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# Unofficial Mark Scheme OCR MEI C3 21st June2016 (Methods for Advanced Mathematics) watch

1. 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
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.

2. Thank you so much for this only dropped marks on the proof and last part i think that required explanation haha
3. Great work! Thanks for doing this.

All my answers are the same so, yay.
4. 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.
5. 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*
6. Yeah he's right, mine resembled ^ more, but still great job for a first sweep. Especially with just two guys.
Deserves more rep.
7. 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.
Attached Images
8. inductiveproof.pdf (111.0 KB, 157 views)
9. can anyone point me in the direction of the question paper for these solutions ? MEI OCR C3. THANKS
10. (Original post by logfun)
can anyone point me in the direction of the question paper for these solutions ? MEI OCR C3. THANKS
did you find it
11. Does anyone have the question paper for this?

Thanks

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