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

[Paula V]

I dont remember exactly but the question said that the curve didnt intersect right?

I think the question was something like " find the set of values of K for which the curve lies above the x-axis", I'm not sure if I remember it correctly though

 

[ASSmart]

I think the question was something like " find the set of values of K for which the curve lies above the x-axis", I'm not sure if I remember it correctly though

yes yeah that was the question, so in that case wasn't it determinant<0 because u know it lies above x-axis

 

[Paula V]

yes yeah that was the question, so in that case wasn't it determinant<0 because u know it lies above x-axis

I thought the determinant was supposed to be >0, but I'm not 100% sure. Wait a sec, I'll check my copybook because I know we did something similar in class a while ago.

 

[ASSmart]

I thought the determinant was supposed to be >0, but I'm not 100% sure. Wait a sec, I'll check my copybook because I know we did something similar in class a while ago.

sure np

 

[Paula V]

yes yeah that was the question, so in that case wasn't it determinant<0 because u know it lies above x-axis

found it, you're right, it should be discriminant<0.

 

[Mohammad2018]

Any idea what threshold will be ?

 

[ASSmart]

Any idea what threshold will be ?

Around 95-100 out of 125 im guessing

 

[Paula V]

Around 95-100 out of 125 im guessing

Yeah, I agree, probably something like that, it depends on the Statistics/ Mechanics paper too

 

[Mohammad2018]

I still dont understand why n is not 13
In question they asked to find amount of salt of 12th week after change

 

[Paula V]

I still dont understand why n is not 13
In question they asked to find amount of salt of 12th week after change

It's because the first term (8000) was the first week after the change, not the week of the change, so the 12th week is the 12th term.

 

[mirolev]

(i) Area of circle = area of shaded region. Find pi*r^2 = r^2*tan(theta) - (2theta*r^2)/2 where length of triangle is found by tan(theta) = AT/r (AT is one of 2 tangents to the circle).

(ii) Formula used is theta*r = arc length. 19.2 = 8*2theta, theta = 1.2

(iii) Shaded area is found by 8*8tan1.2 - (2theta*8^2)/2.
Area = 87.8

 

[ThePacifics]

(i) Area of circle = area of shaded region. Find pi*r^2 = r^2*tan(theta) - (2theta*r^2)/2 where length of triangle is found by tan(theta) = AT/r (AT is one of 2 tangents to the circle).

(ii) Formula used is theta*r = arc length. 19.2 = 8*2theta, theta = 1.2

(iii) Shaded area is found by 8*8tan1.2 - (2theta*8^2)/2.
Area = 87.8

there were 'two' triangles, this was a 6 marks question. You had to add up 2 triangles and subtract area of sector from it.

 

[Mohammad2018]

Anyone remebers the question after drawing graph

How many marks was it

 

[Paula V]

Anyone remebers the question after drawing graph

How many marks was it

yeah, it was two marks

 

[Mohammad2018]

And how many marks was f(x)=k has no solution

 

[Paula V]

there were 'two' triangles, this was a 6 marks question. You had to add up 2 triangles and subtract area of sector from it.

oooooooh, it was so easy , I wanted to do a way more complicated method

 

[Paula V]

And how many marks was f(x)=k has no solution

I think 3 or 2, I'm not completely sure

 

[mirolev]

there were 'two' triangles, this was a 6 marks question. You had to add up 2 triangles and subtract area of sector from it.

That's true. Area of 2 triangles should be 8*8tan(theta).

 

[ASSmart]

And how many marks was f(x)=k has no solution

3 marks and i didnt understand that question at all

 

[Bai fong]

its probably k > 9 , because if u send b^2 ( b can never be negative )on the RHS, you need to divide by a -ve number to make it become positive, .... ( b^2 - 4ac <0)
also i remember getting k = 11 (b^2-4ac =0)