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Edexcel A2 C4 Mathematics June 2015 - Official Thread

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Original post by Nurishment
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Can someone explain b) to me. I don't quite fully understand why you take out a factor of 1000 :s-smilie:. Why wouldn't you rearrange 7100 to = 8-9x and solve for x and sub that back in?

BTW theres an error in the MS for that Question. Its supposed to be 10³√7.1 not 10³√71. Try typing on the calc.
Now to explain i "think" x should always be less than 1. Someone correct me.

Dont know why but if subbing 7100 gives you the wrong answer compared to when you type ³√7100 on calculator try turing it to surds.
³√7100 = ³√7.1 × ³√1000 = 10³√7.1
Equate 10³√(9-8x) = 10³√7.1
9-8x = 7.1
x = 0.1
Original post by jamesoneil98
Really confused. trying to do this integral: 1/(3x+3). Pretty sure the answer is (1/3)*ln(3x+3) but surely if you take out 1/3 you get 1/(x+1) and then as the top is the differetial of the bottom it would be (1/3)*ln(x+1). Any help appreciated. Good luck everyone!


ln(3x + 3) = ln3 + ln(x + 1)
All you're doing with the first case is adding a specific constant bit
Who reckons this paper will be at the same difficulty or harder as the c3 paper?
Reply 2043
0π4πx2(1+sin2x) dx\displaystyle\int^\frac{\pi}{4}_0 \pi x^2(1+sin2x)\ dx

For integration by parts to work in this question would you need to expand out the brackets first then solve them as two separate integrals? Like this:

0π4πx2 dx+0π4πx2(sin2x) dx\displaystyle\int^\frac{\pi}{4}_0 \pi x^2\ dx + \displaystyle\int^\frac{\pi}{4}_0 \pi x^2(sin2x)\ dx

I tried to to this question last night without expanding first and got most of the answer correct but the first term was missing, guessing this is why...
Original post by Bealzibub
for what shapes do we need t know how to work out volume, area etc.??



volume of polyhedron

and circumference of mars

standard procedure


This is what i did....Untitled.png
Reply 2046
What are the "R" papers by the way? I always seem to find them harder, how come?
Original post by grn
0π4πx2(1+sin2x) dx\displaystyle\int^\frac{\pi}{4}_0 \pi x^2(1+sin2x)\ dx

For integration by parts to work in this question would you need to expand out the brackets first then solve them as two separate integrals? Like this:

0π4πx2 dx+0π4πx2(sin2x) dx\displaystyle\int^\frac{\pi}{4}_0 \pi x^2\ dx + \displaystyle\int^\frac{\pi}{4}_0 \pi x^2(sin2x)\ dx

I tried to to this question last night without expanding first and got most of the answer correct but the first term was missing, guessing this is why...


You would need to integrate by parts twice for that wouldn't you?
Original post by 1 8 13 20 42
ln(3x + 3) = ln3 + ln(x + 1)
All you're doing with the first case is adding a specific constant bit


but what have i done wrong? one of the 2 ways is wrong as they dont equal each other.
Original post by jamesoneil98
but what have i done wrong? one of the 2 ways is wrong as they dont equal each other.


It's not wrong, it's just a specific case. Normally you would put a constant of integration, right? If you take out the constant in the 3x + 3 version, you can then amalgamate that with some arbitrary constant to get another arbitrary constant; i.e. if you write the integral as

(1/3)ln(3x + 3) + c this is 1/3(ln(3) + ln(x + 1)) + c = (1/3)ln3 + (1/3)ln(x + 1) + c

Now this can be written as (1/3)ln(x + 1) + k because the c and the (1/3)ln3 are constants

If you put the answer (1/3)ln(x + 1) this is more correct to be honest (although having a plus c is always the most correct)
When a gradient is parallel to the x axis and you have eg dy/dx = (3x+ y)/(2y+1) is the 3x + y = 0 or then 2y+1? thanks
Original post by jamesoneil98
Really confused. trying to do this integral: 1/(3x+3). Pretty sure the answer is (1/3)*ln(3x+3) but surely if you take out 1/3 you get 1/(x+1) and then as the top is the differetial of the bottom it would be (1/3)*ln(x+1). Any help appreciated. Good luck everyone!

Both are equivalent because the 1/3ln(3x+3) is 1/3(ln3+ln(x+1))
When you put your limits in you have ln3 and -ln3 so the answer is the same for both
Original post by WillGood
BTW theres an error in the MS for that Question. Its supposed to be 10³√7.1 not 10³√71. Try typing on the calc.
Now to explain i "think" x should always be less than 1. Someone correct me.

Dont know why but if subbing 7100 gives you the wrong answer compared to when you type ³√7100 on calculator try turing it to surds.
³√7100 = ³√7.1 × ³√1000 = 10³√7.1
Equate 10³√(9-8x) = 10³√7.1
9-8x = 7.1
x = 0.1


I'm beginning to grasp why this works now. How Edexcel thinks about putting questions like these in, hurt me.
Reply 2053
Original post by WillGood
BTW theres an error in the MS for that Question. Its supposed to be 10³√7.1 not 10³√71. Try typing on the calc.
Now to explain i "think" x should always be less than 1. Someone correct me.

Dont know why but if subbing 7100 gives you the wrong answer compared to when you type ³√7100 on calculator try turing it to surds.
³√7100 = ³√7.1 × ³√1000 = 10³√7.1
Equate 10³√(9-8x) = 10³√7.1
9-8x = 7.1
x = 0.1


But if you solve for x you get 0.2375?
Original post by Anon-
What are the "R" papers by the way? I always seem to find them harder, how come?


they are built and made for the asians

Spoiler

Original post by bg9876
When a gradient is parallel to the x axis and you have eg dy/dx = (3x+ y)/(2y+1) is the 3x + y = 0 or then 2y+1? thanks


If the gradient is equal to 0, the top line (numerator) is equated to 0. If a line is parallel to the y-axis then the gradient is undefined, and you only get an undefined value when 0 is on the bottom of a fraction, therefore you equate 2y+1= 0 (denominator)
Original post by sj97
But if you solve for x you get 0.2375?


probably meant 8 - 9x
Original post by 1 8 13 20 42
It's not wrong, it's just a specific case. Normally you would put a constant of integration, right? If you take out the constant in the 3x + 3 version, you can then amalgamate that with some arbitrary constant to get another arbitrary constant; i.e. if you write the integral as

(1/3)ln(3x + 3) + c this is 1/3(ln(3) + ln(x + 1)) + c = (1/3)ln3 + (1/3)ln(x + 1) + c

Now this can be written as (1/3)ln(x + 1) + k because the c and the (1/3)ln3 are constants

If you put the answer (1/3)ln(x + 1) this is more correct to be honest (although having a plus c is always the most correct)


got it thanks!
what are all the formulas we need to know for vectors?
Original post by johnh545
Both are equivalent because the 1/3ln(3x+3) is 1/3(ln3+ln(x+1))
When you put your limits in you have ln3 and -ln3 so the answer is the same for both


got it thanks!

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