Q16 (2 Viewers)

[freeeeee]

whats the answer 16a(ii)?

 

[RealiseNothing]

whats the answer 16a(ii)?

 

[Frostkruncher]

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

 

[duckysd]

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

Makes sense, nice thinking. How about c)i).

 

[HeroicPandas]

c)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 !!

AHHHHHHHHH thats where u get the r>1/2 hmmm nice work

The way i did it was by cheating on reaaranging c>1/2 and found r>1/2 xD and i didnt know where to find r>1/2 lol

again nice work!

 

[brittallen1994]

what were the coordinates of N????

 

[Timske]

what were the coordinates of n????

n ( -9303, 948.444.44)

 

[Frostkruncher]

Makes sense, nice thinking. How about c)i).

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

 

[Saksyyna]

@Frostkruncher I did it the exact same way as you! Think we'd get the mark? This is for part c) ii)

 

[maths_genius]

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

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

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.

 

[markGee]

i got part a correct
then somehow bossed my way through part b
i was so happy i got part b that i completely forgot there was a part c and didn't even attempt it
but there was only five minutes or so to go, so i went back and fixed up a question, so i probably still got the same amount of marks as i would have

 

[JChou]

do u think there would be 1 mark awarded for equating the two equations in 16 c) i, but not using the dsicriminant

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 :S

 

[brianphamm]

For 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!

 

[HeroicPandas]

For 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!

Delta Goodrem is better than Calculus in this question

 

[captainpigs]

i found q16 quite easy

 

[jnigga07]

For question 16 a I proved that angle BFE and angle BCD are corresponding but instead of proving another angle is equal is said that the 2 sides (surrounding the angle) were in equal ratio, that is, BE/EA and BF/FC, and since FC=ED, therefore BF/ED. I stated that they were in equal ratio because they were cut by a line which is parallel to the base (FE), meaning the sides are cut into equal ratios. So I proved they were similar on the test of two sides in equal ratio and included angle equal. Anyone confirm if I'll get the 2 marks?

 

[jesli16]

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

 

[Strangeoranges]

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

Is your name Jordan?

 

[jesli16]

nope

 

[ZombieApocalypse]

lol I wanted to die


I could do 16a


but I absolutely was stumped at b and c


then again, I never attempt q10 of the past hsc papers


was to be expected i'd fail, really