Methods Exam 1: Discussion!

[BLACKCATT]

Ughhh I'm so pissed, got the last 3 parts of question 10 wrong.. that's 4 marks gone. And it was coz of running out of time. fml fml. Could've easily got 39 or 40

same here why didnt i think of similar triangles ffs

[Brunette15]

What does everyone think the A/ A+ cut off will be this year?

[Sayf44]

Could a 35 be an A+??

[Mayesy]

m was 5ln(2) but would ln(32) be accepted or does it have to be simplified

That should be fine!

[Kaleidoscope]

Do you guys reckon it'll be marked up as much as last year? Or does it really depend on the difficulty of tomorrow's exam?

[Mieow]

Here are my preliminary solutions guys

I didn't check them, so I may have a few mistakes, but based on conversations with guys who did the exam they seem correct.

P.S. I have a blank copy of the exam too, so I'll upload that next.

32/40 for me. Hopefully that'll be a decent grade on the grade distribution 

[Aman476]

If you lose one mark on exam 1 do you reckon it's still possible to get a 50 raw?

[nikegod]

If you lose one mark on exam 1 do you reckon it's still possible to get a 50 raw?

I believe so.

[drake]

If you lose one mark on exam 1 do you reckon it's still possible to get a 50 raw?

definitely. i lost 1 mark in exam 1 and 1 mark in exam 2 and got 50. but it will depend on tomorrow's difficulty. if exam 2 is hard, then that is good for you. but if it easy, just losing one mark can be costly.

[Brunette15]

32/40 for me. Hopefully that'll be a decent grade on the grade distribution 

I got the same....still 80%

[katiesaliba]

omfg I forgot to find the coordinates  Just wrote the x-intercepts instead omg

[jumcakes]

Question 10)b)i) could have also been done by finding the gradient in terms of u, and then subbing this into the equation y=mx+v using the point (2, 4) to find v.

[DanielJ]

same here why didnt i think of similar triangles ffs

I let the gradient of (0,v) to (2,4) equal the gradient (2,4) to (u, 0). Ended up with the same thing.

[speedy]

I let the gradient of (0,v) to (2,4) equal the gradient (2,4) to (u, 0). Ended up with the same thing.

Yup, that's what I did

[Aman476]

Question 10)b)i) could have also been done by finding the gradient in terms of u, and then subbing this into the equation y=mx+v using the point (2, 4) to find v.

Yep that's how i did it haha, i just hope +-1/2 (not just +1/2) was an acceptable answer for the second question, cos i don't know if it meant principal square root or just square root (no restrictions?) I drew the hyperbola and pointed to where each gradient would apply.........if i lost that gonna be so pissed haha