The Student Room Group

Scroll to see replies

It crosses at y=4

Posted from TSR Mobile
Original post by Chazley123
many people actually missed q 9, a few did in my school lol

I just thought they'd given way too much working out space. *facepalm*
Reply 2502
Anyone get 1/24pi anywhere in any question?
Need the unofficial markscheme ASAP.. Mixed emotions about that exam.. Good paper but with some weird parts and amount of marks allocated to some questions didn't seem fitting.. Literally no idea how I've done!
Original post by Chazley123
many people actually missed q 9, a few did in my school lol


what was question 9??!!
****, made the most stupidest mistake when converting sec2A + tan2A...

why the **** did I use Tan2A = 2tanA/1-Tan^A
I should have used sin2A/Cos2A

thats 4 marks down the ****ing drain...
what did everyone get for their solutions for part b)?
Original post by Neilg99
Not at all


How did u do the differentiating question which had to be in the form f(x)e^-2 ??
For the range of g question:

Did factorise the cubic as to produce dy/dx=0 and then solve it, then substitute in the 2 x values to find the 2 y values?
I seem to remember substituting in [something+(root7)]/2 as one of my x values.
Original post by Kill3er
Anyone get 1/24pi anywhere in any question?

I got it. Not sure what question though
The graph question (think it was question 2) definitely said state the asymptotes on both graphs so i put y=5 for the second graph even though i thought it was wrong?

Was there an asymptote and what was it?
Reply 2510
Original post by Ripper Phoenix
Absolutely correct but u had to put it in y


Posted from TSR Mobile


For the finding the coordinates of A, I only managed to get up to finding the gradient of the tangent as 1/20pi. Is that correct, and if so, how many marks do you think I'd be able to get? Also for the K question, I accidentally didn't give a range and simply said that k = + or - square root 20.
9b and c and asymptote for the left bit
Original post by big fat poo
right for the range question I had a cubic: 2x^3 - 3x^2 + 3x
x(2x-1)(x-2) (I think)
Anyone else?


Yep I got that. But not sure about it.

Posted from TSR Mobile
Original post by Isymega
Guys why couldn't you do the inverse of the function??!!??


Not a 1-1 mapping :smile:
Reply 2514
Original post by ArtemMakushin
I'm not sure man, but I got a y coordinate being massive... Because the gradient of the line was 24pie for me... Did you get same?


Wasn't it 1/24pi

You worked out dx/dy and then had to do 1 over
Original post by jf1994
An asymptote is defined as a straight line that a curve approaches but never touches.

The curve in this case was the mod of an exponential function

Clearly the curve crosses y=5 so it is not an asymptote


So did you draw on an asymptote for that one?
Original post by Kill3er
Anyone get 1/24pi anywhere in any question?


Yep, for the gradient of the line where you had to find where it crosses the y-axis i think
Reply 2517
Anyone else get 4 answers for 8b?? (The one using the proof where you got tan(2theta) = -1/3
Original post by dhamm
k was negative, meaning if you subbed a negative number into the numerator and denominator of the gradient function, they would always be positive. So gradient always positive, and the function is increasing.


Would i get marks for trying to find the y intercept of the graph, which was -0.5 i got. then find the turning point which was a maximum, and i prooved it by differentiating again. then sketching the graph out roughly and then saying its decreasing?

also, for the sketch the graphs question i used pencil would it scan through properly ?
Original post by dhamm
k was negative, meaning if you subbed a negative number into the numerator and denominator of the gradient function, they would always be positive. So gradient always positive, and the function is increasing.


I am so happy you said that. That's exactly what i put

Latest

Trending

Trending