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March 28, 2024, 09:07:39 pm

Author Topic: VCE Methods Question Thread!  (Read 4802224 times)  Share 

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Syndicate

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Re: VCE Methods Question Thread!
« Reply #15000 on: June 26, 2017, 10:34:03 pm »
+2
(Image removed from quote.)
Q16 and 17 pls

Q16) d/dx(5xcos(2x) = 5cos(2x) -10xsin(2x)

Therefore int(5cos(2x) - 10xsin(2x) = int(d/dx(5xcos(2x))
int(5cos(2x) +5int(-2xsin(2x) = 5xcos(2x) + c
5int(-2xsin(2x)) = 5xcos(2x) - int(5cos(2x) + c
therefore int(-2xsin(2x)) = xcos(2x) - 5/2sin(2x)

Q17 can be done in a similar method
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MisterNeo

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Re: VCE Methods Question Thread!
« Reply #15001 on: June 26, 2017, 11:07:08 pm »
+3
(Image removed from quote.)
Q16 and 17 pls


Use product rule.
Let u=5x
Let v=cos2x








Sub π/4 into the LHS and subtract 0 because subbing 0 would give 0.
∴The final answer would be:



But the question asks to evaluate NEGATIVE -2xsin2x, so the answer would be...

I think that's correct. Answers would be nice. :)
(I'll do q17 next)
« Last Edit: June 26, 2017, 11:36:02 pm by MisterNeo »

MisterNeo

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Re: VCE Methods Question Thread!
« Reply #15002 on: June 26, 2017, 11:34:47 pm »
+4
(Image removed from quote.)
Q16 and 17 pls

For Q17...

Use product rule for xsinx.
Let u=x.
Let v=sinx.






Sub the 1 and 0, then subtract...



Hope this helps :)
« Last Edit: June 26, 2017, 11:36:59 pm by MisterNeo »

lzxnl

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Re: VCE Methods Question Thread!
« Reply #15003 on: June 27, 2017, 01:19:12 am »
+2

Use product rule.
Let u=5x
Let v=cos2x








Sub π/4 into the LHS and subtract 0 because subbing 0 would give 0.
∴The final answer would be:



But the question asks to evaluate NEGATIVE -2xsin2x, so the answer would be...

I think that's correct. Answers would be nice. Hope this helps :)
(I'll do q17 next)


Use \sin{(x)} instead of sin x; it'll look like
Also, all integrals need the 'dx' to make sense.
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Quantum44

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Re: VCE Methods Question Thread!
« Reply #15004 on: June 27, 2017, 07:23:08 am »
+2
If the gradient graph (as in the derivative) crosses the x-intercept, that means there is a local maximum or a local minimum on the actual graph. So when f'(x) = 0, there is a local max/min on f(x).

I am assuming no other information is given (I am also going to assume that this is a positive cubic function). Therefore the local min is at x = 1, as the max occurs more to the left of the x-axis.

I'm a bit confused. If the gradient function has a maximum value of 6, wouldn't it mean the gradient function is a negative quadratic and hence the actual function must be a negative cubic?
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Syndicate

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Re: VCE Methods Question Thread!
« Reply #15005 on: June 27, 2017, 11:38:15 pm »
+4
I'm a bit confused. If the gradient function has a maximum value of 6, wouldn't it mean the gradient function is a negative quadratic and hence the actual function must be a negative cubic?

Yea, you're right. I misread the question (I thought the maximum value of f(x) was 6 :P).

Since f'(x) has got a local maximum, it is a negative quadratic, which implies that f(x) is a negative cubic. This means that the local min occurs at x=-2.

(Thanks for correcting me)
« Last Edit: June 27, 2017, 11:46:32 pm by Syndicate »
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TheCommando

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Re: VCE Methods Question Thread!
« Reply #15006 on: June 28, 2017, 09:47:09 am »
+2
If f(x) is the function and it touches 2y+6x=15 at (0,15/2) and has a stationary point at (3,-6). Then it means that f(0)=15/2, f(3)=-6, f'(3)=0 and f'(0)=gradient of the line. From that you get simultaneous equations and solve for a,b,c and d. Hope that helps.  :)
Not sure, are u plugging thoose value into the equation of 2y+6x=15
Where u areange it for y =mx+c?

Edit: kkk nvm i got it . Thanks man :)
« Last Edit: June 28, 2017, 09:52:46 am by TheCommando »

Perryman

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Re: VCE Methods Question Thread!
« Reply #15007 on: June 28, 2017, 04:33:03 pm »
0
heyy......i need help on question 3 and 5....
have noo idea where to start...any guidance will be great

TheCommando

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Re: VCE Methods Question Thread!
« Reply #15008 on: June 28, 2017, 04:43:10 pm »
0
Mdo u solve for x withgout a cas
Ythe ggott x =-2,-3 and 0

tryinng to upload pictures directly  from here is so shitt. Doesnt work
http://imgur.com/a/8Li37

zhen

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Re: VCE Methods Question Thread!
« Reply #15009 on: June 28, 2017, 04:45:15 pm »
+2
heyy......i need help on question 3 and 5....
have noo idea where to start...any guidance will be great
I'm just going to assume these exist, but when you do it properly the range of the inside function must be a subset of the domain of the outside function
For 3 here's the start of it


I hope this is right. Someone should check just in case I stuffed something up.
For 5 expand out the matrix and get x and y in terms of x' and y' and then just substitute it into the equation. So you get and equation in terms of x' and y' instead of y and x and then rearrange to get y' by itself.
Hope that helps.  :)
« Last Edit: June 28, 2017, 04:50:21 pm by zhen »

zhen

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Re: VCE Methods Question Thread!
« Reply #15010 on: June 28, 2017, 04:52:37 pm »
+1
Mdo u solve for x withgout a cas
Ythe ggott x =-2,-3 and 0

tryinng to upload pictures directly  from here is so shitt. Doesnt work
http://imgur.com/a/8Li37
For that x=0 and -2. Becuse if x=0 0(0+2)^2=0 and when x=-2 -2(-2+2)^2=-2(0)^2=0
I don't think that x=-3 is right.
« Last Edit: June 28, 2017, 04:54:24 pm by zhen »

TheCommando

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Re: VCE Methods Question Thread!
« Reply #15011 on: June 28, 2017, 09:37:04 pm »
0
For that x=0 and -2. Becuse if x=0 0(0+2)^2=0 and when x=-2 -2(-2+2)^2=-2(0)^2=0
I don't think that x=-3 is right.
But how do u know to sub in x=0 and x=-2 without knowing the answer
Dont u just set x and (x+2)^2 to both equal 0 and sove for x
Also they had -3 as an answer as well i think

zhen

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Re: VCE Methods Question Thread!
« Reply #15012 on: June 28, 2017, 09:45:52 pm »
+2
But how do u know to sub in x=0 and x=-2 without knowing the answer
Dont u just set x and (x+2)^2 to both equal 0 and sove for x
Also they had -3 as an answer as well i think
Yea, just set it up so x or (x+2)^2 equal 0, because that will make the whole thing equal 0.

TheCommando

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Re: VCE Methods Question Thread!
« Reply #15013 on: June 28, 2017, 09:49:25 pm »
0
Yea, just set it up so x or (x+2)^2 equal 0, because that will make the whole thing equal 0.
It says on the answers apparently x =-3
Wtf?

zhen

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Re: VCE Methods Question Thread!
« Reply #15014 on: June 28, 2017, 10:02:05 pm »
+2
It says on the answers apparently x =-3
Wtf?
They are subbing in x=-3 into the equation of the derivative and say that at that point the derivative is <0. They also sub in x=-1 after. I suspect that they're doing this to find the nature of the stationary point (whether it's a local minimum/local maximum/point of inflection). So x=-3 isn't a solution of the equation of the derivative when it equals zero.
« Last Edit: June 28, 2017, 10:04:50 pm by zhen »