Q16 (1 Viewer)

[GoldyOrNugget]

For ci), there must be exactly 2 solutions to the equation representing the intersection of the two shapes. Because this equation is a quadratic over x² and the intersections are symmetric along the x-axis, there must be one solution to x²=0 which will provide 2 intersections with +/-.



For cii) the observation that had to be made was that if c=r, then there's only one solution to the equation -- x=0, because the circle is resting at the base of the parabola. This situation must be avoided, so c > r.

 

[gorab]

For q b)i) with showing the equation of the tangent, what if I used the coordinate Q (on the line y = -1) to find the equation??

Since they gave that Q's y coordinate was 1, i used the equation to verify it's x coordinate and then used the point gradient formula. Would that still give full marks?

 

[shumit]

discriminant = 0

 

[RealiseNothing]

For q b)i) with showing the equation of the tangent, what if I used the coordinate Q (on the line y = -1) to find the equation??

Since they gave that Q's y coordinate was 1, i used the equation to verify it's x coordinate and then used the point gradient formula. Would that still give full marks?

I don't think so. You used the end result to prove the end result.

 

[El Torra]

For 16a, will I get marked down if I didn't state that opposite sides of a rhombus are parallel before proving through corresponding angles?

 

[4lifee]

I reckon that Q 16 was like 3 unit, I just don't get what the question is saying???? with little info WTF?

 

[RishBonjour]

I reckon that Q 16 was like 3 unit, I just don't get what the question is saying???? with little info WTF?

I can assure you It was not 3 unit. But I think question 16 was a tad harder than past HSC Q 10s.

 

[sebel]

i think it's all about luck (dont hate if im wrong!)

but for me it kinda was as i only did two past papers haha .......................

what i did was go over the core principles of every topic

say for question 16 (c) it was about the parabola (discriminant etc.)

and quickly did it without thinking TOO much.. (what my teacher told me was, thinking too hard for a question will actually lead you to no where)

and in the end i got through all of question 16 so i'm happy and relieved at the same time!! woohoo

 

[Rawf]

I can assure you It was not 3 unit. But I think question 16 was a tad harder than past HSC Q 10s.

Funny you say that cause my friend who dropped from 4u to 3u said shes seen a question similar to that in 4u o_o

 

[qwe_r1]

(c) (i) simultaneous Eqn then use discriminant to find where the intercepts/roots of new equation, are = (ie equal 0), rearrange and get it in the form needed.

(ii) lim r-> 0 gave c>1/4, since there are two distinct points of interception with y=x^2, c>1/2. otherwise the two points become one when the equation specifies 2, so you have to double it.


Painful as i can do it but choked in exam, over thought it

 

[RishBonjour]

Funny you say that cause my friend who dropped from 4u to 3u said shes seen a question similar to that in 4u o_o

ahah, yeah "similar" except prolly with I^2 and complex numbers etc and stuff that REQUIRE 4 unit knowledge.

but def wasn't similar to other Q10. they probably just did that because the rest of the test was fairly "generic". Bos loves trolling.

 

[GoldyOrNugget]

LOL it was not 4u. It was application of 2u concepts to a question of 3u difficulty, same as every other last question in a HSC 2u maths paper. A question like that would be given as the introduction to a 4u question, with 6 other parts that follow, each harder than the previous one, culminating in "prove e is irrational" or some bullshit like that.

 

[Rawf]

I do 3u and I had no idea what to do for that question lol. I basically just stared at it for 45 minutes

 

[heirach]

Re: Q16 part c

Well, here's how I did it;

c/
(i) x^2 + (y - c)^2 = r^2 ----(1)
y= x^2 ----(2)
sub (2) into (1)
y + y^2 - 2cy + c^2 - r^2 = 0
y^2 + (1 - 2c)y + (c^2 - r^2) = 0
delta = 0, because it 'touches', not cuts
delta = (1 - 2c)^2 - 4(c^2 - r^2)
= 4r^2 - 4c + 1
= 0
i.e. 4c = 4r^2 + 1

(ii) Now 4c = 4r^2 + 1
since c > r
4c = 4r^2 + 1 < 4c^2 + 1
4c < 4c^2 + 1
(2c - 1)^2 > 0
2c > 1
c > 1/2

 

[Shippa63]

For c (i), use simultaneous equations to get a quadratic in y. As the y ordinates for both points of contact are the same, there must only be one solution to the quadratic. Hence, the discriminant must be zero. Simplifying the discriminant gives the result.
For c (ii), for the quadratic to be a perfect square (discriminant=0), then the coefficient of y must be negative (as the solution has a positive value for y, above the axis). In the quadratic, the coefficient of y was -(2c-1). Hence -(2c-1)<0, therefore 2c-1>0, 2c>1, and hence c>1/2 as required.
Hope that helps, but don't forget...you can't change things now so don't worry about it!
Cheers

 

[SIRSIR]

For c (i), use simultaneous equations to get a quadratic in y. As the y ordinates for both points of contact are the same, there must only be one solution to the quadratic. Hence, the discriminant must be zero. Simplifying the discriminant gives the result.
For c (ii), for the quadratic to be a perfect square (discriminant=0), then the coefficient of y must be negative (as the solution has a positive value for y, above the axis). In the quadratic, the coefficient of y was -(2c-1). Hence -(2c-1)<0, therefore 2c-1>0, 2c>1, and hence c>1/2 as required.
Hope that helps, but don't forget...you can't change things now so don't worry about it!
Cheers

solving using simultaneous equations given discriminate = 0 gave the answer to 16ci) as above and it turns out that the y ordinate is y=-1/2+C.

Also since the parabola is y=x^2 i.e. concave up with vertex at (0,0) then it follows that the y-ordinate y=-1/2+C has to be >0 -------- i.e. -1/2+C>0
C>1/2

 

[hayabusaboston]

question: Idk how to judge, in all of ur opinions, was this mathematics test harder than most/all others before? for the hSC i mean

 

[iBibah]

question: Idk how to judge, in all of ur opinions, was this mathematics test harder than most/all others before? for the hSC i mean

No. The question were all very straightforward. Q16 things got a bit more difficult, with 16(c) proving difficult for most. Overall it was easier than earlier years, but the final question was just as hard as some older papers.

 

[RealiseNothing]

I did question 16C a different way to everyone else (like completely different). Not sure if what I did was right though :/

 

[Peter Zhang]

My solutions to 16(c), which probably make a whole lot more sense because its JUST quadratic formula...
You eventually derive

y^2 + (1 - 2c)y + (c^2 - r^2) = 0 after substituting x = y^2

Now, y = (2c - 1 +-sqrt(1 - 4c + 4c^2 - 4c^2 + 4r^2 ))/2

-> y = (2c - 1 +-sqrt(1 - 4c + 4r^2))/2

Knowing that the y value is equal and repeated, 1 - 4c + 4r^2 must equal zero , therefore,

1 - 4c + 4r^2 = 0

4c = 4r^2 + 1

For part (ii), it becomes ludicrously easy having shown the full equation for y with the quadratic formula (apparently quadratic formula is too mainstream and people made the question much harder by only using the discriminant)

As the discriminant is then zero, from part (i),

y = (2c - 1 +-sqrt(1 - 4c + 4r^2))/2

becomes

y = (2c - 1)/2

y = c - 1/2
and knowing that the equation of the parabola is y = x^2 which means y is always positive (concave up parabola), we deduce that y.. must be positive

therefore,

c - 1/2 > 0
and then c > 1/2

Very easy, this took me about 5 minutes in class (I'm in Yr 11 atm) but under pressure I could DEFINITELY understand how people would shit bricks