OCR MEI C3 Maths June 2015

1. (1.107, 4.094) [6]
2. (3(2x-1)^(4/3))/8 + c [4]
3. u = 2x-1, (1/16)(64ln2 - 15) [5]
4. r=h, 1/20pi = 0.0159 [5]
5. gradient = ½ [6]
6. i) x = ½ [2]
ii) x = sqrt(2)/2 [2]
7. i) f(f(x)) = x via substituting in. f^-1(x) = f(x) = (1-x)/(1+x) [3]
ii) g(x) is even since g(-x) = g(x). Symmetrical about the y-axis. [3]
8. i) f^'(x) = 1 - 4/x², f^''(x) = 8/x³ which is less than zero when x=-1 and so Q is a maximum point. [7]
ii) Area was 3 take away integral which gave ½(15-8ln4) [6]
iii) f(x-1)-1=g(x) so g(x) = (x²-3x)/(x+1) [3]
iv) This is the negative of the previous answer, and yes it's negative, because it asked for the value. [2]
9. i) x = ln3 [2]
ii) x = ln2, y = -1 [4]
iii) The integral came out as negative, but this question asked for the area, so the given answer must be positive; 4 - 3ln3 [5]
iv) f^-1(x) = ln(sqrt(x+1)+2), with the domain as x >= -1 and range as y >= ln2. The sketch of the graph was a reflection in the line y = x, but subject to the 'new' domain and range (i.e. not below ln2). [7]

Please quote me if I have made a mistake...

8.ii) I'm pretty sure answer was 1/2(15-8ln2) not ln4 😊

I got 15/2 - 8ln2... I think?

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Mark scheme done

http://www.thestudentroom.co.uk/showthread.php?t=3332257&page=26&p=56931025#post56931025

I got 15/2 - 8ln2... I think?

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I got 15/2 - 8ln2... I think?

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The answer was 15/2 - 4ln4

which is the same as what you got

Yeah that's right 😀

I'm fairly certain it was 15/2 - 4ln2

EDIT: Infact no it was 3ln3!!! I think, I will work it out again in a sec 😊

arcsinx=arccosx --> tanx=1 -->x=pie/4

I forgot about the rectangle in q.8 completely worked out the area perfectly for the integral of f(x) just forgot to to minus it from the area of the rectangle. I didn't do any calculations for the rectangle either, or write out how to obtain the area. I also could work out what it was on about for g(x), so put it into my graphical calculator and go -1.95... the decimal answer. And I also messed up my graph but I did a some sort of sketch (but then wrote reflected in the line y=x. I think these were my only mistakes, that I know of.

How many marks would I lose for this?

arcsinx=arccosx --> tanx=1 -->x=pie/4

I think it might be sqrt2/2 although I put pi/4 as well. :/

Yep, refer to this:

$\sin ^{-1} x = \cos ^{-1} x[br]\theta = \sin ^{-1} x, \, \theta = \cos ^{-1} x[br]x = \sin \theta, \, x = \cos \theta[br]\mathrm{- divide through (solve them simultaneously)}[br]1 = \tan \theta[br]\theta = \dfrac{\pi}{4}[br]x = \sin \dfrac{\pi}{4} = \dfrac{1}{\sqrt{2}} ( = \dfrac{\sqrt{2}}{2})$

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 ~
$6. \,\, 6 \sin ^{-1} x - \pi = 0[br]\sin ^{-1} x = \dfrac{\pi}{6}[br]x = \sin \dfrac{\pi}{6} = \dfrac {1}{2} \,\, [2][br]\mathrm{method:}[br]\sin ^{-1} x = \cos ^{-1} x[br]\theta = \sin ^{-1} x, \, \theta = \cos ^{-1} x[br]x = \sin \theta, \, x = \cos \theta[br]\mathrm{- divide through (solve them simultaneously)}[br]1 = \tan \theta[br]\theta = \dfrac{\pi}{4}[br]x = \sin \dfrac{\pi}{4} = \dfrac{1}{\sqrt{2}} ( = \dfrac{\sqrt{2}}{2}) \,\, [2]$


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


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

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


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

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


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

e^x - 2 = \pm 1
e^x = 0, 2 \,\, \mathrm{(0 \ is \ impossible)}

I'm sorry if this has already been corrected, but in the solution of 9) you state that 2+1=2 and 2-1=0.