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# Edexcel A2 C4 Mathematics June 2016 - Official Thread watch

1. (Original post by edothero)
Can you do ?
I meant 0^even number
2. (Original post by Middriver)
I meant 0^even number
Oh right, well it wouldn't make much of a difference whether if it were even or odd.

for all x
3. (Original post by edothero)
Oh right, well it wouldn't make much of a difference whether if it were even or odd.

for all x
is undefined.
4. (Original post by Zacken)
is undefined.
5. (Original post by Middriver)
I meant 0^even number
I just realised what you were asking me sorry
6. Hi,
Can someone please explain why, when solving a differential equation, you don't put constants of integration on both sides? Thanks in advance

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7. (Original post by SeanFM)
The line l intersects the curve but they do not necessarily have the same gradients at that point (eg think of the intersection of y = x^2 and y = x+2. They don't have the same gradient where they intersect.
Ohh I see, thank you!
8. (Original post by Clovers)
Hi,
Can someone please explain why, when solving a differential equation, you don't put constants of integration on both sides? Thanks in advance

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say your solution is of the form y = f(x) + c for an arbitrary constant c
Well you could write y + b = f(x) + c, where b and c are both constants. But then y = f(x) + (c-b). since c and b are both arbitrary (they can be any constant you can think of), c-b can be any constant you can think of. So putting constants on both sides doesn't alter the situation.
10. Hi guys any hard integrals which i should do in preparation?
11. how would I draw the graph of
y=3/2x-1
12. (Original post by imran_)
how would I draw the graph of
y=3/2x-1
Basically just like you'd draw the graph of 1/x but shift the asymptote accordingly
13. (Original post by 13 1 20 8 42)
Basically just like you'd draw the graph of 1/x but shift the asymptote accordingly
what I did first was draw 3/(x-1)

when I multiply it by the 2 it becomes 1/2 multiplied by 1 right>?
14. (Original post by imran_)
how would I draw the graph of
y=3/2x-1
Vertical asymptote at x=1/2

Horizontal asymptote at y=0
15. (Original post by imran_)
how would I draw the graph of
y=3/2x-1
If you are ever stuck with graphs, plug in 'nice' values as well as test the limits of them.

Eg what happens when x is less than 1/2 but increases to 1/2? What happens when x is greater than 1/2 and decreases 1/2?

Answers are that when x is just under 1/2 then you are dividing by a very large negative number (eg put in x = 0.499999999) so 3/small -ve number = very large -ve number and that tells y is negative to the left ox f = 1/2, and similarly if you put in x = 0.50000001 then you are dividing by a very lage +ve number so y is a very large +ve number. And then as x tends to +ve infinity y tends to 0, same with - infinity.
16. (Original post by imran_)
what I did first was draw 3/(x-1)

when I multiply it by the 2 it becomes 1/2 multiplied by 1 right>?
I'm not sure what you mean..
Probably better to think in smaller steps. How would you draw 1/x? Then how would you draw 1/2x? Then 1/(2x - 1) is just a horizontal shift of this right? And then you multiply by 3
Although to be honest the scale factors (going from 1/x to 1/2x and from 1/(2x - 1) to 3/(2x - 1)) don't really matter if you're just drawing this graph by itself

Edit: Ah yeah well the horizontal shift is by a 1/2 if that's what you're referring to
17. Does anyone have the IAL c34 question paper only (no model answers just want to do the questions first)
18. (Original post by metrize)
Does anyone have the IAL c34 question paper only (no model answers just want to do the questions first)
Should be on physicsandmathstutor

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19. Can someone help on jan 13 c4 question 8?

This is my method:
-ln(3-theta all over A)=t/125
So 3-theta over A=e^(-t/125)
So 3-theta = Ae^(-t/125)
So theta = 3-Ae^(-t/125)

But the answer is meant to be theta = 3+Ae^(-t/125)

20. Spent so long on a still couldn't do it

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