Tough paper. yep its root 52 and alpha is 0.588.

For the last q, i can remember 14 years. And the graphing modulus thing is like a W shape.

hmm, for no.1 the trig question, there are 3 solutions, and one of them is 0. That's all I can remember =o

Reply 4

theisticandy

11

TheWolf

Tough paper. yep its root 52 and alpha is 0.588.

For the last q, i can remember 14 years. And the graphing modulus thing is like a W shape.

hmm, for no.1 the trig question, there are 3 solutions, and one of them is 0. That's all I can remember =o

yes I got the exact same answers as you on all of those question!!!! :biggrin:

, makes me feel good to know that at least lol

Could you do part 5d though? I really couldn't, it was the last part of that √52 question that asked you to hence solve something

A good paper imo however, I made a couple of stupid mistake, but i so far only can think of 8 marks that I lost (possibly 11)

Reply 5

Becca28

1

meh i did really badly on that test, i was only confident on about half the questions, think i might quite a few silly mistakes :frown:

Reply 6

TheWolf

15

theisticandy

yes I got the exact same answers as you on all of those question!!!! :biggrin:

, makes me feel good to know that at least lol

Could you do part 5d though? I really couldn't, it was the last part of that √52 question that asked you to hence solve something

A good paper imo however, I made a couple of stupid mistake, but i so far only can think of 8 marks that I lost (possibly 11)

part d was tough, i had to substitute the R equation = 3. I think thats what you do anyways =o

couldnt do the last question (1mark q)

Reply 7

Hash

2.12 radians for the long trig question (as well as 0 and pi)

Reply 8

KP Nuts

1

For that 5 part d, I used all of the answers to the previous questions and substituted them into the equation it gave you. So you got something like (8 = theta, V = root) cos8 (V52 sin(8-0.588) - 3) = 0.

Therefore, cos8 = 0 or the bit in the brackets = 0

Yeah i also got 0, pi (which is 1.something radians) and 2.12 radians

Reply 9

Becca28

1

what did people get for the question with the graph where you had to find the coordinate values of a and b?

Reply 10

TheWolf

15

For one of the q you have to find when [modulus(x-1) -2] intersects 5x

it would only intersect with -(x-1) -2

so its -x + 1 -2 = 5x

6x = -1
x = -1/6

i cant remember the q totally correctly, so this might be wrong

Reply 11

DarkWeaver

I got -1 and -2, dunno if that's right tho

b=-1 and a =-2

TheWolf

For one of the q you have to find when [modulus(x-1) -2] intersects 5x

it would only intersect with -(x-1) -2

so its -x + 1 -2 = 5x

6x = -1
x = -1/6

i cant remember the q totally correctly, so this might be wrong

I believe that is correct :smile:

Reply 14

joesharp

1

pretty hard I thought, missed out the last two parts on the modulus question and I couldn't do the 1 mark proof question on Q7, think I managed a B though.