C3 Jan 12 Edexcel - Past Papers, Model answers, tips

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If you mean from 05 to 11, then I'm nearly there

I will do the last one tomorrow, june 11

ive done em too but not in exam conditions, did you do yours in exam conditions?
and what have you been getting in them? :smile:

Reply 61

Jukeboxing

I've done all the current spec papers, i love the difficult papers, always score highly in them.

I've moved onto old spec, Solomon and Elmwood papers.

There's so much time left till the exam.

Reply 62

master_blaster66

Original post by Jukeboxing

I've done all the current spec papers, i love the difficult papers, always score highly in them.

I've moved onto old spec, Solomon and Elmwood papers.

There's so much time left till the exam.

lool wowww nice onee

Reply 63

master_blaster66

Original post by fudgesundae

Why do past papers so early lol. Still over 3 weeks till exam, what are you going to do until then if you run out now?

ermmm revise for other exams :s-smilie:

Reply 64

This Honest

Original post by pakizrulez

ive done em too but not in exam conditions, did you do yours in exam conditions?
and what have you been getting in them? :smile:

I didn't do them in exam conditions because I don't feel pressurised in maths exams, I did D1 in exam conditions because some of the questions are lengthy.

I don't really mark them, i just look where i went wrong. Although june 09, i got only 1 part wrong and that was it.

Reply 65

This Honest

Can anyone help me solve for x:

e^x + e^-x = 2

I got as far as
e^-x2 = 2
-x^2=2
-x^2 = ln2

but then the anser says zero

I know you cant sqrt a negative ln so where did i go wrong?

Reply 66

Jukeboxing

Original post by This Honest

Can anyone help me solve for x:

e^x + e^-x = 2

I got as far as
e^-x2 = 2
-x^2=2
-x^2 = ln2

but then the anser says zero

I know you cant sqrt a negative ln so where did i go wrong?

Why don't you try multiplying all terms by e^x

so you get:
e^2x - 2e^x +1 = 0, this leaves you with a quadratic.

now factorise:
(e^x - 1)(e^x - 1)

e^x = 1, so x = ln(1) = 0

Reply 67

fudgesundae

Did my first past paper. Jan 06, got 71/75. Decent but question 4b really ****ed me up. I was just lost

Reply 68

This Honest

Original post by Jukeboxing

Why don't you try multiplying all terms by e^x

so you get:
e^2x - 2e^x +1 = 0, this leaves you with a quadratic.

now factorise:
(e^x - 1)(e^x - 1)

e^x = 1, so x = ln(1) = 0

i was told to always times the powers when the bases i.e (e) were the same

So e^x multiply e^-x =1

i get it now, reciporcal of e^-x
:yes:

but how do u get zero from using my method

(edited 12 years ago)

Reply 69

Jukeboxing

Original post by This Honest

i was told to always times the powers when the bases i.e (e) were the same

So e^x multiply e^-x =1

but how?

Using rules of indices
x^2 X x^4 = x^(2+4) = x^6

so same applies here:

e^x X e^-x = e^(x-x) = e^0 = 1

Reply 70

tiny hobbit

Original post by This Honest

Can anyone help me solve for x:

e^x + e^-x = 2

I got as far as
e^-x2 = 2
-x^2=2
-x^2 = ln2

but then the anser says zero

I know you cant sqrt a negative ln so where did i go wrong?

Ouch!

When multiplying powers of the same thing, you add the indices. You seem to have reversed this.

Reply 71

fudgesundae

Original post by This Honest

Can anyone help me solve for x:

e^x + e^-x = 2

I got as far as
e^-x2 = 2
-x^2=2
-x^2 = ln2

but then the anser says zero

I know you cant sqrt a negative ln so where did i go wrong?

Where is this Q from?

e^x + e^-x = 2
(e^x + e^-x)/2 = 1

(e^x + e^-x)/2 = cosh x

cosh x = 1
x = cosh^-1 (1) = 0

Reply 72

This Honest

Original post by tiny hobbit

Ouch!

When multiplying powers of the same thing, you add the indices. You seem to have reversed this.

I thought when you're multiplying the same base, you add the powers

when you add the same bases, you multiply the powers, due to the log rule well that's what i've been doing so far

ooops, jeez im talking nonsense here sorry, you're right