Mr M's OCR (not OCR MEI) Core 2 Answers May 2016

For the last question 9 iii I left my two answers as (1/3)pi/a and (4/3)pi/a - will I still get the full 4 marks?

Do you know what the question was for 4ii? Thanks

Mr M's OCR (not OCR MEI) Core 2 Answers May 2016


1. (i) 10 cm (2 marks)

(ii) 5.04 cm (2 marks)


2. (i) $\displaystyle \frac{3 \pi}{10}$ (2 marks)

(ii) $r=20.4$ (3 marks)


3. (i) $\displaystyle 27 +27kx+9k^2x^2 +k^3x^3$ (4 marks)

(ii) $k=\pm \sqrt{3}$ (2 marks)


4. (i) $\displaystyle \log_3 \frac{x^2}{x+4}$ (2 marks)

(ii) $x = 12$ (4 marks)


5. (a) $\displaystyle \frac{x^4}{2} -x^3 +2x^2-6x+k$ (3 marks)

(b) (i) $\displaystyle \frac{2}{a^2} - \frac{6}{a} + 4$ (4 marks)

(ii) 4 (1 mark)


6. (i) $k=91$ (3 marks)

(ii) $\displaystyle S_{16} = 978$ (2 marks)

(iii) $N=38$ (6 marks)


7. (i) Quotient $x^2-4x+3$ and remainder 0 (3 marks)

(ii) $x=1$ or $x=-1$ or $x=3$ (3 marks)

(iii) Show (2 marks)

(iv) $\displaystyle \frac{512}{15}$ (4 marks)


8. (i) Translation by 2 units in the positive x direction (2 marks)

(ii) Stretch parallel to the y axis scale factor $\displaystyle \frac{1}{9}$ (2 marks)

(iii) Sketch showing $(0, \frac{1}{9})$ as the only point of intersection (2 marks)

(iv) $x=6.73$ (3 marks)

(v) 9.60 (3 marks)


9. (i) $\displaystyle \frac{2 \pi}{a}$ (1 mark)

(ii) $\displaystyle a=\frac{5}{3}$ and $\displaystyle k=\frac{\sqrt{3}}{2}$ (3 marks)

(iii) $\displaystyle x=\frac{\pi}{3a}$ and $\displaystyle x=\frac{4 \pi}{3a}$ (4 marks)


Sorry guys I made a mess of 9(ii) at the start - I didn't notice k had to be positive. It's right now.

They might ignore subsequent working.

Solve $2 \log_3 x - \log_3 (x+4)=2$

6iii)
I did the sum of Un and the sum to infinity to Wmn.. and I ended up with 3n^2+17n-1200>0 or 3n^2+17n-1200<0 (i forgotten)... getting n=17.4
Did I do the entire question wrong or do I get any method marks?

Wasnt the question to find the minimum points on the graph, not the x-intersects?
(Sorry for double posting and sorry if I'm wrong )

Can someone please explain how to do 4ii, I know ive made a mistake somewhere but im not sure where

I thought q 8 iii and 9 iii were 3 marks each?

N needs to be an integer so you round up and your quadratic is wrong.

You should have obtained $3N^2+17N-4800$.

You would get 3 marks I expect.

For 7iv I integrated between -1 and 1 instead of -1 and 3 do you think I'll get any marks?

$\frac{x^2}{x+4}=9$

$x^2-9x-36=0$

$x=12$ or $x=-3$ but the second solution has to be rejected as it does not satisfy the original equation.

I'm almost certainly wrong, but how comee this doesn't work:

Sin(api/5) =sin(a*2pi/5)

Thus (double angle theorem)

Sin (api/5) =2*sin(api/5)*cos(api/5)

Thus 0.5 = cos(api/5)

Thus pi/3 = api/5

Thus 5/3 = a (which is smaller than 5 so is surely the first solution if it is right?

Would I lose a mark for including -3?

Question 9ii - the question said that k must be a positive constant and 0 isn't a positive constant?

Yes I fixed it some time ago.

How come when you substitute in 4pie/3a into sin(ax) = root(3)cos(ax) they are not equal - it gives -ve value on the cos side.