VCE Methods Question Thread!

[Qwerty1234qwerty]

Thanks   
In binomial and discrete situations for Pr(x>1) the lower bound is 2
Bit in continuous and normal situations for Pr(x>1) the lower bound is 1
Am I correct?

Yes

[sailinginwater]

Yes

Thanks
Do you know of finding antiderivative of things like y = 1/(1-x) is still on the current study design?

[S_R_K]

Thanks
Do you know of finding antiderivative of things like y = 1/(1-x) is still on the current study design?

Yes, finding anti-derivatives of any function of the form f(x) = 1/(ax+b), where a and b are any real numbers, is assessable

[Lear]

Yes, finding anti-derivatives of any function of the form f(x) = 1/(ax+b), where a and b are any real numbers, is assessable

I have noted VCAA has removed the Absolute value sign usually written when we antiderive a function in the form of 1/(ax+b) in the formula sheet. Looking at the formula sheet of, say, 2013 the anti derivative of 1/x is written as loge(|x|) + c whereas last year's is simply loge(x) +c. Any idea what is up with that? If we are required to write the anti derivative of 1/x do we include the absolute sign or not?

[Sine]

I have noted VCAA has removed the Absolute value sign usually written when we antiderive a function in the form of 1/(ax+b) in the formula sheet. Looking at the formula sheet of, say, 2013 the anti derivative of 1/x is written as loge(|x|) + c whereas last year's is simply loge(x) +c. Any idea what is up with that? If we are required to write the anti derivative of 1/x do we include the absolute sign or not?

they removed absolute function from the study design - you used to have to be able to sketch and use them in years before 2016.
So you don't need to put the absolute sign now and probably could just put a domain in to make it accurate.
You probably won't lose any marks if you put the absolute sign in imo

[sailinginwater]

How would you find the antiderivative of 1/(2x+1)?

[Bri MT]

How would you find the antiderivative of 1/(2x+1)?

g(x) = ln(f(x))
g'(x) = f'(x)/f(x)
[ f(x) is of the form mx+c ]

so if we say that f(x) = 2x+1
we would get ln(2x+1) ->   2/(2x+1)
What can number can we multiply  2/(2x+1) by to get 1/(2x+1) ?

[fun_jirachi]

You manipulate the equation like so:

Then you go like normal
Answer in the spoiler

Spoiler

EDIT: miniturtle beat me to it by a few seconds because my internet was too slow

[sailinginwater]

Do we still need to put the absolute sign?

[Yertle the Turtle]

Do we still need to put the absolute sign?

I believe that it isn't a part of the study design, instead you are meant to be able to recognise whether it is positive or negative, due to the graph of ln(x), which is never negative.

[fun_jirachi]

I'm not sure since I'm not Victorian and I dunno what you guys study But in NSW its every time . Imagine you're trying to take the integral of a hyperbola (which is what the 1/(2x+1) is) to the left of the y-axis, it would be undefined without the absolute value on the log in the primitive function, since you cant have the log of a negative. The absolute value on the log removes this problem and that's why it's necessary.

Hope this helps

[sailinginwater]

Do we need to know how to solve cos(x)+cos(3x)=1/2 in exam 1?

[minhalgill]

pls help

[Lear]

pls help

What particularly are you struggling with? Recall that the average value is 1/x2-x1 multiplied the area under a function. Try applying it here.

[Azim.m]

If a graph  has only a stationary point of inflection is it still considered to be one to one?