Maths Extension 1 Predictions/Thoughts (6 Viewers)

igor9 · Wednesday at 6:02 PM

bro what if someone gets like 44 internals and 45 externals suppose, that is not e4 right, that would be sad

because i'm preparing for the worst since my internals look kinda dogshit rn

la_sportive · Wednesday at 6:06 PM

igor9 said:
bro what if someone gets like 44 internals and 45 externals suppose, that is not e4 right, that would be sad because i'm preparing for the worst since my internals look kinda dogshit rn

they round up to 45
dw my internals aren't great either...

igor9 · Wednesday at 7:07 PM

la_sportive said:
they round up to 45
dw my internals aren't great either...

would that still count as an e4 though, like on hsc ninja or whatever?

la_sportive · Wednesday at 8:08 PM

igor9 said:
would that still count as an e4 though, like on hsc ninja or whatever?

yep!

mohitmo · 2024-10-31T04:18:58+1100

leclerc16 said:
how much would a 59/70 align to??

95 I think. You would need to break 60 in this exam to achieve 96.

uart · 2024-10-31T23:19:04+1100

I just took a quick look at this test today. Some interesting question, and parts of Q14 are definitely pretty difficult.

Q14d has an interesting kink that I don't think any of the current solutions have captured. The question doesn't state that $0 \le \theta \le \pi/2$ , though most of the solutions seem to assume that. I think we should take $-\pi/2 \le \theta \le \pi/2$ , as there is nothing in the question to suggest that downward projection of the particle is not allowed (and the other two quadrants are uninteresting as they are identical except for leftward motion instead of rightward).

Why this is interesting is that the solution to
$\sin^2(\theta) < 8/9$
is
$-\sin^{-1}(\sqrt{8/9})< \theta < \sin^{-1}(\sqrt{8/9})$

However this is not correct, as for negative $\theta$ (downward projected particle) the distance from the origin is an increasing function of t for all $-\pi/2 \le \theta < 0$ .

The trick here is that the resultant equation,
$g^2 t^2 - (3 g V \sin \theta) t + 2 V^2$
must be positive definite (>0) for the case of positive $\theta$ , as the axis of symmetry (-b/(2a)) is positive and $t \ge 0$ .

However, for the case of $\theta < 0$ the axis of symmetry is negative, so it's not necessary for the quadratic to be positive definite (>0 for all t), as it only needs to be positive for $t \ge 0$ . And all that's required to prove that is to note that for $\sin \theta < 0$ that all the coefficients are positive and therefore the quadratic is positive for all $t \ge 0$

This gives the overall solution $-\pi/2 \le \theta < \sin^{-1}\sqrt{8/9}$

Hope that helps.

Maths Extension 1 Predictions/Thoughts (6 Viewers)

igor9

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la_sportive

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igor9

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la_sportive

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mohitmo

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uart

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