than the other modulus 'x' value (which is: 0.3) and as such, it satisfies both equations.

if you're still confused:
the (1+4x) ^ -1 equation has: *'modulus x modulus' = 1/4*. this means that it has a numerical x value (on an x-axis in a graph) that starts from: -0.25 to 0.25.

the other one, the (1+3x) ^ -1 equation has: *'modulus x modulus' = 1/3*. this means that it has a numerical x value (on an x-axis in a graph) that goes from: -0.33 to 0.33.

therefore i say go for the (1+4x) ^ -1, as that value goes from -0.25 to 0.25. that is, these are x values that would appear in both x-axis ranges. you wouldn't choose the other equation as,

is that clear or have i made a complete fool of myself either with the working out or with the explanation? :P
Lol I know, i don't even do edexcel, I do OCR(mei) but couldn't find a thread for it! And that made so much sense, thankyou, your amazing

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3 years ago
#568
help 5c
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3 years ago
#569
(Original post by Supermanxxxxxx)
Cause it's a c4 question this is a c3 thread

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Lol back up your own gf maybe, douche.

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3 years ago
#570
(Original post by kprime2)
The sine function inside the brackets has been squared in the question, which means H(ϑ) cannot be negative. Therefore, the smallest value it could be would be 0, since all the negative values would become positives.

Then you equate H(ϑ) = 0, and solve normally.

you my sir, are a genius. my entire class (teacher included); spent about 20-30 mins on this particular question; and to this day, are unsure about an exact way of solving it other than praying to God that it doesn't come up in the exams; and committing Japanese samurai ritual if it does.

thank you So much..
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3 years ago
#571
(Original post by theDanIdentity)
you my sir, are a genius. my entire class (teacher included); spent about 20-30 mins on this particular question; and to this day, are unsure about an exact way of solving it other than praying to God that it doesn't come up in the exams; and committing Japanese samurai ritual if it does.

thank you So much..
seriously? you guys need a new teacher haha
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3 years ago
#572
(Original post by broconomist)
seriously? you guys need a new teacher haha

looooool.. it's a bit late for that.. :P

besides, she's leaving anyway so there's that; and our class generally worked better without the teacher being there (we were forced to - half the time she was 'teaching', she wasn't in the ******* class -_-)
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3 years ago
#573
(Original post by studentwiz)
help 5c
You know what your A value has to be. Can you choose a suitable b, bearing in mind what they've asked in the question?
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3 years ago
#574
(Original post by SeanFM)
You know what your A value has to be. Can you choose a suitable b, bearing in mind what they've asked in the question?
ive done cos(60+15)/sin(60+15) but get the wrong answer
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3 years ago
#575
(Original post by studentwiz)
ive done cos(60+15)/sin(60+15) but get the wrong answer
I don't think you can have cos on top - it doesn't look like you've used part b.

You know on the LHS of the equation you need A = 75, because whatever B is, on the RHS you will have cotA, which is cot75.

Now you just need a suitable value for b, so that some of the terms like cos(A-B) and sin(A-B) = cos60 and sin60 respectively...
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3 years ago
#576
(Original post by SeanFM)
I don't think you can have cos on top - it doesn't look like you've used part b.

You know on the LHS of the equation you need A = 75, because whatever B is, on the RHS you will have cotA, which is cot75.

Now you just need a suitable value for b, so that some of the terms like cos(A-B) and sin(A-B) = cos60 and sin60 respectively...
thanks !
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3 years ago
#577
(Original post by theDanIdentity)
you my sir, are a genius. my entire class (teacher included); spent about 20-30 mins on this particular question; and to this day, are unsure about an exact way of solving it other than praying to God that it doesn't come up in the exams; and committing Japanese samurai ritual if it does.

thank you So much..
20 minutes haha really?
And btw my previous comment should have said 'H(ϑ) = 4 then solve normally'

no worries
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3 years ago
#578
(Original post by Dylann)
Bloody marvelous! Makes sense, cheers
haha no problem. Btw my previous comment should have said 'equate H(ϑ) = 4 then solve normally' if that makes sense.
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3 years ago
#579
(Original post by kprime2)
20 minutes haha really?
And btw my previous comment should have said 'H(ϑ) = 4 then solve normally'

no worries

no joke.. i still remember the white board screen as we all tried to make sense of the disgusting question; and the repulsive, depressed feeling one gets when they (& the whole damn class) can't do a question.
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3 years ago
#580
(Original post by theDanIdentity)
no joke.. i still remember the white board screen as we all tried to make sense of the disgusting question; and the repulsive, depressed feeling one gets when they (& the whole damn class) can't do a question.
haha I get it if the students can't do it, but I find it worrying that the teacher couldn't do it either!
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