VCE Methods Question Thread!

[minhalgill]

in this question p has been made the subject of the formula. could we do that for q as well?

[minhalgill]

can someone please help me with part (b)

[S_R_K]

How does one go about solving this in a manner that is realistically doable under test conditions on a tech active exam? My solution was to apply the formula used to calculate the confidence interval bounds in relation to p̂ (in this case 1.96*sqrt(p̂(1-p̂)/n) ) and rearrange, however with the values presented I just can't believe this is the easiest method.

Thanks in advance!

The method you describe is the easiest way (it's just a straightforward use of the formula). The numbers aren't too bad. Probably the trickiest thing under exam conditions is recognising that 147 = 3 * 49.

can someone please help me with part (b)

This is anti-diff by recognition. From part (a) you should find that the derivative of x*log((x^2)/4) is log((x^2)/4) + 2. Hence, the antiderivative of log((x^2)/4) is equal to the antiderivative of log((x^2)/4) + 2 – 2, which is x*log((x^2)/4) – 2x. Once you know the antiderivative, you can find the definite integral between x = –2 and x = –1.

[sailinginwater]

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

Can you please give an example?
Thanks

[Brittank88]

Hello again!

Another question, this time understanding what I believe is a question that must be solved via integration by recognition.

In part a) you must find the derivative of x * cos(2x), which I believe is cos(2x) - 2x * sin(2x). Part b) states "hence, evaluate.." the integral from 0 to Pi/6 of 2x * sin(2x) dx.

Part b) is where I can understand the link to part a) whereby the derivative contains 2x * sin(2x), however I don't actually know how to go on from there.

Thanks!

[minhalgill]

in this question it states that pr(A'∩ B')= Pr(A') x Pr(B').
can i apply this to any question for independant events? (is this always the case for independant events or is this only relevant to this question?)

thanks in advance

[DBA-144]

in this question it states that pr(A'∩ B')= Pr(A') x Pr(B').
can i apply this to any question for independant events? (is this always the case for independant events or is this only relevant to this question?)

thanks in advance

Yeah as long as they are independent.

This is an excellent way to memorise these values. Thanks for your contribution
Another thing that helps me is to know that Tan(x) is simply the numerator of sin(x) divided by the numerator of cos(x)
Example - Tan(pi/6)
Sin(pi/6) = 1/2
Cos(pi/6) = sqrt3/2
Therefore Tan(pi/6) is simply 1 (numerator of sin) divided by sqrt3(numerator of cos)
 = 1/sqrt3

Even for Tan(pi/2) it is simply 1 / 0 which is undefined

Just a reminder that we need to be expressing nswers in simplest form. This means rationalizing the denominator when dealing wth surds. I imagine it would be quite easy to lose marks for this.

[minhalgill]

how would we find the coordinates of P on this graph? im having trouble solving tan(x) = √3cos(x)?

[minhalgill]

Yeah as long as they are independent.
 

Just a reminder that we need to be expressing nswers in simplest form. This means rationalizing the denominator when dealing wth surds. I imagine it would be quite easy to lose marks for this.

do you know how this pr(A'∩ B')= Pr(A') x Pr(B') comes about?

[fun_jirachi]

Hello again!

Another question, this time understanding what I believe is a question that must be solved via integration by recognition.

In part a) you must find the derivative of x * cos(2x), which I believe is cos(2x) - 2x * sin(2x). Part b) states "hence, evaluate.." the integral from 0 to Pi/6 of 2x * sin(2x) dx.

Part b) is where I can understand the link to part a) whereby the derivative contains 2x * sin(2x), however I don't actually know how to go on from there.

Thanks!

Hi, just gonna jump in and answer this



So you can see that the thing on the far right is essentially the same as the thing on the far left.

\


Should be correct, would be great if someone verified

Hope I helped

[minhalgill]

This is anti-diff by recognition. From part (a) you should find that the derivative of x*log((x^2)/4) is log((x^2)/4) + 2. Hence, the antiderivative of log((x^2)/4) is equal to the antiderivative of log((x^2)/4) + 2 – 2, which is x*log((x^2)/4) – 2x. Once you know the antiderivative, you can find the definite integral between x = –2 and x = –1.

im still very confused about this....could you please elaborate?
thanks in advance

[fun_jirachi]

how would we find the coordinates of P on this graph? im having trouble solving tan(x) = √3cos(x)?

Just gonna go ahead and assume you know that the curve that cuts through the origin is tan and the other one is the other one

Using the quadratic formula you get that:

Since the first solution is out of the bounds of the function sin x, theres only one solution, which is:


Sub it into either tan x or root 2 cos x and you have your answer, which should be (pi/4, 1)

EDIT:

im still very confused about this....could you please elaborate?
thanks in advance

Since every derivative has a corresponding integral, eg like you differentiate something, integrate it back, and you get the same thing plus a constant. So basically what SRK's done there is just manipulated the integral given so it looks a little like the derivative of the thing you were given in part a. Since there was a +2 in the derivative, but everything remains the same, he just adds two to get the same thing, then subtracts two to keep the integral the same. From there, he uses the thing I first talked about, going straight to the thing you were asked to differentiate in part a), minus the 2x added on when manipulating the equation. So you've found the integral of the thing in part b, and with boundaries you should be able to solve from there.

[minhalgill]

in part (b) how do you obtain the dilation? i tried making an equation and got y=-(2/7)*(x+2)^2 -3,
which is a dilation of 2/7, yet they have a dilation of 2.
can someone please explain how they got a dilation of 2.

[DBA-144]

in part (b) how do you obtain the dilation? i tried making an equation and got y=-(2/7)*(x+2)^2 -3,
which is a dilation of 2/7, yet they have a dilation of 2.
can someone please explain how they got a dilation of 2.

Hopefully, you know what/how to do completing the square. If you dont, then it is highly important you learn how to do this. If you do:

When completing the square we get: y= 2 (x+2)^2  - 3.

From this, we know there is a dilation of 2 from the x axis, reflection in x axis and translation 2 units left 3 units down.
 

In case you dont understand how to get to the turning point form,

y= -2x^2 -8x - 11
= -2 ((x^2 +4x +4) -4 + 11/2)
= -2( (x+2)^2 + 3/2)
= -2(x+2)^2 -3

Done. Hope this helps

[S_R_K]

do you know how this pr(A'∩ B')= Pr(A') x Pr(B') comes about?

Recall the formula for conditional probability: Pr(E | F) = Pr(E & F)/Pr(F).

The idea behind independent events is that the occurrence of one event does not affect the probability of the other event. Hence, if E and F are independent events, then Pr(E | F) = Pr(E) (that is, the probability of E occurring is not affected by the occurrence of F).

Hence, if E and F are independent, the formula for conditional probability gives: Pr(E | F) = Pr(E) = Pr(E & F)/Pr(F). And then rearranging gives Pr(E)Pr(F) = Pr(E & F).