UNOFFICIAL MARK SCHEME:
1. Integrate 1 + cos(0.5x) between limits of 0 and pi/2. Answer is pi/2 + root2
2. Solve for x. Answer is x = ln2
3. Can't remember the context. Answer is 4ln4 - 4 - alternatively may be written as 4(ln4 -1) or ln256 - 4 or any equivalent form.
4. |2x+1| = -x. First case, 2x+1 = -x, gives 3x+1 = 0, 3x = -1, x = -1/3
Second case, -(2x+1) = -x, negate both sides to give 2x+1 = x, x+1 = 0, x = -1
Alternative approach, square both sides, 4x2 + 4x + 1 = x2, 3x2 + 4x + 1 = 0, (x+1)(3x+1) = 0, x= -1, x = -1/3 as before
5i. V = 4root(h3+1) - 4. dV/dh using chain rule gives 4*1/2(h3+1)-1/2 * 3h2 = 6h2 * (h3+1)-1/2. Substituting h=2 gives dV/dh = 8 (feel free to check this for yourself).
5ii. dh/dt = dh/dV * dV/dt = 1/8 * 0.4 = 0.4/8 = 0.05
6i. Can't remember the context. Answer is root3 /3
6ii.Transformation is first translate 1 unit to the right, THEN stretch parallel to y-axis, scale factor 1/2
7. x2n-1 = (xn+1)(xn-1). Substituting for x=2, 2n has to be even, so it's a product of two consecutive odd numbers, so one of those must be divisible by 3, so their product is divisible by 3. That's why I put anyway, I've seen someone say it's not right, so not completely sure about this one.
8i. Differentiate using quotient rule, then multiply your result by 2root(x+4), to obtain dy/dx in the form given.
8ii. Using equation for y, asymptote is x= -4, so x coordinate of P is -4. Use dy/dx for x=0, to show that tangent is y = 0.5x (y-intercept c is 0). Then y = 0.5 * -4 = -2, so P is (-4,-2)
8iii.Area under the curve is found by integration to be 14/3. Area of triangle with vertices O,Q and (5,0) is 1/2 * 5 * 5/2 = 25/4
So area of required region is 25/4 - 14/3 = 19/12
9i. y = e2x + ke-2x. When x=0, y = 1 + k*1 = 1+k
9ii. Answer is 2rootk
9iii. Answer simplifies to (0.5k - 1/2) - (1/2 - 0.5k) = 0.5k - 1/2 - 1/2 + 0.5k = k - 1
9ivA. Replace x with (x+ 0.25lnk), to give g(x) = rootk(e2x + e-2x).
9ivB. g(-x) = g(x), so g(x) is an even function.
9ivC. f(x) is symmetrical about the line x = 0.25lnk. This can be deduced because g(x) is f(x) translated 0.25lnk units to the left. g(x) is even, so is symmetrical about the y-axis i.e. symmetrical about line x=0. f(x) is g(x) translated 0.25lnk units to the right, therefore its line of symmetry is x = 0.25lnk. Full reasoning will probably be required for all 3 of these marks.
These are the answers of my friend, who does Additional Further Maths, so I think they should all be right, or most of them anyway. Of course, feedback is more than welcome, if I've made a mistake please point it out.
I think boundaries will be 88 or 89 for 100 UMS (out of 90, for exam and coursework). A* should be 80 or 81, and A should be 72 or 73. Difference between each grade will be about 7 or 8 marks.
Thanks for reading
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Unofficial Mark Scheme OCR MEI C3 21st June2016 (Methods for Advanced Mathematics) watch
- Thread Starter
Last edited by fazee; 21-06-2016 at 13:54.
- 21-06-2016 13:49
- 21-06-2016 13:53
Thank you so much for this only dropped marks on the proof and last part i think that required explanation haha
- 21-06-2016 13:56
Great work! Thanks for doing this.
All my answers are the same so, yay.
- 21-06-2016 13:57
Additionally with you grade predictions 88 marks for 100 UMS and 80 marks for 80 UMS doesnt add up, the highest i have ever seen for an A is 75 marks.
- 21-06-2016 13:57
Edit question 7 to this, stop with giving people false hope
x^2n - 1 = (x^n - 1)(x^n + 1)
When x = 2
=>2^2n - 1 = (2^n - 1)(2^n + 1)
2^n only has a prime factor of 2, meaning it is not divisible by 3.
Additionally, since, for any 3 consecutive integers, one of them must be divisible by three (since every 3rd number is a multiple of three)
We can conclude that either 2^n - 1 OR 2^n OR 2^n + 1 is divisible by 3.
Since we have proven 2^n is not divisible by 3.
Either 2^n - 1 or 2^n + 1 must be divisible by 3
Meaning (2^n - 1)(2^n + 1) = 2^2n - 1 is always divisible by 3, as required.
*You can rewrite it nicely if you wish, but what you have written might only get 2/4*
- 21-06-2016 14:11
Yeah he's right, mine resembled ^ more, but still great job for a first sweep. Especially with just two guys.
Deserves more rep.Last edited by WhiteBison; 21-06-2016 at 14:19.
- 21-06-2016 17:28
This isn't how I solved question 7 in the exam, I had to make assumptions, due to the lack of proof by induction on the C3 spec, but I did manage this rigorous proof when I got home. Please let me know if you find any errors.
- 20-08-2016 12:48
can anyone point me in the direction of the question paper for these solutions ? MEI OCR C3. THANKS
- 13-12-2016 15:27
- 28-02-2017 15:24
Does anyone have the question paper for this?