LittlerCandy

New Member
Mine:

c
b
a
d
c
a
a
c
b
Can't remember what I put, hopefully d

Last edited:

C
b
c
d
a
b
a
c
b
c

1 c
2 d
3 c
4 d
5 c
6 b
7 a
8 d
9 b
10 c

Hahaha

1) c
2) d
3) c
4) d
5) a
6) a
7) a
8) d
9) b
10) c

lulwut

New Member
I'm hoping I got all 10 correct (aren't we all):

1)c
2)d
3)c
4)d
5)a
6)b
7)a
8)d
9)b
10)c

RealiseNothing

what is that?It is Cowpea
I'm hoping I got all 10 correct (aren't we all):

1)c
2)d
3)c
4)d
5)a
6)b
7)a
8)d
9)b
10)c
Fairly sure question 6 is A.

Magical Kebab

Member
Fairly sure question 6 is A.
I think it is a...
I just though b cos I assumed the whole expression would be divided by 2.

lulwut

New Member
6a was the general formula one right?

2x should = the general formula i think, so i figured x= (general formula) / 2

Yep 6 would be A

Mitsunami

Member
1) c
2) d
3) b
4) d
5) a
6) d
7) a
8) d
9) b
10) c

Realized 6 was wrong after the examiners say the exam's over. *sigh*

RealiseNothing

what is that?It is Cowpea
6a was the general formula one right?

2x should = the general formula i think, so i figured x= (general formula) / 2
The general formula has a $\bg_white 2n\pi$ though, so when you divide by the 2 you just get $\bg_white n\pi$

lulwut

New Member
Yep 6 would be A
oh wow, just saw the sin-1(a) not a... LOL aw man how did i miss that

alexandred

New Member
6 is b. Consider the graph of y=sinx; for any given vertical line y=a (|a|<=1) the solutions for intersections are: x = 2kpi + arcsin(a), x = (2k+1)pi - arcsin(a), or equivalently x = kpi + (-1)^k*arcsin(a). Divide this by 2 for 2x

RealiseNothing

what is that?It is Cowpea
Cambridge 3U year 12, page 33:

"The general solution of $\bg_white \sin(x)=a$ is $\bg_white x=\sin^{-1}(a) + 2n\pi$"

Divide by 2 gets A doesn't it?

Note the (-1)^k

lulwut

New Member
The general formula has a $\bg_white 2n\pi$ though, so when you divide by the 2 you just get $\bg_white n\pi$
general formula is just pi not 2pi for sin though. and its sin^-1(a) not just a, so sin^-1(a) represents 2x which when divided by 2 is equal to x, which works with the general formula where you have to have the angle at the end. I'm pretty sure it's A) now.

Randox

Member
qn 2 is b isnt it?

lulwut

New Member
general formula is just pi not 2pi for sin though. and its sin^-1(a) not just a, so sin^-1(a) represents 2x which when divided by 2 is equal to x, which works with the general formula where you have to have the angle at the end. I'm pretty sure it's A) now.
Im back to thinking it's B) again.. LOL

sin(2x)=a so arcsin(a)=2x

but lets do it in general for sinx first. Sinx=sina so x=a

x=npi+(-1)^n*a

sub in x as 2x and a as arcsin(a)

2x=npi+(-1)^n*arcsin(a)

so x=(npi+(-1)^n*arcsin(a)) / 2 which is answer b)

kewon

Member
Im back to thinking it's B) again.. LOL

sin(2x)=a so arcsin(a)=2x

but lets do it in general for sinx first. Sinx=sina so x=a

x=npi+(-1)^n*a

sub in x as 2x and a as arcsin(a)

2x=npi+(-1)^n*arcsin(a)

so x=(npi+(-1)^n*arcsin(a)) / 2 which is answer b)
pls be b...
in the test i was thinking dam could be a trick and could be a but i cbf spending more time on it so i just hoped for the best...