Never mind I'm wrong  wish I'd had time to check!!!
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Further Maths GCSE AQA Paper 1 (Unofficial mark scheme)
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 21
 20062016 16:08

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 22
 20062016 16:10
(Original post by Torticci)
Can someone explain the limiting factor one and how it was 3/5? I though it was 2.4 cos then the denominator would equal 0 and the fraction would be undefined
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Start by dividing everything by the highest power of n (in this case it was just n)
You'd have got 3/5+(12/n). As n increases, 12/n gets smaller and smaller (closer and closer to 0) so it can be ignored.
Hence 3/5.Post rating:1 
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 23
 20062016 16:11
(Original post by ErinMei)
That is just an A* distinction i think depending on grade boundaries. I think i got 55 marks at the least
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(Original post by Redcoats)
Which question was the x = 55 ?
Solve for x
(Original post by NiamhM1801)
For the inequality, t asked for integer solutions. Wouldn't you therefore have had to put 9, 10 and 11?
Other than that and question 6 (which I knew I'd completely messed up) I think I've done very well
(Original post by NiamhM1801)
Oh, also I didn't get +75x for the cubic, I got +25x. I'm probably wrong though haha 
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 24
 20062016 16:12
(Original post by NiamhM1801)
Never mind I'm wrong  wish I'd had time to check!!! 
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 25
 20062016 16:13
(Original post by Torticci)
Can someone explain the limiting factor one and how it was 3/5? I though it was 2.4 cos then the denominator would equal 0 and the fraction would be undefined
Posted from TSR Mobile
Factor out n : n(3)/ n(5+12/n)
Cancel out the n's: 3 / (5 + 12/n)
Substitute infinity: 3/ (5 + 12 / infinity)
12 / infinity tends towards 0 so: 3 / 5 + 0
Therefore the limiting value is 3 / 5 
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 26
 20062016 16:14
(Original post by Chittesh14)
I thought you got like 65 or something xD!
It was on the question where you had the brackets and then to the power of a third.
Solve for x
Yeah, I read the question wrong... I'm so struggling these days with just reading questions correctly .
It was (x5)^3 I think and the answer is +75x so I think you got it wrong .Last edited by Redcoats; 20062016 at 16:16. 
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 27
 20062016 16:15
(Original post by Chittesh14)
I thought you got like 65 or something xD!
It was on the question where you had the brackets and then to the power of a third.
Solve for x
Yeah, I read the question wrong... I'm so struggling these days with just reading questions correctly .
It was (x5)^3 I think and the answer is +75x so I think you got it wrong .(Original post by Chittesh14)
Don't worry, I thought I was running out of time  so I rushed it  only to realise I had 30 or so minutes to spare.
And I know, our clock didn't get reset so it finished on 2:46, really hard to judge your timings like that! And at least it wasn't the other way round! 
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 28
 20062016 16:16
(Original post by Redcoats)
Pretty sure it was x = 45 for the power of a third question : (3 sqrt(x))^1/3 = 2 or something 
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 29
 20062016 16:16
Thanks for the limiting factor help!
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 30
 20062016 16:18
(Original post by Redcoats)
Pretty sure it was x = 121 for the power of a third question : (3 sqrt(x))^1/3 = 2 or something
All I remember is 9  64 = x lol. 
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 31
 20062016 16:19
(Original post by NiamhM1801)
I definitely got 55, not sure about the positive or negative though.
(3  sqrt(x))^{1/3} = 2
3  sqrt(x) = 8
 sqrt(x) = 11
sqrt(x) = 11
Therefore x = 11^{2}
So x = 121 
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 32
 20062016 16:21
(Original post by Redcoats)
Sorry not 45, x = 121. Check for yourself, it works!
(3  sqrt(x))^{1/3} = 2
3  sqrt(x) = 8
 sqrt(x) = 11
sqrt(x) = 11
Therefore x = 11^{2}
So x = 121 
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 33
 20062016 16:22
(Original post by Chittesh14)
No idea man, we can only tell once we know the exact question again .
All I remember is 9  64 = x lol.
(3  sqrt(x))^{1/3} = 2, solve for x.Last edited by Redcoats; 20062016 at 16:25. 
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 34
 20062016 16:23
(Original post by NiamhM1801)
Mathway says the same haha, what did I do so wrong lmao 
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 20062016 16:25
My method:
(3√x)¹/₃ = 2
3√x = 8
9x= 64
649=x=55
x=55 
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 36
 20062016 16:26
(Original post by Redcoats)
Don't worry! What do you think your mark will be out of 70? 
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 37
 20062016 16:31
(Original post by NiamhM1801)
Mathway says the same haha, what did I do so wrong lmao
(Original post by Redcoats)
Sorry not 45, x = 121. Check for yourself, it works!
(3  sqrt(x))^{1/3} = 2
3  sqrt(x) = 8
 sqrt(x) = 11
sqrt(x) = 11
Therefore x = 11^{2}
So x = 121
(3  sqrt(x))^{1/3} = 2
(3^{1/3}  x ^{1/6}) = 2
9  x = 64
x = 55
(Original post by Redcoats)
Are you sure your not mixing up questions, i.e. are you thinking of the cos theta or sqrt(125) + sqrt(20) = sqrt(80) + sqrt(x) where x = 45 question? The exact question, I'm pretty sure, is
(3  sqrt(x))^{1/3} = 2, solve for x. 
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 38
 20062016 16:32
First, x cannot equal a negative number as you're square rooting it.
Secondly, when you squared both sides of the equation in 3  sqrt(x) = 8, you square rooted incorrectly:
Square rooting both sides gives: (3sqrt(x))^{2} = (8)^{2}
This is : (9  6sqrt(x) +x) = 64 as it is a double bracket, you can't just square each term as you have done.
Really sorry to tell you this! I'm sure you've still got your A^  remember the boundaries for such are really low 
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 39
 20062016 16:33
(Original post by Chittesh14)
We both ****ed up lol.
That's weird though, what I did and probably NiamhM too:
(3  sqrt(x))^{1/3} = 2
(3^{1/3}  x ^{1/6}) = 2
9  x = 64
x = 55
Nah I got 45 for that, and 55 for this  I'm wrong, you're right. I'm dead.
And no I didn't do it that way, as you can see above. But I got the same answer as you.
Lmao same here. Although I'm looking forward to Friday's now 
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 40
 20062016 16:33
I'm just going down every post lol, I think I've lost A*^ now.
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Updated: June 28, 2016
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