Question 16 c (ii) 2012 paper (1 Viewer)
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RealiseNothing
what is that?It is Cowpea
If  \geq P(k-1))
then we substitute part (i) in for both and solve:
!k}{n^k(n-k)!} \geq \frac{(n-1)!(k-1)}{n^{k-1}(n-k+1)!})
#
Do you need help with part (iii) and (iv)?
Do you need help with part (iii) and (iv)?
RealiseNothing
what is that?It is Cowpea
I'm going to assume you do since you said the last question:
iii) We are given
Now since 'n' and 'k' are integers, then the LHS is greater than the RHS by atleast 1. Hence we can get rid of the
on the LHS and the strict inequality will still hold:
^2)
(note that 
and so the RHS>0 and we can take the square root without any worries). #
iii) We are given
Now since 'n' and 'k' are integers, then the LHS is greater than the RHS by atleast 1. Hence we can get rid of the
Kurosaki
True Fail Kid
I think he's asking about 2U maths lol
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RealiseNothing
what is that?It is Cowpea
Holy shit, I thought I was in the MX2 section.
RealiseNothing
what is that?It is Cowpea
I've probably just scared the fuck out of him.
Kurosaki
True Fail Kid
Shazer2
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You scared the fuck out of me.
louielouiee
louielouielouielouielouie
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I'm certain that almost every HSC student last year just looked at that question and just thought to themselves, fuck that.
I know I did
I know I did
AHAHAHA u kinda did lol
And ye its 2u
Btw I never knew the 4u exam had 16 questions
It's the same format at the 2U one
Their sample answer didnt make sense to me.
Sy123
This too shall pass
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From the previous question, remember our quadratic equation.Hey guys
Can someone please explain how to do the very last question of the 2012 paper, which is Question 16 c(ii). Sorry I cant show you the question because im on my phone.
Thanks
The solutions to this equation are the y-coordinates of the intersection of the parabola and circle. But it is evident that at both intersections they have the same y-coordinate, and hence discriminant is zero (this is how you do the first part).
It is asking us to prove a restriction that c must adhere to, in order to satisfy 'the circle touches the parabola at exactly two points'.
These intersection points must be positive, our solutions to the quadratic equation must be greater than zero, so if we solve the quadratic in y.
This solution must be greater than zero.
But that isn't quite the answer yet, we need to examine what happens when c=1/2, we see that if this is the case, then y=0. But this can't be the case, because then the intersection is at the origin only, which doesn't satisfy the condition in the question 'touches the parabola at exactly 2 points'. So we must discount c=1/2
Hence c > 1/2
You know me too well.