Mr M's OCR FSMQ Additional Mathematics Answers May 2012

First, of all, thanks a lot! However, just wondering whether you could have a copy of the paper to scan in - if so it'd be really helpful!

YES. OH GOD YESSS

But you bearing one is wrong. Degree is correct but bearing has to go clockwise from north, i.e. 180 degrees to get to south, 90 degrees more the get to east and then the original degree added to get the bearing.

So 270 degrees+your degree that you found was the right answer

[Mr M's OCR FSMQ Additional Mathematics Answers May 2012


4. Constant acceleration

Distance = 190 metres and acceleration = 0.6 metres per second squared (4 marks)

[Mr M's OCR FSMQ Additional Mathematics Answers May 2012


1. Quadratic inequality

(i) $1 \leq x \leq 3$ (3 marks)

(ii) Two shaded circles on or above 1 and 3 with a line joining them (1 mark)


2. Biased die

(i) $\frac{2101}{3125}$ (2 marks)

(ii) $\frac{32}{625}$ (4 marks)


3. Factors and remainders

(i) Show a = -7 (2 marks)

(ii) $f(x)=(x-1)(x-2)(x+3)$ (3 marks)


4. Constant acceleration

Distance = 190 metres and acceleration = 0.6 metres per second squared (4 marks)


5. Trigonometric equation

(i) Show trigonometric equation (2 marks)

(ii) Theta = 41.8 degs, 138.2 degs, 270 degs (4 marks)


6. Stationary point

(i) Show S.P. when x = 2 (4 marks)

(ii) Maximum (2 marks)


7. Yachtsman

(i) 3.11 km (3 marks)

(ii) Angle ABC = 61.6 degs and bearing = 152 degs (4 marks)


8. Integral

(i) Show value of integral = 2/3 (3 marks)

(ii) Part of the area is below the x axis and so would have a negative value (1 mark)

(iii) Area = 4 square units (3 marks)


9. Fairground

(i) 2 metres (1 mark)

(ii) 12 metres (2 marks)

(iii) 14 seconds (4 marks)


10. Circle

(i) Midpoint (4, 6) (1 mark)

(ii) $(x-4)^2 + (y-6)^2 = 25$ (4 marks)

(iii) Show B is on circle (1 mark)

(iv) Show lines are perpendicular (3 marks)

(v) (0, 3) (3 marks)


11. Shaded region

(i) 2y = x + 6 (5 marks)

(ii) (3, 4.5) (3 marks)

(iii) Area = 125/12 (4 marks)


12. Highway Code

(i) Show formula (5 marks)

(ii) 38.75 feet (3 marks)

(iii) 23.2 mph (4 marks)


13. Chords

(i) a=12, b=6, c=1 (3 marks)

(ii) Show gradient of chord (3 marks)

(iii) $12+6h+h^2$ (2 marks)

(iv) 12 (1 mark)

(v) Gradient = $32 +24h+8h^2+h^3$

As h gets close to zero, the gradient becomes ever closer to 32. (3 marks)

Are you sure you didn't do this in a rush! I think it is a minimum point, bearing was around 118 degrees and distance was 130m? Can you please check these, though I might be wrong as well.

It's not my property to scan. Under the terms of our agreement with OCR, I may only use past papers with my own students in my Centre.

Are you sure you didn't do this in a rush! I think it is a minimum point, bearing was around 118 degrees and distance was 130m? Can you please check these, though I might be wrong as well.

It is a minimum (that is a typo) - I'll change it.

Bearing and distance is right.

I remember distinctly looking at the distance question for Q4, and getting 130m as well (I even double-checked!), so a little confused and worried.

My arithmetic error - it is 130m!

Sorry!!! My excuse is I did do it in 25 mins.

My arithmetic error - it is 130m!

Sorry!!! My excuse is I did do it in 25 mins.

No Worries! More marks for me eh

This is great stuff though! I am hoping I have got a half decent mark
Was the calculus proof for the minimum point re differentiating dy/dx to get d²y/dx² and putting the number in and as that was > 0 = Minimum?