[SUGGESTED SOLUTIONS] Specialist Exam 1 2013 Solutions

[belsebob]

who got the temperature question right

[Infinity Plus]

Hey guys I drew the entire graph of Hyperbola instead of just the first quadrant (#stupidexammistakes), how many marks do you think they'll take off? Also my working for the part before was that sec^2(x)-tan^2(x)=1 therefore, since sec(x) = x/4 and tan(x) = y/2    x^2/16 - y^2/4 = 1. Is this enough working for the answer? Sorry this looks so crappy but I don't know how to use Latex.

[BubbleWrapMan]

Hey guys I drew the entire graph of Hyperbola instead of just the first quadrant (#stupidexammistakes), how many marks do you think they'll take off? Also my working for the part before was that sec^2(x)-tan^2(x)=1 therefore, since sec(x) = x/4 and tan(x) = y/2    x^2/16 - y^2/4 = 1. Is this enough working for the answer? Sorry this looks so crappy but I don't know how to use Latex.

That working looks fine, and you'll probably get a mark for the graph.

[Infinity Plus]

Thanks 

[Limista]

what's been the A+ cut off for the past few years for exam 1? Curious. & thanks!

[Damoz.G]

what's been the A+ cut off for the past few years for exam 1? Curious. & thanks!

Re-posting....

SPECIALIST MATHEMATICS

Exam 1
2006: 73/80
2007: 72/80
2008: 64/80
2009: 75/80
2010: 73/80
2011: 77/80
2012: 69/80

Average: 71.9/80 ~36 out of 40

Exam 2
2006: 136/160
2007: 137/160
2008: 132/160
2009: 140/160
2010: 137/160
2011: 143/160
2012: 147/160

Average: 138.9/160 ~69/70 out of 80

[lzxnl]

2008. Those scores.

[BubbleWrapMan]

VCAA. What happened.

[AsiandudeJZ]

I got Q1 for question 7b too, but reconsidering since it had a -ve sign at the front. Does that mean it's actually Q4?

It should be in the first quadrant as the domain for t is [0,pi/2] if i remember correctly, so the range for the y component (y=2tan(x)) is [0,infinity], and as x is also greater or equal to 4 (work it out the same way), the graph has to be in the first quadrant. As for the negative sign you mentioned, if you're talking about the one in front of y^2, it just gives the shape of the hyperbola, nothing to do with domain and range restrictions.