VCE Methods Question Thread!

[S200]

Methods exam 1 2010

2.b. find the antiderivative of 1(1-x) and evaluate it from 3 to 2

How would we find the antiderivative without using absolute values?

If \(\int{\frac{1}{x}}=\log_e{x}\), shouldn't \(\int{\frac{1}{1-x}}=\log_e{(1-x)}\)?
Hence, \(-(\log_e{(1-3)}) + \log_e{(1-2)}=\log_e{\frac{1-2}{1-3}} \therefore p=\frac{1}{2}\)...

Edit: Hmmm. The exam report disagrees...

Double edit; It doesn't now...

[VinnyD]

If \(\int{\frac{1}{x}}=\log_e{x}\), shouldn't \(\int{\frac{1}{1-x}}=\log_e{(1-x)}\)?
Hence, \(\log_e{(1-3)} - \log_e{(1-2)}=\log_e{\frac{1-3}{1-2}} \therefore p=2\)?

Edit: Hmmm. The exam report disagrees...

You need to divide the entire log function by the coefficient of the x term, in this case it is -1 therefore the function is -ln(1-x)
usually any anti derivative of 1(/ax+b)= 1/a(ln(ax+b))  +c

[S200]

You need to divide the entire log function by the coefficient of the x term, in this case it is -1 therefore the function is -ln(1-x)
usually any anti derivative of 1(/ax+b)= 1/a(ln(ax+b))  +c

Thanks! I'll highlight that in my summary book!

[sailinginwater]

I know how to work out the basic exact values for sin, cos and tan, but i've seen questions on exam 1 involving things like sin(-3pi/4), cos(pi), etc

How would these be done?
no calculator allowed!!

Thanks

[GMT. -_-]

Aren't those just trig identities that you need to understand/memorise

[Sine]

I know how to work out the basic exact values for sin, cos and tan, but i've seen questions on exam 1 involving things like sin(-3pi/4), cos(pi), etc

How would these be done?
no calculator allowed!!

Thanks

as stated above everyone basically memorises them

or you can "derive" them but that's not time efficient in an exam.

[sailinginwater]

Aren't those just trig identities that you need to understand/memorise

Can't really memorise them, because what it we get something like cos(100pi/5) on exam 1?

[Sine]

Can't really memorise them, because what it we get something like cos(100pi/5) on exam 1?

cos(100pi/5) = cos(20pi) = cos(0) = 1

They could give you stuff like that in exam 1 e.g. cos(33pi), sin(29.5pi)

[sailinginwater]

How do you know that cos(20pi) = cos(0)?

[S200]

How do you know that cos(20pi) = cos(0)?

Because \(20\pi\) is just 10 whole revolutions around the unit circle, arriving back at the start (i.e: 0)...
1 rev = \(2\pi\), 10 rev = \(20\pi\)...

[Lear]

Knowing how the unit circle works really REALLY helps when solving questions such as these.

[sailinginwater]

Is one whole revolution around the unit circle 2pi?

[darkz]

Is one whole revolution around the unit circle 2pi?

Yes, 2pi = 360 degrees

[sailinginwater]

Yes, 2pi = 360 degrees

So that means that cos(2pi), cos(4pi), cos(6pi) etc all equal cos(0)?

[darkz]

So that means that cos(2pi), cos(4pi), cos(6pi) etc all equal cos(0)?

Yes