VCE Methods Question Thread!

[Bri MT]

How would I find the average value over the interval -a to a ?

Find the area by splitting into two shapes (triangle & square) rather than by integrating.
Area=.5a(2a) + a^2 = 2a^2

Then divide by a-(-a) which equals 2a

This gives you the answer of a

[integrationbyrecognition]

For the "odd" and "even" area questions, did it specify to have the answer in terms of n? If I used n = 0 and n = 1 for even and odd respectively would i get 1/2 for each, since i obviously have the wrong answer, but still used the correct method of evaluating a definite integral?
For the 3 mark area question, if i just stated top terminal - bottom terminal (function) dx, but didn't evaluate the definite integral, would there be a chance of 1/3?
I'm hoping the cutoff is 33-34 for an a+
Hoping a better exam 2 will follow today, good luck everyone

[integrationbyrecognition]

bump
For the "odd" and "even" area questions, did it specify to have the answer in terms of n? If I used n = 0 and n = 1 for even and odd respectively would i get 1/2 for each, since i obviously have the wrong answer, but still used the correct method of evaluating a definite integral?
For the 3 mark area question, if i just stated top terminal - bottom terminal (function) dx, but didn't evaluate the definite integral, would there be a chance of 1/3?
I'm hoping the cutoff is 33-34 for an a+
Hoping a better exam 2 will follow today, good luck everyone

[darkz]

bump
For the "odd" and "even" area questions, did it specify to have the answer in terms of n? If I used n = 0 and n = 1 for even and odd respectively would i get 1/2 for each, since i obviously have the wrong answer, but still used the correct method of evaluating a definite integral?
For the 3 mark area question, if i just stated top terminal - bottom terminal (function) dx, but didn't evaluate the definite integral, would there be a chance of 1/3?
I'm hoping the cutoff is 33-34 for an a+
Hoping a better exam 2 will follow today, good luck everyone

Well I'd assume that the mark allocation would be 1 for writing the integral, and 1 for evaluating it. Since you cannot assume that n is a specific constant, since, you could lose both answer marks. However, if you wrote the integral with the endpoints including 'n', you may score a mark for that.

[integrationbyrecognition]

Well I'd assume that the mark allocation would be 1 for writing the integral, and 1 for evaluating it. Since you cannot assume that n is a specific constant, since, you could lose both answer marks. However, if you wrote the integral with the endpoints including 'n', you may score a mark for that.

But how are we meant to know that n can't be a specific constant when the question never specified it?
Also, for the 3 mark area question, if i just stated top terminal - bottom terminal (function) dx, but didn't evaluate the definite integral, would there be a chance of 1/3?
is it possible for the a+ cutoff to be 33-34?
Thanks

[darkz]

But how are we meant to know that n can't be a specific constant when the question never specified it?
Also, for the 3 mark area question, if i just stated top terminal - bottom terminal (function) dx, but didn't evaluate the definite integral, would there be a chance of 1/3?
is it possible for the a+ cutoff to be 33-34?
Thanks

You don't need to know the specific values of n, this is why they've separated the question into two cases. Lets use the example of sin(xpi). For X is even or 0, sin(0), sin(2pi)...etc will always =0. That way, you can evaluate the trig functions. The 3 mark area question was an integration by recognition question. Therefore, I doubt you'll get the mark from simply copying down the answer from the part above, since you cannot actually use that integral to evaluate the area. However, if you had the correct integral which you can evaluate, then you'd get a mark for that

[passbleh24]

Find the area by splitting into two shapes (triangle & square) rather than by integrating.
Area=.5a(2a) + a^2 = 2a^2

Then divide by a-(-a) which equals 2a

This gives you the answer of a

Hey mini turtle,

Thats exactly what I did.. But the answer is a/2 sorry I should have stated that.

[passbleh24]

Hey could someone please help me understand the difference between options C and D. The answer is D but I thought that they both would lead to the second graph.

Thanks

[integrationbyrecognition]

You don't need to know the specific values of n, this is why they've separated the question into two cases. Lets use the example of sin(xpi). For X is even or 0, sin(0), sin(2pi)...etc will always =0. That way, you can evaluate the trig functions. The 3 mark area question was an integration by recognition question. Therefore, I doubt you'll get the mark from simply copying down the answer from the part above, since you cannot actually use that integral to evaluate the area. However, if you had the correct integral which you can evaluate, then you'd get a mark for that

omg I didn't even recognize it was integration by recognition haha, evidence that doing an exam under exam conditions is not equal to doing an exam at home, and yeah I just copied the answer from the part above but couldn't go any further, so unlikely to even get 1/3?
And for the odd and even questions since i substituted in values for n i'm unlikely to get any marks here aswell?
is exam 2 likely to be hard as well? since apparently yesterday's exam 1 was similar to nht standard according to my mates who have done nht exams.
a+ cutoff predictions likely to be 33-34?

[darkz]

omg I didn't even recognize it was integration by recognition haha, evidence that doing an exam under exam conditions is not equal to doing an exam at home, and yeah I just copied the answer from the part above but couldn't go any further, so unlikely to even get 1/3?
And for the odd and even questions since i substituted in values for n i'm unlikely to get any marks here aswell?
is exam 2 likely to be hard as well? since apparently yesterday's exam 1 was similar to nht standard according to my mates who have done nht exams.
a+ cutoff predictions likely to be 33-34?

Don't think you'd get the marks if you subbed in the value for 'n' at the very start of the question, and no marks for the integration one unfortunately
In my opinion, since this exam was rather quite creative, I'm sure they'll be some challenging questions on the exam 2 paper.

[S200]

Hey could someone please help me understand the difference between options C and D. The answer is D but I thought that they both would lead to the second graph.

Thanks

Option C has that -3, without any reflection in the x-axis, and that is clearly wrong.
The graph of \(f(x)\) has been reflected in the x-axis...
We can see this from the asymptote changing from \(y=2\) to \(y=-2\).

[integrationbyrecognition]

Don't think you'd get the marks if you subbed in the value for 'n' at the very start of the question, and no marks for the integration one unfortunately
In my opinion, since this exam was rather quite creative, I'm sure they'll be some challenging questions on the exam 2 paper.

by "quite creative" do you mean harder than usual?
I sure hope I'm not the only one who found it challenging

[lifeisgood]

hey could someone help me out with this type of question. they always seem to be popping up.
let f be one to one differntiable funtions such that f(3)=7 f(7)=8  f'(3)=2 and f'(7)=3. the function g is differentiable and g(x)=f^-1(c) for all x

g'(x) is equali to

a) 1/2
b) 2
c) 1/6
d) 1/8
e) 1/3

[MB_]

Hey mini turtle,

Thats exactly what I did.. But the answer is a/2 sorry I should have stated that.

The linear part of the graph has an average value of 0.

[passbleh24]

Option C has that -3, without any reflection in the x-axis, and that is clearly wrong.
The graph of \(f(x)\) has been reflected in the x-axis...
We can see this from the asymptote changing from \(y=2\) to \(y=-2\).

Hey wouldnt the -y be a reflection in the x axis ? According to my notes it is in option C and D.