# Mei C3 Coursework Mark Scheme

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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, 4x^{2}+ 4x + 1 = x^{2}, 3x^{2}+ 4x + 1 = 0, (x+1)(3x+1) = 0, x= -1, x = -1/3 as before

5i. V = 4root(h^{3}+1) - 4. dV/dh using chain rule gives 4*1/2(h^{3}+1)^{-1/2}* 3h^{2}= 6h^{2}* (h^{3}+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. x^{2n}-1 = (x^{n}+1)(x^{n}-1). Substituting for x=2, 2^{n}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 = e^{2x}+ 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(e^{2x}+ 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 readingThank you so much for this only dropped marks on the proof and last part i think that required explanation haha

Great work! Thanks for doing this.

All my answers are the same so, yay.

Вся сцена напоминала некий извращенный вариант представления, посвященного празднику Хэллоуин. Хотя Стратмор и сожалел о смерти своего молодого сотрудника, он был уверен, что ее можно отнести к числу «оправданных потерь».

Фил Чатрукьян не оставил ему выбора. Когда запыхавшийся сотрудник лаборатории безопасности завопил о вирусе, Стратмор, столкнувшийся с ним на лестнице служебного помещения, попытался наставить его на путь истинный.

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