VCE Methods Question Thread!

[paper-back]

When finding the average value of a function, do we take into consideration that there is a portion under the x axis?
E.g. find the average value of for

[lzxnl]

When finding the average value of a function, do we take into consideration that there is a portion under the x axis?
E.g. find the average value of for

No. Take the function y=x. Does it make sense that the average value of this function for -1<x<1 is 0? It looks symmetrical about x=0, that's for sure. If you made amends for the change of sign of the function, your integral would be positive and you wouldn't get zero as your average. Your integral must evaluate to zero.

[LiquidPaperz]

if 0 degrees < a < 360 deg, then the equation sin(a) = cos(a) has 2 solutions, 45 and 225 how??

also

if 0 degrees <equalto a <equalto 180, and sin(a) = cos(a), then a is equal to 45....how?


note: i havent done radians yet just unit circle


thanks

[keltingmeith]

If we have sin(a)=cos(a), we can divide by cos(a) to collect our terms together and then use some circle magic:



Now, solving for a, where gives that . So, can you tell me why the next one is only equal to 45 when we restrict the domain?

[LiquidPaperz]

If we have sin(a)=cos(a), we can divide by cos(a) to collect our terms together and then use some circle magic:



Now, solving for a, where gives that .

i can understand how you got the 1, but how did you get the 45 and 225? i dont quite understand that

So, can you tell me why the next one is only equal to 45 when we restrict the domain?

im not sure, the question was, as stated above, and the answer was C, which was 45 degrees.

[Rishi97]

How can I antidifferentiate cos(x)e^sin(x)  ?
Thanks guys :-)

[keltingmeith]

i can understand how you got the 1, but how did you get the 45 and 225? i dont quite understand that

Exact values, my friend - you need to know them. If , then for .

How can I antidifferentiate cos(x)e^sin(x)  ?
Thanks guys :-)

In specialist, you'd do a substitution - for methods, you can just recognise that . So, what you do is divide by the derivative of the power, and leave the exponential as is:



Using a substitution:

[Rishi97]

Thanks EulerFan
I think I'll stick to the methods way 

[soNasty]

Find the value of a and b so that the tangent to the curve y=ax^2+bx+11 at x=2 has the equation y=10x-1
Answers: a= 3, b=-2

[keltingmeith]

Find the value of a and b so that the tangent to the curve y=ax^2+bx+11 at x=2 has the equation y=10x-1
Answers: a= 3, b=-2

This one's tricky, so follow along:
We have two variables, so we need two points. We know that at x=2, the gradient is 10 (from the tangent line), so that means that if y=f(x), f'(2)=10. But how do we get the other point?

Well, the tangent line touches the curve at exactly one point - at x=2. So, the tangent and the curve must share this point, so sub this into the tangent line to get y = 10(2) - 1 = 19. So, f(2)=19.

We now have two points, and can solve simultaneously:

[soNasty]

Thanks for that! I forgot that at that point the tangent and the curve shared a point!

[soNasty]

Using Z~N(0,1) find k such that
Pr(|Z|>k)=0.05
Pr(|Z|>=k)=0.1

Is there a logical approach in tackling these questions? They confuse me lol
I get the ones that ask for when Z is less than k but I'm not that sure with there ones.

[TrueTears]

Since is standard normal, then follows a half-normal distribution.

[brightsky]

use symmetry properties.

Pr(|Z|>k) = 0.05
2Pr(Z<-k) = 0.05
2(1-Pr(Z<k)) = 0.05
1 - Pr(Z<k) = 0.05/2 = 0.025
Pr(Z<k) = 1 - 0.025 = 0.975

and then use inverse normal function on your cas. same process for the second question.

[soNasty]

Thank you both!! Alright I get it now.