whats the answer 16a(ii)?
Makes sense, nice thinking. How about c)i).Here is my solution to cii (differs from EVERYONE i've seen so far)
from the quadratic: y^2 + y(1-2c) + (c^2 - R^2) = 0
The intersection between the circle and parabola is always positive (y-value)
hence (root1) + (root2) > 0
root1 + root2 = -b/a = (2c-1)
therefore 2c-1>0
2c > 1
c > 1/2
AHHHHHHHHH thats where u get the r>1/2 hmmm nice workc)ii)
4c = 1+4r^2
c=1/4 +r^2
c>0
Therefore, 1/4 +r^2 >0
rearrange and r>1/2.... because r >0
But since c>r then c>1/2 !!
n ( -9303, 948.444.44)what were the coordinates of n????
Since the x values of the intercepts is symmetrical AND y=x^2 is an even functionMakes sense, nice thinking. How about c)i).
Here is my solution to cii (differs from EVERYONE i've seen so far)
from the quadratic: y^2 + y(1-2c) + (c^2 - R^2) = 0
The intersection between the circle and parabola is always positive (y-value)
hence (root1) + (root2) > 0
root1 + root2 = -b/a = (2c-1)
therefore 2c-1>0
2c > 1
c > 1/2
These are both correct answers. The rest of question 16 was fair. 16bi) should have been worth 3 marks though and 16cii) should have been worth 2. Just looking over the paper it's a bit frustrating that the rest of the test (relative to question 16) was incredibly easy. Not a great test to differentiate the good from bad in my opinion.Since the x values of the intercepts is symmetrical AND y=x^2 is an even function
hence the y-value is the same
therefore discrim=0
y^2 + y(1-2c) + (c^2 - R^2) = 0
(1-2c)^2 - 4(1)(c^2 - r^2) = 0
1 - 4c + 4c^2 - 4c^2 + 4r^2 = 0
1 - 4c + 4r^2
1+4r^2 = 4c
definitely 1 mark if you equated them two, showing that you've changed them into a quadratic form-like to show your progress towards the answer. but if u merely equated them both, I don't think u would get a mark for it since it's a 2 marker :Sdo u think there would be 1 mark awarded for equating the two equations in 16 c) i, but not using the dsicriminant
Delta Goodrem is better than Calculus in this questionFor 16 c)i the discriminant isn't the only option, though it is probably better.. I differentiated it, and equated both derivatives (since they touch once i.e. tangents, they have the same gradients at that point). Then after subbing in, Whala!
Is your name Jordan?this is what i did for 16cii which seems to be different from everyone else...
using i) 4c = 1 +4r^2 rearrange for r
r^2 = c - 1/4
c > r
c^2 > r^2
c^2 > c - 1/4
c^2 - c + 1/4 > 0
(c - 1/2)^2 > 0
c - 1/2 > 0
c > 1/2