For (c) (ii), sketch y=(x^2-1)/(x^2+1) and y=kx+1, and you can see no matter what K value you pick (other than k=0, which cannot be the case anyway) you will always have exactly one point of intersection. In other words always one real root. From this, you can also see that when K is large, the real root approaches 0 (will be used later).does anyone know how to do question 15cii and iii?
Oh. Well I substituted four values, namely, -1/k, 1, 0 and -1For (c) (ii), sketch y=(x^2-1)/(x^2+1) and y=kx+1, and you can see no matter what K value you pick (other than k=0, which cannot be the case anyway) you will always have exactly one point of intersection. In other words always one real root. From this, you can also see that when K is large, the real root approaches 0 (will be used later).
For (c) (iii) play around with sums and products of roots, and use the fact that the real root approaches zero to deduce the behavior of the modulus and argument. What ends up happening is that the non-real roots approach i and -i.
Alternatively, can just differentiate P and show that its derivative has no zeros so P is monotonic (and use the fact that P has at least one real root, being a cubic (follows from IVT), so it has exactly one root)). Neither of these methods really use part (i) though, unless you subbed your values in to the expression in part (i), or differentiated the expression in part (i), but these are trivial ways to "use" it.Oh. Well I substituted four values, namely, -1/k, 1, 0 and -1
P(-1/k) = 1-(1/k^2), P(1) = 2(k+1), P(0) = 2, P(-1) = 2(1-k)
We will now consider three cases for k.
Case 1: |k| < 1 means that P(-1/k) is negative. Combine with P(0) and there is a root by the intermediate value theorem.
Case 2: k < -1 means that P(1) is negative. Similar as above.
Case 3: k > 1 means that P(-1) is negative. Similar as above.
Degenerate cases occur if |k| = 1, as the root is trivial.
As you may have deduced already, I despise curve sketching.
...OR you could just divide the entire cubic in it's reduced form by k, and letting k tend to infinity will degenerate the cubic into (x^2+1)x = 0For (c) (iii) play around with sums and products of roots, and use the fact that the real root approaches zero to deduce the behavior of the modulus and argument. What ends up happening is that the non-real roots approach i and -i.
Perhaps saying "Hence, or otherwise", would have been preferable.Alternatively, can just differentiate P and show that its derivative has no zeros so P is monotonic (and use the fact that P has at least one real root, being a cubic (follows from IVT), so it has exactly one root)). Neither of these methods really use part (i) though, unless you subbed your values in to the expression in part (i), or differentiated the expression in part (i), but these are trivial ways to "use" it.
This will find what the roots eventually become, but how would you then describe the behaviour of the roots as k varies? Part of the idea of the question was to get students to picture an 'animation' of what would happen to the roots as K gradually increases in value....OR you could just divide the entire cubic in it's reduced form by k, and letting k tend to infinity will degenerate the cubic into (x^2+1)x = 0
Extension 1.Is this Extension 1 or 2?
I admit Q12 was quite difficult.Extension 1.
11d is pretty much the idea behind newtons method since you approximate roots using tangents.Theres no question 12 C ?
Was question 11d the result of newtons method of approximation ?
I rarely get intimidated. Frequently, however, I find myself flummoxed.Q14 (b) and (c) are not difficult really (once you see the solution), but intimidating perhaps.
Hmm...Here's a question scrapped from 2U because it was too easy.
Wait. How did you know what Q16 was gonna be before when you were in thereMy guesses are Simpsons rule/ trapezoidal rule or the AM-GM inequality
I think it would've been the fact that x=0 is a solution.I was just scrolling.
Hmm...
I'm curious, what was the trap?
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Wait. How did you know what Q16 was gonna be before when you were in there
Wait. How did you know what Q16 was gonna be before when you were in there
Dunno, I made it pop out on it's own.I think it would've been the fact that x=0 is a solution.
It's supposed to be my best topic. I haven't gone into beyond HSC yet but I had 98% of 4U integration in the bag sometime during Year 11 lolIf youre as good as you are in integration i reckon you could state rank
Page 1 of question 16 was to prove AM-GM second was with trap rule lol...