Yes it was 1 because they were recipricals of each other and recipricals multiply to 1
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AQA Maths FP1  15 June 2016 [Exam discussion]
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 15062016 11:38

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 15062016 11:41
(Original post by tomdavis.)
Yes it was 1 because they were recipricals of each other and recipricals multiply to 1 
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 15062016 11:42
I got 7/7 so 1 aswell, and my final answer was 7x^24x+7= 0

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 15062016 11:44
Both the big marks for complex numbers and the matrices were painful tbh, think it'll be like ~58 for an A
Last edited by Chickenslayer69; 15062016 at 11:45. 
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 15062016 11:45
(Original post by Chickenslayer69)
For the alpha & beta question, did anyone get 14/14 = 1 for the product on the last part? 1 just seems like an odd answer lol 
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 15062016 11:46
(Original post by Hjyu1)
Yeah I got that but I cant remember if I put 7 or 4 (by accident) as the cofficent of x^2 
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 15062016 12:03
(Original post by tomdavis.)
I mean part d where the question asked P is imaged at point [0,4] then something to do with the matrix A^2 and reflected on the line X+root3Y=0. Did anyone else get something like 3root2, 2? 
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 15062016 12:13
Can anyone make an unofficial mark scheme?

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 15062016 12:22
I got q=7,3,7,23 the sum of the roots is 20 a w=p+2i so w*=p2i so (p+2i)(p2i)=20 so p=4 or 4 so w=4+2i or w=4+2i. I plugged p+2i in equation (p+2i)(p+2i)+(p+2i)(4+i+qi)+20=0
For real parts i got p^2+4p2q+14=0
For imaginary I got 5p+8+pq=0
Plugging p=4, p=4 I got q=7,3,7,23
That the only way I could see how to do it, was a horrible question nevertheless. 
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 15062016 12:36
For the conics question, solving the possible values of k, did anyone get
2<k<0.5 (or 0.5<k<2 cant remember which way around i had my numbers) 
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 15062016 12:40
Just some big parts where I have just messed up. ~60/75 for an A imo
I think the boundaries will be similar to last year.
 Standard alpha and beta, ended up getting 7x^2  4x + 7=0 something like that
 Linear laws, gradient = 0.4, value of a is 10^intercept, value of b is 10^0.4
 General solution gave npi + pi/2 and npi + pi/3
So tanx can only have pi/3 which is root 3
tan(60)=root 3
 Matrices enlargement sf 4 for A^2 ??? then I ended up doing simultaneous to find the coordinates to plot to (0,4) ??
I hope I've read the question right :/ I think i've got my reflection matrix wrong ffs
It was in the line y=1/root 3 x so y=tan(30)x and which is tan(150) so I had to find cos(300) and sin(300) lol wth so I got 1/2 & root3/2, then simultanous equations to solve for x & y. basically I've wrecked that question
dropped ~3 marks probably on this entire question
 parabola and show a=2 was fine, but second part I've messed up forgot to realise that it was translated (2, 3) ?? something like that y^2=4ax > (y3)^2=4a(x2) and it passes through 4,7. so 16=8a a=2
you had to show b^24ac<0 with ky=x but since its translated k(y3)=(x2)
lol i give up thats like another 5 marks gone.
 series questions show that was fine, second part was to show 4 linear factors
I didn't finish that one ffs but got something with (2n+1)(2n1)
hopefully I've only dropped 3 marks on this idk
Complex question I worked out p=+4 and i worked out sum and product
I got two equations for q lol idk wrecked that
You had to explain why q=1 for two marks, so I said something like because it needs to be a real coefficient or something like that. And when q=1 it gets rid of all the imaginary parts?? :/
z^2 + (4 + i +qi) z + idk =0so if q=1(4+ii)z so 4z something like that
asymptotes y=0, x=2, x=1/2 ??? something like that
sketch, (x1)/(x2)(2x+1) something like that ??? I can't remember
form cubic inequality with intersection then stuck it into my calculator go like x=0, x=1, x=5/2 for my critical values
I can't remember.
Overall id be lucky to get an ALast edited by Pentaquark; 15062016 at 13:24. 
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 32
 15062016 12:48
(Original post by Pentaquark)
Just some big parts where I have just messed up.
 Standard alpha and beta, ended up getting 7x^2  4x + 7=0 something like that
 Linear laws, gradient = 0.4, value of a is 10^intercept, value of b is 10^0.4
 General solution gave npi + pi/2 and npi + pi/3
So tanx can only have pi/3
 Matrices enlargement sf 4 for A^2 ??? then I ended up doing simultaneous to find the coordinates to plot to (0,4) ??
I hope I've read the question right :/ I think i've got my reflection matrix wrong ffs
It was in the line y=1/root 3 x so y=tan(30)x and which is tan(150) so I had to find cos(300) and sin(300) lol wth
so I got 1/2 & root3/2, then simultanous equations to solve for x & y. basically I've wrecked that question
Because my working out is a load of bs.
Dropped idk how many marks, will be error carried forward for second part though?
dropped 5 marks probably on this question
 parabola and show a=2 was fine, but second part I've messed up forgot to realise that it was translated (2, 3) ?? something like that y^2=4ax > (y3)^2=4a(x2) and it passes through 4,7. so 16=8a a=2
you had to show b^24ac<0 with ky=x but since its translated k(y2)=(x2)
lol i give up thats like another 5 marks gone.
 series questions show that was fine, second part was to show 4 linear factors
I didn't finish that one ffs but got something with (2n+1)(2n1)
hopefully I've only dropped 3 marks on this idk
Complex question I worked out p=+4 and i worked out sum and product
I got two equations for q lol idk wrecked that
You had to explain why q=1 for two marks, so I said something like because it needs to be a real coefficient or something like that. And when q=1 it gets rid of all the imaginary parts?? :/
asymptotes y=0, x=2, x=1/2 ??? something like that
sketch,
form cubic inequality with intersection then stuck it into my calculator go like x=0, x=1, x=5/2 for my critical values
I can't remember.
Overall id be lucky to get an A
~59/75 I think
I put that when q=1 it eliminates imaginary parts too but it was literally a guess LOL
btw what did your sketch for the asymptote & line look like? Mine was like one curve in bottom right, one curve top right and one in the middle cutting through x axis or something?? idkLast edited by Chickenslayer69; 15062016 at 12:50. 
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 33
 15062016 12:50
(Original post by Chickenslayer69)
"You had to explain why q=1 for two marks, so I said something like because it needs to be a real coefficient or something like that. And when q=1 it gets rid of all the imaginary parts?? :/"
I put that when q=1 it eliminates imaginary parts too but it was literally a guess LOL 
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 15062016 12:51
(Original post by An1998)
For the conics question, solving the possible values of k, did anyone get
2<k<0.5 (or 0.5<k<2 cant remember which way around i had my numbers)Post rating:1 
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 15062016 12:52
(Original post by A Slice of Pi)
What was the exact question? I might be able to help 
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 15062016 12:55
Apart from a few odd (and very badly worded, I must say) questions the paper was easier in comparison to the last couple of years'.
q was real in the first part of the question, but when 'p is real', is it real as well as or instead of q?! 
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 15062016 12:56
(Original post by Chickenslayer69)
I can't remember exactly :/ But there was an equation with " i + qi " in it and it asked to explain why q must be 1 for it to be real 
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 15062016 12:57
(Original post by A Slice of Pi)
So that the imaginary parts vanish I should imagine. 
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 15062016 12:59
(Original post by smartsy)
Apart from a few odd (and very badly worded, I must say) questions the paper was easier in comparison to the last couple of years'.
q was real in the first part of the question, but when 'p is real', is it real as well as or instead of q?! 
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 40
 15062016 13:03
(Original post by Pentaquark)
Just some big parts where I have just messed up. ~59/75 for an A imo
 Standard alpha and beta, ended up getting 7x^2  4x + 7=0 something like that
 Linear laws, gradient = 0.4, value of a is 10^intercept, value of b is 10^0.4
 General solution gave npi + pi/2 and npi + pi/3
So tanx can only have pi/3
 Matrices enlargement sf 4 for A^2 ??? then I ended up doing simultaneous to find the coordinates to plot to (0,4) ??
I hope I've read the question right :/ I think i've got my reflection matrix wrong ffs
It was in the line y=1/root 3 x so y=tan(30)x and which is tan(150) so I had to find cos(300) and sin(300) lol wth so I got 1/2 & root3/2, then simultanous equations to solve for x & y. basically I've wrecked that question
Because my working out is a load of bs.
Edit: I've just seen in the textbook lol reflection in y=x/root3 is the same as a reflection in y=tan150x
which is (cos300, sin 300) (sin300, cos300) WAIT THERE. MAYBE IM ONTO SOMETHING
Dropped idk how many marks, will be error carried forward for second part though?
dropped 5 marks probably on this question
 parabola and show a=2 was fine, but second part I've messed up forgot to realise that it was translated (2, 3) ?? something like that y^2=4ax > (y3)^2=4a(x2) and it passes through 4,7. so 16=8a a=2
you had to show b^24ac<0 with ky=x but since its translated k(y2)=(x2)
lol i give up thats like another 5 marks gone.
 series questions show that was fine, second part was to show 4 linear factors
I didn't finish that one ffs but got something with (2n+1)(2n1)
hopefully I've only dropped 3 marks on this idk
Complex question I worked out p=+4 and i worked out sum and product
I got two equations for q lol idk wrecked that
You had to explain why q=1 for two marks, so I said something like because it needs to be a real coefficient or something like that. And when q=1 it gets rid of all the imaginary parts?? :/
z^2 + (4 + i +qi) z + idk =0so if q=1(4+ii)z so 4z something like that
asymptotes y=0, x=2, x=1/2 ??? something like that
sketch, (x1)/(x2)(2x+1) something like that ??? I can't remember
form cubic inequality with intersection then stuck it into my calculator go like x=0, x=1, x=5/2 for my critical values
I can't remember.
Overall id be lucky to get an A
I thought that tanx=root 3 as pi(n) +pi/3 are the only possible general solution for it and as tanx had a period of pi you always get root 3. And you matrices question seemed fine as cos/sine have periods of 360 degrees so cos(60)=cos(300) same for sine
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Updated: August 30, 2016
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