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Edexcel A2 C4 Mathematics June 2016 - Official Thread

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Can you do

^{0}\sqrt{4}

I meant 0^even number :frown:

Reply 2141

edothero

Original post by Middriver

I meant 0^even number :frown:

Oh right, well it wouldn't make much of a difference whether if it were even or odd.

0^{x} = \ 0

for all x

Reply 2142

Zacken

Original post by edothero

Oh right, well it wouldn't make much of a difference whether if it were even or odd.

0^{x} = \ 0

for all x

0^0

is undefined.

Reply 2143

edothero

Original post by Zacken

0^0

is undefined.

Yeah my bad, except 0 :biggrin:

Reply 2144

edothero

Original post by Middriver

I meant 0^even number :frown:

I just realised what you were asking me :laugh:

sorry

Reply 2145

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 :smile:

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Reply 2146

Isabella~

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! :smile:

Reply 2147

math42

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 :smile:

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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.

Reply 2148

INdra ff

http://www.examsolutions.net/a-level-maths-papers/Edexcel/Core-Maths/Core-Maths-C4/2008-January/paper.php can somebody please help me with part A

Reply 2149

metrize

Hi guys any hard integrals which i should do in preparation?

Reply 2150

imran_

how would I draw the graph of
y=3/2x-1

Reply 2151

math42

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

Reply 2152

imran_

Original post by 1 8 13 20 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>?

Reply 2153

metrize

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

Reply 2154

Kevin De Bruyne

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.

Reply 2155

math42

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

(edited 7 years ago)

Reply 2156

metrize

Does anyone have the IAL c34 question paper only (no model answers just want to do the questions first)

Reply 2157

Azzer11

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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Reply 2158

metrize

Can someone help on jan 13 c4 question 8?

https://googledrive.com/host/0B1ZiqBksUHNYQXE5T2xiNDBRd2s/January%202013%20QP%20-%20C4%20Edexcel.pdf

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)