OCR MEI C3 Maths June 2015

I think I got ~60 in C3, and 14 in the coursework, and in C4 I think I got 80/90, do you think that's enough for an A*?

I think your C3 exam and coursework both might let you down, unfortunately, considering that it is probably going to be like 77-ish for an A* in c4 (it was a very easy exam)

what happens if in the exam the answer is '4 - 3 ln 3' and you write 4 - ln (3^3)? Technically they are the same, but from their mark schemes they dont shed any light on further unnecessary simplification. I hope I haven't lost stupid marks doing this, as i did it throughout c3 and c4

Bump.
Can anyone help me with this concern?

In mark schemes they usually have "oe" next to the final answer mark; this means "or equivalent". You will get the mark if your answer is numerically identical to the answer in the mark scheme. If your answer is grossly un-simplified though, you may lose the mark.

In this case, you would definitely get the mark.

I'm going to try and do a mark scheme here. Bare in mind that I'm in Year 12 too >.>

Firstly, thanks to Barrel who actually posted the paper (page 15). here are the pages in order:
http://www.thestudentroom.co.uk/attachment.php?attachmentid=426179&d=1434104177
http://www.thestudentroom.co.uk/attachment.php?attachmentid=426181&d=1434104202
http://www.thestudentroom.co.uk/attachment.php?attachmentid=426183&d=1434104269
http://www.thestudentroom.co.uk/attachment.php?attachmentid=426185&d=1434104300


$1. \,\, y = e^{2x} \cos x [br]\dfrac{dy}{dx} = e^{2x} (-\sin x) + 2e^{2x} \cos x[br]\dfrac{dy}{dx} = e^{2x} (2 \cos x - \sin x) \,\, [2][br]\mathrm{- solve for max:}[br]0 = e^{2x} (2 \cos x - \sin x) [br]2 \cos x = \sin x [br]\tan x = 2 \,\, [2][br]x = 1.1 \, \mathrm{to 2 s.f.}[br]\mathrm{- sub into y = e^{2x} \cos x}[br]y = 4.1 \, \mathrm{to 2 s.f.} [br]P(1.1, 4.1) \,\, [2]$


$2. \,\, \displaystyle \int \sqrt [3]{2x - 1} \, dx[br]\mathrm{- solve by inspection of substitution. Substitution would be u = 2x - 1}[br]= \dfrac {3}{8} (2x - 1)^{\frac {4}{3}} + c \,\, [4]$


$3. \,\, \displaystyle \int^2 _1 x^{3}\ln x \, dx[br]\mathrm{- integrate by parts:}[br]\dfrac {dv}{dx} = x^{3}, u = lnx[br]v = \dfrac {x^4}{4}, \dfrac{du}{dx} = \frac{1}{x} \,\, [2][br]\mathrm{- follow through [2], and}[br]= 4 \ln 2 - \frac{15}{16} \,\, [1]$


$4. \,\, V = \dfrac{1}{3} \pi r^{2} h, \, \dfrac{dV}{dt} = 5[br]\mathrm{- draw a triangle:}[br]\tan 45 = \dfrac{r}{h}[br]r = h[br]V = \dfrac{1}{3} \pi h^3 \,\, [2][br]\dfrac{dV}{dh} = \pi h^2[br]\mathrm{- chain rule:}[br]\dfrac{dh}{dt} = \dfrac{dV}{dt} \times \dfrac{dh}{dV} = \dfrac{\frac{dV}{dt}}{\frac{dV}{dh}}[br]\dfrac{dh}{dt} = \dfrac{5}{\pi h^2} \,\, [2][br]h = 10, \, \dfrac{dh}{dt} = \dfrac{5}{100 \pi} = \dfrac{1}{20 \pi} \,\, [1]$


$5. \,\, y^2 + 2x \ln y = x^2[br]\mathrm{- just put (1, 1) in - I'm not writing that out for you... [1]}[br]\mathrm{- differentiate implicitly:}[br]2y \dfrac{dy}{dx} + 2x \dfrac{1}{y} \dfrac{dy}{dx} + 2 \ln y = 2x[br]\dfrac{dy}{dx} 2 \left(y + \dfrac{x}{y} \right) = 2 \left(x - \ln y \right)[br]\dfrac{dy}{dx} = \dfrac{x - \ln y}{y + \frac{x}{y}} \,\, [3][br]\mathrm{- substitute in (1, 1)}[br]\dfrac{dy}{dx} = \dfrac{1 - 0}{1 + 1} = \dfrac{1}{2} \, [1]$


- I wasn't completely sure of the method for the second part, this is what a friend told me the method was (I think I do it correctly). Because of the nature of these two questions, however, I'm not 100% sure you need a method if you get the answer correct. It's only 1 mark each either way ~
$6. \,\, 6 \sin ^{-1} x - \pi = 0[br]\sin ^{-1} x = \dfrac{\pi}{6}[br]x = \sin \dfrac{\pi}{6} = \dfrac {1}{2} \,\, [2][br]\mathrm{method:}[br]\sin ^{-1} x = \cos ^{-1} x[br]\theta = \sin ^{-1} x, \, \theta = \cos ^{-1} x[br]x = \sin \theta, \, x = \cos \theta[br]\mathrm{- divide through (solve them simultaneously)}[br]1 = \tan \theta[br]\theta = \dfrac{\pi}{4}[br]x = \sin \dfrac{\pi}{4} = \dfrac{1}{\sqrt{2}} ( = \dfrac{\sqrt{2}}{2}) \,\, [2]$


$7. \,\, \mathrm{- for \ the \ first \ part \ you \ just \ substitute \ f(x) \ in \ wherever \ there \ is \ an \ x \ in \ f(x) \ [2]}[br]\mathrm{- if you really want to see it reply to this}[br]\mathrm{- for the second part, I didn't know this as an actual rule, but:}[br]f^{-1}(x) = f(x) \,\, [1][br]\mathrm{- sub (-x) into g(x), prove that g(x) = g(-x), easy. [2]}[br]\mathrm{- g(x) has a line of symmetry in the y-axis/is reflected in the y-axis owtte [1]}$


$8. \,\, f(x) = \dfrac{(x - 2)^2}{x} = \dfrac{x^2 - 4x + 4}{x}[br]f(x) = x - 4 + 4x^{-1}[br]\mathrm{- no quotient rule for me, thanks}[br]f'(x) = 1 - 4x^{-2} = 1 - \dfrac{4}{x^2} \,\, [2][br]f''(x) = 8x^{-3} = \dfrac{8}{x^3} \,\, [2][br]\mathrm{- solve for Q:}[br]0 = 1 - \dfrac{4}{x^2}[br]\dfrac{4}{x^2} = 1[br]4 = x^2[br]x = \pm 2[br]x = +2 \, \mathrm{is solution for P}[br]x = -2 \, \mathrm{for Q}[br]f(-2) = -8[br]Q(-2, -8) \,\, [2][br]f''(-2) = -1<0 \, \mathrm{so Q is a maximum} \,\, [1]$

$\mathrm{- \ verify \ is \ simple \ [2]}[br]\mathrm{- the \ next \ part \ can \ be \ done \ two \ ways...}[br]\mathrm{- rectangle - integral:}[br]\displaystyle A = 1 \times 3 - \int ^4 _1 f(x) \, dx[br]\mathrm{- or as an integral, top curve - bottom curve:}[br]\displaystyle A = \int ^4 _1 1 - f(x) \, dx[br]\mathrm{- follow either through}[br]\mathrm{- the integration is easy using the expansion of f(x)}[br]A = \dfrac{15}{2} - 4 \ln 4 \,\, [4][br]\mathrm{- next part...}[br]g(x) = f(x + 1) - 1 \, qed. \,\, [3][br]\displaystyle \int ^3 _0 g(x) \, dx = 4 \ln4 - \dfrac{15}{2}[br]= - \mathrm{[your answer from before]} \,\, [1][br]\mathrm{- I \ can't \ really \ say \ what \ the \ words \ for \ this \ answer \ are} \,\, [1]$


$9. \,\, f(x) = (e^x - 2)^2 - 1[br]0 = (e^x - 2)^2 - 1[br]1 = (e^x - 2)^2[br]e^x - 2 = \pm 1[br]e^x = 1, 3 \,\, \mathrm{(ln1 \ is \ 0 \ which \ is \ for \ O)}[br]x = \ln 3 \,\, [2][br]f'(x) = 2e^{x} (e^x - 2)[br]0 = 2e^{x} (e^x - 2)[br]e^x = 2[br]x = \ln 2 \,\, [3][br]f(\ln 2) = -1 \,\, [1][br]Q(\ln 2, -1)[br]\mathrm{- as I said earlier, it wants the 'enclosed' area...}[br]\displaystyle \int ^{\ln 3} _0 (e^x - 2)^2 - 1 \, dx[br]\displaystyle= \int ^{\ln 3} _0 e^{2x} - 4e^{x} + 3 \, dx[br]\mathrm{- follow through [4]}[br]A = 3 \ln 3 - 4 \,\, [1]$

$\mathrm{- \ swap \ y \ and \ x \ and \ rearrange \ etc}[br]f^{-1}(x) = \ln {( \sqrt {x + 1} + 2)} \,\, [3][br]domain: \, x \geq -1 \,\, [1][br]range: \, f^{-1}(x) \geq \ln 2 \,\, [1][br]\mathrm{- graph is reflection in y = x [2]}[br]\mathrm{- starts at (-1, ln 2) intercept at ln3, then curve}[br]\mathrm{- should cross with f(x) at y = x}$


Anything I missed or got wrong just ask. If you think it would be marked differently then say so too.