# Ocr mei further maths b core pure 2019 unofficial mark schemeWatch

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#1
Unofficial markscheme:
Sum to 100 question, use the method of differences to find the sum formula, 1.5-(1/n+1) - (1/n+2)

Equation with roots 2a-3 and 2b-3, sub in (w+3)/2 to get 3w^2+16w+29

M^-1 done on calculator, matrix = (0, 1, -1, 2, -8, 5, -1, 5, -3) as a 3x3

Significance of determinant being 0 is that it maps all points onto a single line (colinear) because the area scale factor is 0

Angle between planes = angle between normal vectors, cos (x) =a•b/|a||b| so the angle = 40.2 degrees

Given the matrix B^-1 A^-1, AB is the inverse so (1/3, -1/3, 1/3, 2/3) as a 2x2 matrix.
B = (0, -3, 1/3, 2/3)

Possible values of lambda, greater than 0 or less than -4.

There are no invariant lines of the transformation because there are no real roots to 2m^2 + 2m +1.

Points of intersection on argand diagram are: (sqrt(7)/sqrt(2)) + (4+(sqrt(7)/sqrt(2)))i along with the negative of those square roots.

Roots of quartic equation i (given), -i (conjugate pair) and substitution gives 1+i and 1-i
Last edited by Ed_e12; 1 week ago
0
1 week ago
#2
I didnt get your intersection for the complex number one. A lot of people in my college got the same answer. It was like +-(sqrt14)/2, +-(8+sqrt14)/2, or something (not sure about the y coordinate, but I'm sure the x is right) and the question about the determinant was the significance if >0
(Original post by Ed_e12)
Unofficial markscheme:
Sum to 100 question, use the method of differences to find the sum formula, 1.5-(1/n+1) - (1/n+2)

Equation with roots 2a-3 and 2b-3, sub in (w+3)/2 to get 3w^2+16w+29

M^-1 done on calculator, matrix = (0, 1, -1, 2, -8, 5, -1, 5, -3) as a 3x3

Significance of determinant being 0 is that it maps all points onto a single line (colinear) because the area scale factor is 0

Angle between planes = angle between normal vectors, cos (x) =a•b/|a||b| so the angle = 40.2 degrees

Given the matrix B^-1 A^-1, AB is the inverse so (1/3, -1/3, 1/3, 2/3) as a 2x2 matrix.
B = (0, -3, 1/3, 2/3)

Possible values of lambda, greater than 0 or less than -4.

There are no invariant lines of the transformation because there are no real roots to 2m^2 + 2m +1.

Points of intersection on argand diagram are: (sqrt(7)/sqrt(2)) + (4+(sqrt(7)/sqrt(2)))i along with the negative of those square roots.

Roots of quartic equation i (given), -i (conjugate pair) and substitution gives 1+i and 1-i
Last edited by BrandonShulver; 1 week ago
1
1 week ago
#3
how did you work out the intersection? I found the equation of the line and the circle but didn't know if i was going along the right track
(Original post by BrandonShulver)
I didnt get your intersection for the complex number one. A lot of people in my college got the same answer. It was like +-(sqrt14)/2, +-(8+sqrt14)/2
0
1 week ago
#4
You knew the angle of tan pi/4 (or 45 degrees) and the length from the origin to the centre was 4 since the half line started at -4, so you can use trig to find the height of the triangle, to then find the gradient of the half line. Then plug in (-4,0) into y=x+c and get y=x+4, then sub into circle of (x-1)^2+(y-3)^2=9 and solve for x and y
(Original post by mudmuggleblood)
how did you work out the intersection? I found the equation of the line and the circle but didn't know if i was going along the right track
1
#5
Substitute x+4, into the equation of the circle
(Original post by mudmuggleblood)
how did you work out the intersection? I found the equation of the line and the circle but didn't know if i was going along the right track
Last edited by Ed_e12; 1 week ago
1
#6
It's the same you've just rationalised the denominator

(Original post by BrandonShulver)
I didnt get your intersection for the complex number one. A lot of people in my college got the same answer. It was like +-(sqrt14)/2, +-(8+sqrt14)/2, or something (not sure about the y coordinate, but I'm sure the x is right) and the question about the determinant was the significance if >0
0
1 week ago
#7
yh i started to do this towards the end after being puzzled for so long but didn't get time to find the roots. hopefully i get method marks
(Original post by BrandonShulver)
You knew the angle of tan pi/4 (or 45 degrees) and the length from the origin to the centre was 4 since the half line started at -4, so you can use trig to find the height of the triangle, to then find the gradient of the half line. Then plug in (-4,0) into y=x+c and get y=x+4, then sub into circle of (x-1)^2+(y-3)^2=9 and solve for x and y
0
1 week ago
#8
Oh right lol its just what my calculator gave me as the answers to the quadratic

(Original post by Ed_e12)
It's the same you've just rationalised the denominator
0
1 week ago
#9
I disagree with some of your answers to question 6. For 6ai) If DET M > 0 for all values of lambda, then the quadratic you obtain you use b^2 - 4ac < 0 as this means there are no real roots (the graph is always > 0). If you do this you get -4 < k < 0. For 6aii) you need to explain the significance of DET M > 0 for the transformation, not equal to 0. FULL PAPER :
Last edited by Somerandomdud3; 1 week ago
0
1 week ago
#10
(Original post by Somerandomdud3)
I disagree with some of your answers to question 6. For 6ai) If DET M > 0 for all values of lambda, then the quadratic you obtain you use b^2 - 4ac < 0 as this means there are no real roots (the graph is always > 0). If you do this you get -4 < k < 0. For 6aii) you need to explain the significance of DET M > 0 for the transformation, not equal to 0.
Can you send the rest of the paper?
0
1 week ago
#11
Full paper edited in now.
(Original post by MrRhino)
Can you send the rest of the paper?
0
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