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C4 today...Summer 2013

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Any one got the integration using substitution part right? the show part (part a)
I think I will lose a mark because I got integral of 2u/u(u-1) du
but then I used the 2/u(u-1) for solving part b
 
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Any one got the integration using substitution part right? the show part (part a)
I think I will lose a mark because I got integral of 2u/u(u-1) du
but then I used the 2/u(u-1) for solving part b

I remember having to use partial fractions.
I wasn't sure on whether to change the limits? After part (a) they gave the equations again and the limits were 9&1 but the limits were on the equation with x's in it (LHS). I thought that you may need to change limits as we used substitution?
 
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107
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28
I remember having to use partial fractions.
I wasn't sure on whether to change the limits? After part (a) they gave the equations again and the limits were 9&1 but the limits were on the equation with x's in it (LHS). I thought that you may need to change limits as we used substitution?
all that is in part b; I used partial fractions and changed limits
 
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I remember having to use partial fractions.
I wasn't sure on whether to change the limits? After part (a) they gave the equations again and the limits were 9&1 but the limits were on the equation with x's in it (LHS). I thought that you may need to change limits as we used substitution?

The limits were given with u so you donthve to change them
 
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The limits were given with u so you donthve to change them

Yes but they were given to you on the Left Hand Side of the equations, which had the original x's in it.
If the question started with the integration of an equation which had limits and x's, then wouldn't you use substitution and change the limits?
Part (a) was just doing the substitution but didn't touch the limits.
 
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Yes but they were given to you on the Left Hand Side of the equations, which had the original x's in it.
If the question started with the integration of an equation which had limits and x's, then wouldn't you use substitution and change the limits?
Part (a) was just doing the substitution but didn't touch the limits.

This, the limits should of changed from 9 and 1 to 3 and 1.
 
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How did you prove the binomial part a? i tried expanding both and then dividing them, but could only get the first two terms. Wasn't sure how to do it. Also the vectors part B was odd.

Besides from that all my answers are the same as Harolds, so i'll assume theyre correct
 
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Thought it was a very good exam overall! I think I only messed up 7, part b which I tried dx/dy = 0 instead of dy/dx = 0. Oh well.
 
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Agree with most of that.
7b) I got (-3,6)
8a) I got p=-1
8b) I got somewhere along the calculations 8tan45 = 8. Then you know B was either on the left hand side or right hand side of A,
so I added 8 to the i coordinate of position A and subtracted 8 from the i coordinate of position A.
i agree for 7b) i also got (-3,6)
and 8a) p= -1
but am n0t sure about 8b
 
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Thought it was a very good exam overall! I think I only messed up 7, part b which I tried dx/dy = 0 instead of dy/dx = 0. Oh well.
it is correct dx/dy = 0 since the tangent is parallel to the y axis gradient is undefined so if you flip it it wil be dx/dy = 0
 
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i agree for 7b) i also got (-3,6)
and 8a) p= -1
but am n0t sure about 8b

yeah i got the same for 7b its when the lower hald of dy/dx=0 so when dx/dy=0

8b i just realised after the exam, i think the answer was (7,2,4) and (-1,-6,8)
 
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