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AQA Maths FP1 - 15 June 2016 [Exam discussion] watch

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    Yes it was 1 because they were recipricals of each other and recipricals multiply to 1
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    (Original post by tomdavis.)
    Yes it was 1 because they were recipricals of each other and recipricals multiply to 1
    Ok, thanks.
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    I got 7/7 so 1 aswell, and my final answer was 7x^2-4x+7= 0
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    Both the big marks for complex numbers and the matrices were painful tbh, think it'll be like ~58 for an A
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    (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
    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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    (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
    You should be fine, I got 7s and 4s in places too lool
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    (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?
    Omg i got that!!!!!!
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    Can anyone make an unofficial mark scheme?
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    I got q=-7,-3,7,23 the sum of the roots is 20 a w=p+2i so w*=p-2i so (p+2i)(p-2i)=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+4p-2q+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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    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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    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 -> (y-3)^2=4a(x-2) and it passes through 4,7. so 16=8a a=2
    you had to show b^2-4ac<0 with ky=x but since its translated k(y-3)=(x-2)
    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)(2n-1)
    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+i-i)z so 4z something like that

    -asymptotes y=0, x=2, x=-1/2 ??? something like that
    -sketch, (x-1)/(x-2)(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
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    (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 -> (y-3)^2=4a(x-2) and it passes through 4,7. so 16=8a a=2
    you had to show b^2-4ac<0 with ky=x but since its translated k(y-2)=(x-2)
    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)(2n-1)
    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
    "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

    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?? idk
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    (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
    What was the exact question? I might be able to help
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    (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)
    I think I got -2<k<0.5
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    (Original post by A Slice of Pi)
    What was the exact question? I might be able to help
    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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    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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    (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
    So that the imaginary parts vanish I should imagine.
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    (Original post by A Slice of Pi)
    So that the imaginary parts vanish I should imagine.
    Yeah, so I thought... I think it's right but the question was worded weirdly so I have no idea.
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    (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?!
    No idea, that question was horrible. IMO this paper was easier than other papers with the exception of the large mark questions. I personally think grade boundaries will be fairly low due to the large mark questions being difficult.
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    (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 -> (y-3)^2=4a(x-2) and it passes through 4,7. so 16=8a a=2
    you had to show b^2-4ac<0 with ky=x but since its translated k(y-2)=(x-2)
    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)(2n-1)
    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+i-i)z so 4z something like that

    -asymptotes y=0, x=2, x=-1/2 ??? something like that
    -sketch, (x-1)/(x-2)(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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