VCE Methods Question Thread!

[Orb]

How are you expected to be able to graph this graph without a calculator.
How would you find the endpoints and xintercepts and yintercepts

First, we see that the amplitude is 5, because the max and min of any non-dilated cosine graph is [-1,1]

The period is calculated by 2pi/n, so the period in our case is 2pi/2 (the number before x) = pi

That means we're going through two revolutions of the graph through the domain [-pi,pi]

To find the endpoints, we sub in x= -pi and x = pi into the equation, then solve.
f(pi) = 5cos(2(4pi/3)) = 5cos(8pi/3) = 5cos(2pi + 2pi/3) = 5cos(2pi/3) = -5/2

To find the x-intercepts, we sub in y=0 into the equation, so 5cos(2(x+pi/3)) = 0
cos(2(x+pi/3))=0
inverse cos (0) = 2(x+pi/3) = pi/2, 3pi/2, 5pi/2, -pi/2
x+pi/3 = pi/4, 3pi/4, 5pi/4, -pi/4
x= -pi/12, 5pi/12, 11pi/12, -7pi/12

To find the y-intercepts we sub in x=0 into the equation and do the same process

And then to graph you just have to know how to draw a regular cos graph and use the endpoints and intercepts as a guide!

[knightrider]

First, we see that the amplitude is 5, because the max and min of any non-dilated cosine graph is [-1,1]

The period is calculated by 2pi/n, so the period in our case is 2pi/2 (the number before x) = pi

That means we're going through two revolutions of the graph through the domain [-pi,pi]

To find the endpoints, we sub in x= -pi and x = pi into the equation, then solve.
f(pi) = 5cos(2(4pi/3)) = 5cos(8pi/3) = 5cos(2pi + 2pi/3) = 5cos(2pi/3) = -5/2

To find the x-intercepts, we sub in y=0 into the equation, so 5cos(2(x+pi/3)) = 0
cos(2(x+pi/3))=0
inverse cos (0) = 2(x+pi/3) = pi/2, 3pi/2, 5pi/2, -pi/2
x+pi/3 = pi/4, 3pi/4, 5pi/4, -pi/4
x= -pi/12, 5pi/12, 11pi/12, -7pi/12

To find the y-intercepts we sub in x=0 into the equation and do the same process

And then to graph you just have to know how to draw a regular cos graph and use the endpoints and intercepts as a guide!

Thanks hamo 94

[knightrider]

How would you do these questions?

Find an antiderivative of

  and

[lenaaa_]

Hi, I have a quick question based on the topic of binomial distributions probability

It is found that there are 10 houses on the round for which the paper boy’s accuracy is given exactly by the distribution above. Therefore the probability of hitting the front step of any one of these 10 houses three or four times a week is 0.4.

Find the probability (correct to 4 decimal places) that out of the 10 houses he hits the front step of exactly four particular houses three or four times a week.

[Orb]

How would you do these questions?

Find an antiderivative of

  and

What I do is just expand the function then find the antiderivative

so for


You antiderive each one, one-by-one

so

becomes
becomes
and 7 becomes 7x

Remember there will always be a '+c' at the end when you're finding the antiderivative!

[Phy124]

Remember there will always be a '+c' at the end when you're finding the antiderivative!

Just a reminder that if you're asked to find "an" antiderative the "+C" is not necessary.

Any function that when derived gives f(x) is a suitable solution to the question, the simplest of which would not contain any constant. (e.g. is an antiderivative of f(x), but is a simpler antiderivative of f(x), so it is standard to write the latter)

[Orb]

Just a reminder that if you're asked to find "an" antiderative the "+C" is not necessary.

Any function that when derived gives f(x) is a suitable solution to the question, the simplest of which would not contain any constant. (e.g. is an antiderivative of f(x), but is a simpler antiderivative of f(x), so it is standard to write the latter)

Yes, my bad, I didn't see the 'an' in the original question

[Reus]

Could someone please go through this question? I stuffed up towards the end resulting in a different answer .__.

[Zealous]

Could someone please go through this question? I stuffed up towards the end resulting in a different answer .__.

For two lines to be parallel, they need to have the same gradient. So the answer will be any linear line such that it's gradient is , in other words .

[IndefatigableLover]

Hi, I have a quick question based on the topic of binomial distributions probability

It is found that there are 10 houses on the round for which the paper boy’s accuracy is given exactly by the distribution above. Therefore the probability of hitting the front step of any one of these 10 houses three or four times a week is 0.4.

Find the probability (correct to 4 decimal places) that out of the 10 houses he hits the front step of exactly four particular houses three or four times a week.

I'm kind of unsure of my answer here (as I am with any probability question LOL) but using the value you just gave (0.4), that is the probability of hitting the front step of any one of these 10 houses three or four times a week. It can also be said that the probability of NOT hitting the front step of any one of these 10 houses three or four times a week is 0.6.

Because they're just looking for four particular houses, you can just go: (disregard combinations since you're not looking for any four houses but just particular ones).




Correct to 4 decimal places, your final answer would be 0.0012

On the contrary, if you were to be looking for any four houses, then you would need to put add something else into the equation (that is combinations). You can see that you have 10 houses but only 4 are required so basically if you were to be finding the probability of any house then you would do:







(Just to show the difference between particular and any I guess )

[theBRENDAN97]

Sorry, I'm having trouble with a few differentiation questions (quite basic), i just don't know how to do them:
Find coordinates at gradient:
y=x^3-6x+4; dy/dx=-12

f: R\{2} ->R, where f(x)=4+4/2-x, f '(x)>0 for

Find derivative:
-2/x^1/3
(looked at back of book and had: 2/3x^4/3)

Find derivative:
x^2+x^5 (all over:)
/x^4=
I tried and was no where near correct answer.

Any help with any questions would be greatly appreciated.
Second last day of holdiays thought I'd do my methods homework, been trying to do the entire bio course hasn't worked out again (got family over).

[AirLandBus]

Sorry, I'm having trouble with a few differentiation questions (quite basic), i just don't know how to do them:
Find coordinates at gradient:
y=x^3-6x+4; dy/dx=-12

f: R\{2} ->R, where f(x)=4+4/2-x, f '(x)>0 for

Find derivative:
-2/x^1/3
(looked at back of book and had: 2/3x^4/3)

Find derivative:
x^2+x^5 (all over:)
/x^4=
I tried and was no where near correct answer.

Any help with any questions would be greatly appreciated.
Second last day of holdiays thought I'd do my methods homework, been trying to do the entire bio course hasn't worked out again (got family over).

Find coordinates at gradient:
y=x^3-6x+4; dy/dx=-12

First find the derivative.
dy/dx=3x^2-6
Find x when dy/dx (the derivate - which is the gradient) = -12
3x^2-6 = -12
3x^2 = -6
No x value as you cant have the square root of a negative number.
Sub x=0 into y=x^3-6x+4
y=4
m=-12 at the point (0,4) Which is incorrect. When graphed m=-6 at (0,4)

Are you sure the question is right. Can you confirm?

[keltingmeith]

Sorry, I'm having trouble with a few differentiation questions (quite basic), i just don't know how to do them:
Find coordinates at gradient:
y=x^3-6x+4; dy/dx=-12

f: R\{2} ->R, where f(x)=4+4/2-x, f '(x)>0 for

Find derivative:
-2/x^1/3
(looked at back of book and had: 2/3x^4/3)

Find derivative:
x^2+x^5 (all over:)
/x^4=
I tried and was no where near correct answer.

Any help with any questions would be greatly appreciated.
Second last day of holdiays thought I'd do my methods homework, been trying to do the entire bio course hasn't worked out again (got family over).

Can't answer these while on my phone, but I cannot stress this enough:

Please, please, PLEASE use brackets when writing out questions if your not going to use TeX typesetting - there are about 3 different ways I can interpret each of these questions right now.

[theBRENDAN97]

Find coordinates at gradient:
y=x^3-6x+4; dy/dx=-12

First find the derivative.
dy/dx=3x^2-6
Find x when dy/dx (the derivate - which is the gradient) = -12
3x^2-6 = -12
3x^2 = -6
No x value as you cant have the square root of a negative number.
Sub x=0 into y=x^3-6x+4
y=4
m=-12 at the point (0,4) Which is incorrect. When graphed m=-6 at (0,4)

Are you sure the question is right. Can you confirm?

Yeah sorry, wrote the question out wrong, it was supposed to be y=x^3-6x^2+4, dy/dx=-12
Thanks for trying, I undestand how you got the answer though now. Thanks. Answer was (2,-12)

Can't answer these while on my phone, but I cannot stress this enough:

Please, please, PLEASE use brackets when writing out questions if your not going to use TeX typesetting - there are about 3 different ways I can interpret each of these questions right now.

Yeah looking back the questions were ambiguous, I didn't even know about the TeX typesetting, I'll use brackets next time. Thanks.
The other questions were supposed to be:
f: R\{2}->R, where f(x)= (4+4)/(2 - x), f '(x)>0

Derivative of -2/(x^(1/3)

Derivative of (x^2+x^5)/x^4

I hope i used the brackets right.

[keltingmeith]

Yeah sorry, wrote the question out wrong, it was supposed to be y=x^3-6x^2+4, dy/dx=-12
Thanks for trying, I undestand how you got the answer though now. Thanks. Answer was (2,-12)

Yeah looking back the questions were ambiguous, I didn't even know about the TeX typesetting, I'll use brackets next time. Thanks.
The other questions were supposed to be:
f: R\{2}->R, where f(x)= (4+4)/(2 - x), f '(x)>0

Derivative of -2/(x^(1/3)

Derivative of (x^2+x^5)/x^4

I hope i used the brackets right.

Much better - slightly wrong use of brackets, but I'm pretty sure I know what you're talking about here:

Okay, the first graph doesn't even require a derivative. You can derive it if you want, and check for when f'(x)>0, but it's so much easier to just graph it: https://www.google.com.au/webhp?sourceid=chrome-instant&rlz=1CASMAE_enAU568AU568&ion=1&espv=2&es_th=1&ie=UTF-8#q=4%2B(4)%2F(2%20-%20x)
Look at the graph from left-to-right - if the curve moves up, the f'(x)>0 (as this means it's increasing). You should see that the graph increases for all x - EXCEPT when it jumps from positive infinity to negative infinity. The graph actually isn't defined at this point - it's our asymptote, the one point not defined in the domain they give you.

Second question, just flip the x using exponents:


For the last one, it's tempting to do something like the quotient rule - instead, split up the fraction and go from there: