OCR C2 exam Wed 25th May

anyone know the exact question for 7 ii 4 marks

sorry i meant the actual equation you had to integrate for 4 marks from -1 to 3 or something

Do people think i will get any marks on q9 for doing 180/a - (what one of the values was)= other value
(right method but used 180/a instead of pi/a the whole way throuhg..)

Probably like 2 method marks?

What was 8v about

do u remember what it was out of?

i think i got something like 38, i'm probably wrong

I am gonna start an unofficial mark scheme - I think this is about correct!

1i) 10cm (2m)

1ii) 5.04cm (2m)

2i) 3/10π (2m)

2ii) 20.4cm (3m)

3i) 27+27kx+9k^2x^2+k^3x^3 (4m)

3ii) +/-√3 (2m)

4i) log base 3 x^2/x+4 (2m)

4ii) x=12 (4m)

5a) x^4/2-x^3+2x^2-6x (3m)

5bi) -6/a+2/a^2+4 (4m)

5bii) 4 (1m)

6i) k=91 (3m)

6ii) Sum of terms = 978 (2m)

6iii) N=38 (6m)

7i) R=0, quotient = x^2-4x+3 (3m)

7ii) (x+1), (x-1), (x-3) (3m)

7iii) Prove (2m)

7iv) 512/15 (4m)

8i) Translation 2 units in positive x direction (2m)

8ii) Stretch scale factor 1/9 in y direction (2m)

8iii) Sketch, crosses y at (0,1/9) (2m)

8iv) x=6.73 (3m)

8v) Area =9.60 (3m)

9i) 2π/a (1m)

9ii) a=5/3 and k=√3/2 (3m)

9iii) π/3a and ? (4m)

4ii) I got 12 and other root as well. Are we supposed to leave another root?
9ii) the given values of x were pi/5 and 2pi/5 and hence periodicity is pi/5. Equating this to 2pi/a gives a=10 and k=0. Can anyone explain to me flaw in the argument.
9iii) pi/3a and 4pi/3a

My other answers match with yours as far as I remember.

4ii) I got 12 and other root as well. Are we supposed to leave another root?
9ii) the given values of x were pi/5 and 2pi/5 and hence periodicity is pi/5. Equating this to 2pi/a gives a=10 and k=0. Can anyone explain to me flaw in the argument.
9iii) pi/3a and 4pi/3a

My other answers match with yours as far as I remember.

4ii) I got 12 and other root as well. Are we supposed to leave another root?
9ii) the given values of x were pi/5 and 2pi/5 and hence periodicity is pi/5. Equating this to 2pi/a gives a=10 and k=0. Can anyone explain to me flaw in the argument.
9iii) pi/3a and 4pi/3a

My other answers match with yours as far as I remember.

You can't log a negative number (-3)

4ii) I got 12 and other root as well. Are we supposed to leave another root?
9ii) the given values of x were pi/5 and 2pi/5 and hence periodicity is pi/5. Equating this to 2pi/a gives a=10 and k=0. Can anyone explain to me flaw in the argument.
9iii) pi/3a and 4pi/3a

My other answers match with yours as far as I remember.

It isn't a negative number if you apply the multiplying 2 as an indice before logging