? Point P was on the asymptote I thought - given asymptote is x=-4 and y=1/2x was equation of tangent, I thought it was (-4, -2)... Oh dear, I really hope I haven't ****ed that one up as well...

I thought the line was y=0.5x too. The majority are saying (-4,-2) I think we're safe :P

Reply 146

gemdarkstone

Original post by daisyn97

4ln4 - 4?

hmmm, i got -4ln4-4 (which i know realise should be 4ln+4...)
not sure who's right tho

Reply 147

Danny_L

Original post by Maligaha

I got 0.5k-0.5

I got k-1

Reply 148

che-mystery

I thought it was a nice paper except question 9 was a bit weird. I definitely messed up on a few bits of that, i think i kept getting my es and ins confused

Reply 149

ScalyJake

Original post by treeporn

How the hell did you guys find out the equation of the tangent?

The line was a tangent to the curve at the origin. So you sub x=0 into the dy/dx of the curve giving you a gradient=1/2

The line went through the origin so C=0, so y=0.5x

Reply 150

Cadherin

Original post by gemdarkstone

hmmm, i got -4ln4-4 (which i know realise should be 4ln+4...)
not sure who's right tho

It was 4ln4 - 4 I'm pretty certain - it evaluated to (4ln4-8)-(-4).

Anyone else get 0.5k - 0.5 for ak + b?? I'm pretty sure that's right as you had to take a factor of 0.5 out before you made the substitution...

Reply 151

Cadherin

Original post by ScalyJake

The line was a tangent to the curve at the origin. So you sub x=0 into the dy/dx of the curve giving you a gradient=1/2

The line went through the origin so C=0, so y=0.5x

Yeah, I was a bit rushed... Will I get full marks on that part for correct coordinates and writing down y=0.5x without actually showing how I got it, as I didn't have time? :frown:

Reply 152

dreadfortmormont

Original post by daisyn97

4ln4 - 4?

Same, but I put it as ln256 - 4. Would that still get the marks?

Reply 153

username1915885

very last q, about geometrical properties of f(x), what did you guys put

dr/dt 0.03?

Original post by VokeA

So did I oh wow. I was nervous about that one

Posted from TSR Mobile

yea me to, for 5 marks i was worried....

Reply 156

Siama26

You calculate dy/dx at the point x=0 (because it was at the origin) and then you use y-y1=m (x-x1) to find the equation of the tangent. X1 and y1 are (0,0) and m is the value of dy/dx.

Reply 157

Rawsonj

Answers that I got:
1)integrating gives area = pi/2+sqrt2
2) ln(2+e^x)=2x is a quadratic in e^x with roots 2 and -1. e^x is always greater than 0 so e^x=2. x=ln2
3)Area = ln(256) -4 or 4(ln(4) -1) by parts
4)draw the graphs, intersect at x=-1 and x=-1/3
5)dv/dh=8 dh/dt=0.4/8=0.05 m per minute
6)dy/dx=1/2cos2y gradient at x=1.5 y =pi/12 is sqrt(3) /3 For second part y=1/2 arcsin(x-1) a translation of 1 in the positive x direction followed by an enlargement of scale factor 1/2 parallel to the y axis.
7)x^2n-1 = (x^n+1)(x^n-1) (2^n+1)(2^n-1) as n is greater than but not equal to 0, 2^n is even and so the brackets are consecutive odd numbers. Therefore one is a multiple of 3. Therefore the product of the two brackets is divisible by 3.
8)first part was a show that. gradient was 1/2 so point was (-4,-2) third part was integration by substition which gave 19/12.
9) p=1+k show that minimum is at x=1/4ln(k) and y value at this point is 2*sqrt(k). Integrating gives area =k-1. Next part was to show what g(x) is equal to. Show it is even g(x)=g(-x). Last part: line of symetry in x=1/4ln(k). explanation is that as when translated by 1/4ln(k) in the negative x direction, we see that the origin is a line of symetry (line of reflection) as it is even. Therefore when we translate g(x) back to f(x) we move the line of symetry to x=1/4ln(k)