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AS math 9709/May June/2018/ Paper 12 Hard Or Easy?

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p and q were 2 and 6. i just found the points and showed their gradients were same to prove y=x idk if i'll get ecf for this method
Wait wait...the gradients were not same
How can the lines meet if the gradients were same...
 
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wtf? the papers were leaked!? no wonder specific questions were trickier otherwise we would have gotten pattern similar to march, since they keep an extra question paper for such instances, they might have used it and we got those woah
 
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p and q were 2 and 6. i just found the points and showed their gradients were same to prove y=x idk if i'll get ecf for this method
they cant be the same... the gradient of one should be the -ve reverse of the other.... and yes their points of intersection meet on the line y=x , which when you equate both, gives you an x that equals to the y
 
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2y + 3x = k
2y = -3x + k
m of perpendicular bisector = -(3/2)
m of line AB = -(3/2) x m (ab) = -1
M ab = 2/3

5h -h/ (4h+6) -h = 2/3
4h /3h +6= 2/3
12h =6h +12
h = 2

A = ( 2,2) B ( 12 ,10)
Mid of AB = (7 , 6)
2(6) + 3(7) =k
k =33
 
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y = 4
y = x/2 + 6 /x
4 = x/2 + 6/x = > x^2 -8x + 12
(x-6)(x-4) = 0
x= 6 or x=4

dydx = 0.5 - 6/x^2
m at x = 6, is 1/3
m at x =2 is -1
point p ( 2,4) point q (6,4)
equation of tangent passing through p , y - 4 / x - 2 = -1 => y= -x + 6
equation of tangent passing through q, y -4 / x -6 = 1/3 => y= [x+6]/3
equation both, -3x + 18 = x +6 => -4x =-12 , x = 3
when x = 3 , y = -3 + 6 = 3
hence shown y =x
 
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feel free to do the volume part lol, im tired , u got all the info necessary, limits 6 and 2
[4 - (x/2 + 6/x )]^2
 
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binomial, a = 3
k<9 and k = 11

a = 8000 , r = 1.02

h= 1 k =22

a = 7 b= -4 , for no solution k > 7 , and k < 3

x = 146 , 323
range for which 2cos > -3sin , 0 >x> 146
wasnt it k>9 as the inequality sign reversed cuz of dividing it by a negative value?
 
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