VCE Specialist Maths 2 - 11/11/19 - Discussion/Questions/Solutions

[Massimooo123]

Extended response 6a. I think your calculation gives the chance of getting at least one sample with mean mass less than 370 or greater than 375. I did 1 – (1 – Pr(370 ≤ X ≤ 375))^2.

Otherwise, thanks again.

I did (Pr(370 ≤ X ≤ 375))^2 + 2*(Pr(370 ≤ X ≤ 375))*(1-Pr(370 ≤ X ≤ 375)) and got the same answer that your expression gives.

[MubMurshed]

With q2c about -1 < d < 5, doesn’t d have to be be greater than 2? I did 2 < d < 5.

Because we have sqrt(2d-4), and we have already taken out i from the expression. Therefore we can’t make it negative otherwise we create an another i and the restriction that it has to be lower than sqrt(6)/2 no longer applies since it is not an imaginary value. Therefore d should be > 2.

Would I be correct in thinking that?

[S_R_K]

With q2c about -1 < d < 5, doesn’t d have to be be greater than 2? I did 2 < d < 5.

Because we have sqrt(2d-4), and we have already taken out i from the expression. Therefore we can’t make it negative otherwise we create an another i and the restriction that it has to be lower than sqrt(6)/2 no longer applies since it is not an imaginary value. Therefore d should be > 2.

Would I be correct in thinking that?

If 2 < d ≤ 5, then the solutions have non-zero imaginary component, and lie somewhere on the vertical line segment joining –1 – i*sqrt(6)/2 and –1 + i*sqrt(6)/2.

If d = 2, the solution is –1.

If –1 ≤ d < 2, the solutions have zero imaginary component, and lie on the real axis between -1 – sqrt(6)/2 and -1 + sqrt(6)/2.

So –1 ≤ d ≤ 5 gives all values of d that satisfy |z + 1| ≤ sqrt(6)/2.

[AlphaZero]

"All complex solutions... have non-zero... imaginary parts"
So

will yield an imaginary number, which would make q complex, where q should instead be a real number.

While \(\dfrac{\sqrt{b^2-4ac}}{2a}\) might be non-real, \(\left|\dfrac{\sqrt{b^2-4ac}}{2a}\right|\) is most certainly non-negative and real. Your answer is equivalent to mine, and the modulus brackets take care of it.

Multiple choice, Question 4: i^1! = i, and both i^2! and i^3! = -1, then i^n! = 1, for all n ≥ 4, since n! is a multiple of 4 for all n ≥ 4. This gives i – 2 + 97*1.

I knew this doing the question, but I couldn't be bothered carefully counting when it's an MCQ.

Extended response 3a ii, I'm confused about your inequalities. I got, when 0 ≤ b < a, we have r > s (so that a – b > 0 and r/s > 1).

\(k\) is also positive when \(a-b<0\) and \(0<r/s<1\), making the log negative as well. There are two possibilities, and I think both should be stated.

Extended response 6a. I think your calculation gives the chance of getting at least one sample with mean mass less than 370 or greater than 375. I did 1 – (1 – Pr(370 ≤ X ≤ 375))^2.

Yep, I screwed this up. My bad. In my defence, I had my Human Structure and Function exam 1 just before I did the solutions (so I was pretty tired) and I had my Differential Equations exam this morning, so I wasn't exactly taking my time with the solutions

[MubMurshed]

If 2 < d ≤ 5, then the solutions have non-zero imaginary component, and lie somewhere on the vertical line segment joining –1 – i*sqrt(6)/2 and –1 + i*sqrt(6)/2.

If d = 2, the solution is –1.

If –1 ≤ d < 2, the solutions have zero imaginary component, and lie on the real axis between -1 – sqrt(6)/2 and -1 + sqrt(6)/2.

So –1 ≤ d ≤ 5 gives all values of d that satisfy |z + 1| ≤ sqrt(6)/2.

Ahh I see, completely forgot about that it could be just real too haha. Thanks!

[S_R_K]

\(k\) is also positive when \(a-b<0\) and \(0<r/s<1\), making the log negative as well. There are two possibilities, and I think both should be stated.

Agreed, but if a – b > 0, then we require r/s > 1, which gives a > b and r > s; in your solutions you've got a > b and s > r. Similarly, if b > a we need s > r, but you've got r > s.

Also: If a > b, can we have (and if not, why not) r < s < 0? This would also give r/s > 1. I don't see any reason to rule out negative values of P (eg. perhaps P is a monetary balance).

[AlphaZero]

Agreed, but if a – b > 0, then we require r/s > 1, which gives a > b and r > s; in your solutions you've got a > b and s > r. Similarly, if b > a we need s > r, but you've got r > s.

Also: If a > b, can we have (and if not, why not) r < s < 0? This would also give r/s > 1. I don't see any reason to rule out negative values of P (eg. perhaps P is a monetary balance).

*sigh yuck too many errors (2 now rip) in the solutions. I'll fix Q3a.ii now.

In the previous part, we required \(P>0\) to arrive at the result, so I think it's safe to assume \(P>0\) then.