The Student Room Group

Higher Maths 2012

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Reply 120
Guys, any full explanation would be useful for this question. Heinemann Textbook EX15G Q4. All bases are 10

"The difference, n, in decibels, between two sound intensities is given by the formula n=10log(S1/S2), where S1 and S2 are the sound intensities in phons, S1>S2. Shuffling a pack of card has a sound intensity of 40 phons. If a clock alarm is 4 decibels greater than shuffling cards what is the sound intensity of the clock alarm?"

So tough. If anybody would be so kind as to give a detailed description I would be hugely grateful!
Reply 121
Original post by Bonzo10
Guys, any full explanation would be useful for this question. Heinemann Textbook EX15G Q4. All bases are 10

"The difference, n, in decibels, between two sound intensities is given by the formula n=10log(S1/S2), where S1 and S2 are the sound intensities in phons, S1>S2. Shuffling a pack of card has a sound intensity of 40 phons. If a clock alarm is 4 decibels greater than shuffling cards what is the sound intensity of the clock alarm?"

So tough. If anybody would be so kind as to give a detailed description I would be hugely grateful!


4=10log(S1/40)
0.4=log(S1) - log(40)
0.4 + 1.60 = log(S1)
10^2 = S1
S1 = 100

I think this works? :smile: Basically you substitute into the equation and then solve as normal.
Original post by LLL17
See in number 2? .. Why is it 1/3?


Original post by hollieeilloh
yeah I don't actually know why I just know you need to use these formulae


Well, differentiating e.g. y = sin(ax) using the chain rule. Let u = ax, then dy/du = cos(u) and du/dx = a which gives us dy/dx = a.cos(ax). To find the antiderivative, you reverse this process, so you divide.

EDIT: dy/dy should have been dy/du, obviously. Fixed.
(edited 11 years ago)
Original post by TheUnbeliever
Well, differentiating e.g. y = sin(ax) using the chain rule. Let u = ax, then dy/dy = cos(u) and du/dx = a which gives us dy/dx = a.cos(ax). To find the antiderivative, you reverse this process, so you divide.


Thank you!, that makes sense :smile:
Reply 124
Original post by (:emily.
4=10log(S1/40)
0.4=log(S1) - log(40)
0.4 + 1.60 = log(S1)
10^2 = S1
S1 = 100

I think this works? :smile: Basically you substitute into the equation and then solve as normal.


Thanks. Also, going to sound daft...2009 Paper 1 q11? Can't figure out why it's 3? Any explanations?
Reply 125
Original post by JaggySnake95
Well the pressure is off for me because I performed well in my 1st three exams and I will hopefully do well in Biology as well so I think I have As for those 4 subjects. So my aim is to get a B in maths so I could end up with AAAAB which is very good and be able to do advanced higher maths next year.

Although looking at last years paper it appears to me pretty easy. WAY easier than the two prelims we doneso hopefully it'll be okay. I've not really started my revision yet and it's my birthday today. :lol: I'll do a few hours today, same tomorrow and then ramp it up on Sunday.

Good Luck. :biggrin:


Biology should be a good paper, I'm quite looking forward to it :colone:
Ooooft AH maths :eek: why would you :tongue:
Have a good birthday and good luck with maths :smile:
Reply 126
Original post by Bonzo10
Thanks. Also, going to sound daft...2009 Paper 1 q11? Can't figure out why it's 3? Any explanations?


It is three because sinx+1=0 has one solution and 4sinx-root5=0 has two. Does this help or do you need actual numerical answers?
Reply 127
Original post by Bonzo10
Thanks. Also, going to sound daft...2009 Paper 1 q11? Can't figure out why it's 3? Any explanations?


I got stuck with that, but my teacher explained if root 9 = 3 then any number lower than root 9 the equation would have a solution

so (4 sin x /5) would have 2 solutions and (sinx + 1) would be a negative so do the cast diagram and and it will only have one solution cause the negative would not count as a value so in total 3.

Hope that makes sense :P
Reply 128
Original post by Bonzo10
Thanks. Also, going to sound daft...2009 Paper 1 q11? Can't figure out why it's 3? Any explanations?


sinx =(root5)/4 : 2 solutions
sinx = -1 : 1 solution
Reply 129
Original post by daisyclouds
I got stuck with that, but my teacher explained if root 9 = 3 then any number lower than root 9 the equation would have a solution

so (4 sin x /5) would have 2 solutions and (sinx + 1) would be a negative so do the cast diagram and and it will only have one solution cause the negative would not count as a value so in total 3.

Hope that makes sense :P


Thank you very much. It's gone over my head but that's only because ive been up since 6. I'll read again later. Thanks again.

Edit: I get it now. Totally neglected to draw a sketch of the graph. Thanks everyone.
(edited 11 years ago)
Reply 130
How would I do this question?

Solve log2(x + 3) + log2(x2 + 5x 4) = 3.
(worth 5marks)
The red type is mean to be a base number :smile:
I tried it but i couldn't work it out properly, any ideas?
Someone just posted on Twitter that it's %78 this year to get an A in Maths.. Please tell me this isn't true?
Reply 132
Original post by ScottishDan
Someone just posted on Twitter that it's %78 this year to get an A in Maths.. Please tell me this isn't true?


No one has even sat the paper yet, so they don't know how realistic the paper is, therefore surely they cannot judge the boundaries atm?
Original post by lapples
How would I do this question?

Solve log2(x + 3) + log2(x2 + 5x 4) = 3.
(worth 5marks)
The red type is mean to be a base number :smile:
I tried it but i couldn't work it out properly, any ideas?


Is the answer x=-4 and x=-5 ?
Reply 134
Original post by ScottishDan
Is the answer x=-4 and x=-5 ?


According to the answer scheme it is x=1 or x=-4 or x=-5 (Stated explicitly here) Then it says for the final mark it should be x=1 only :s-smilie:
Original post by lapples
Biology should be a good paper, I'm quite looking forward to it :colone:
Ooooft AH maths :eek: why would you :tongue:
Have a good birthday and good luck with maths :smile:


Thanks. Not much else to pick :L
Original post by lapples
According to the answer scheme it is x=1 or x=-4 or x=-5 (Stated explicitly here) Then it says for the final mark it should be x=1 only :s-smilie:


Oh yeah x=1 too! Basically I worked out the log equation by multiplying the brackets by one another, i then changed it to an exponential so brackets = 8 and then used synthetic division using 1 as a factor. :smile:
Reply 137
Original post by lapples
According to the answer scheme it is x=1 or x=-4 or x=-5 (Stated explicitly here) Then it says for the final mark it should be x=1 only :s-smilie:


If you have got to x= 1,-4,-5 what you have had to do is convert from logs to a quadratic equation. The solutions 1,-4,-5 are all correct for the quadratic but placing -4,-5 back into the original log equation, the first log ends up being log2a negative. The log graph doesn't cover any value of x equal to or less than 0 so both of these(-4,-5) cannot be correct, so only x=1
(edited 11 years ago)
Reply 138
Original post by ScottishDan
Oh yeah x=1 too! Basically I worked out the log equation by multiplying the brackets by one another, i then changed it to an exponential so brackets = 8 and then used synthetic division using 1 as a factor. :smile:


Ah right that makes sense :smile: the part a used synthetic division but i didn't think that was part of this bit
Thanks :smile: :smile:
Reply 139
Original post by benten17
If you have got to x= 1,-4,-5 what you have had to do is convert from logs to a quadratic equation. The solutions 1,-4,-5 are all correct for the quadratic but placing -4,-5 back into the original log equation, the first log ends up being log2a negative. The log graph doesn't cover any value of x equal to or less than 0 so both of these(-4,-5) cannot be correct, so only x=1


Thank you :smile:

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