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How was Mathematics 9709/12 ?

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the paper was relatively easy though. The last part of the paper i think came in 2011 or 2012 papers or something like that because i have done a similar question like that before.
 
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do we get grades according to M1 and Pure or individually like suppose I lost 11+ marks in pure can I still get A?
well it depends on gt ,if A would be at 60 somehwere then u will get around 85 up if u get 65 around . and check the syllabus , m1 contributes 40% of As level and p1 60% . so accordingly u are graded . :D
 
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how did you all get the shaded area in that radian question? I got it by( 2 x area of trapezium- area of sector)
 
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how did you all get the shaded area in that radian question? I got it by( 2 x area of trapezium- area of sector)
Area of rectangle - (Area of sector - Area of triangle AOB)
If you subtract area of triangle from area of sector you get the unshaded part of the rectangle. So you can use the rectangles area to find the shaded part,
 
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How did you guys find p and q in the parallel vector question??

you had to find what was the factor that related the two vectors, in terms of j you didn't have either a p or a q so if you did 4j/2j=2 then you'd just equal AB to 2OC (i think those were the vectors), find p and then find q i think i got p=5/2 and then q i don't remember
 
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Guys I screwed up.. Might score a 45/75..
Do you know how much do I need to get in the other papers so I can score an A/B (Taking A level math)
 
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Guys I screwed up.. Might score a 45/75..
Do you know how much do I need to get in the other papers so I can score an A/B (Taking A level math)
you can check the thresholds of may june 2016 paper & see around how much you'll need for an A or B for the component youre giving.
but dont stress, the paper is over now, work real hard for the others and aim for a high A in all of them
 
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co-efficient of x as 80, then in (ii) it was -40

The equation of the line AB was 3y = -2x + 14

The co-ordinates of B I got (4,2)

The angles I got were 19.5o and 160.5o.

The area of the shaded region was 22.7 cm2

The equation of the tangent line was y = 3/4 x - 9/2

The rate of change of y with time was 0.12 units per second

The volume of the shaded region when rotated about the x axis was 9pi units3.

The minimum number of terms needed to exceed 20,000 was 70

The sum to infinity was 64.8

The angle AOB was 68.2o

p = 2.5 and q = 4

The stationary point was (4,8)

The curve met the line y=6 when the x co-ordinates were 9 and 1

The range of values of k which for which the curve did not intersect the line k>8

The value of x when f(x)+4 = 0 to one dp was x=-1.2

The domain of the inverse of f was the same as the range of f which was -5<= x <= 1.
 
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co-efficient of x as 80, then in (ii) it was -40

The equation of the line AB was 3y = -2x + 14

The co-ordinates of B I got (4,2)

The angles I got were 19.5o and 160.5o.

The area of the shaded region was 22.7 cm2

The equation of the tangent line was y = 3/4 x - 9/2

The rate of change of y with time was 0.12 units per second

The volume of the shaded region when rotated about the x axis was 9pi units3.

The minimum number of terms needed to exceed 20,000 was 70

The sum to infinity was 64.8

The angle AOB was 68.2o

p = 2.5 and q = 4

The stationary point was (4,8)

The curve met the line y=6 when the x co-ordinates were 9 and 1

The range of values of k which for which the curve did not intersect the line k>8

The value of x when f(x)+4 = 0 to one dp was x=-1.2

The domain of the inverse of f was the same as the range of f which was -5<= x <= 1.

Thanks!! Do you remember what the stationary point question was?
 
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