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2017 HSC Mathematics Extension 2 paper thoughts? (1 Viewer)

calamebe

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As written in the question, a can be negative.

However, once you get the positive solutions for a, getting the rest of the solutions is trivial since you just negate the positive ones.

Since the question was only 2 marks, in my opinion, they should just give the full marks if you get the correct positive values (since getting the negative values isn't really the hard part or point of the question).
Yeah, that was my logic as well, hence why I did negative. But, yeah as you said, I don't see any real reason to deduct marks for neglecting the negatives.
 

Green Yoda

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What would you guys reckon the e4 cut off will be? And the cut off for 80?
 

janzrn

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|a|e - |a| = |a|(e-1) = 1

Therefore |a| = 1

a= 1 or a = -1

Oh and also you can do the other vertex:

|a|e + |a| = |a|(1+e) = 1

Therefore |a| = 1/3

a = 1/3 or a = -1/3
Can a>0?
I thought since on a graph, one would write "a" and "-a" on the positive and negative x axis
 

InteGrand

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Can a>0?
I thought since on a graph, one would write "a" and "-a" on the positive and negative x axis
Yes, a can be positive or negative as written in the question. The equation for the hyperbola doesn't change if you replace a by -a, so you can just find the positive solutions first and negate them if you want the full solution set.
 

janzrn

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Did people get 6.6666...u^3 for that one triangle prism volumes qn
 

calamebe

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Can someone post the answers for Questions 8, 9 and 10 in MCQ? I got B, D and C respectively
Correct me if I'm wrong, but here's what I get:

Yeah 8 is B which is just by subbing in (-x) into the equation.

Then 9 is C. So dy/dt = dy/dt * dt/dx = -x/y * 1/y = -x. Then, consider that in the first quadrant, y is positive so dx/dt is positive, and hence x is increasing. So it seems like it goes clockwise. You can do this for other quadrants, e.g. in the third quadrant y is negative so x is decreasing, again going clockwise.

Then 10 is B, as it is a concave up function by the definition. So draw a line from (0,f(0)) to (1,f(1)) and see that the graph will always lie under this, and hence the intergral will be less than the midpoint if that makes sense. A counter example to A is x^2, a counter example to C is x(x-2), and a counter example to D is x(2-x).
 

pikachu975

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Correct me if I'm wrong, but here's what I get:

Yeah 8 is B which is just by subbing in (-x) into the equation.

Then 9 is C. So dy/dt = dy/dt * dt/dx = -x/y * 1/y = -x. Then, consider that in the first quadrant, y is positive so dx/dt is positive, and hence x is increasing. So it seems like it goes clockwise. You can do this for other quadrants, e.g. in the third quadrant y is negative so x is decreasing, again going clockwise.

Then 10 is B, as it is a concave up function by the definition. So draw a line from (0,f(0)) to (1,f(1)) and see that the graph will always lie under this, and hence the intergral will be less than the midpoint if that makes sense. A counter example to A is x^2, a counter example to C is x(x-2), and a counter example to D is x(2-x).
I agree with these
 

jjHasm

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What would you guys reckon the e4 cut off will be? And the cut off for 80?
dw man, as long as you PASSED, im pretty sure thats an e3 cutoff
as for e4, not too sure, personally im hoping for a 66~ as the cutoff
 

zammijz

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OH BTW,

Was it just me, or did you guys feel very suspicious during the exam that the maximum height question was so standard and easy for 4 whole marks?
I thought the exact same thing!
 
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