OCR MEI C3 Maths June 2015

In the integration in Q8 where you had to do the area of the rectangle - the area under the graph, was the area of the rectangle 3(units^2)?

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

guys for volume ques; if i put r = hsin 45 (stupid me :/ ) and done it but i got a differnet dV/dr would i get any follow through marks? also would you get a mark just writing dV/dt or something = 5 ?

is someone able to draw the graph please think mines wrong

Can you explain this stage of your working please?

How many marks was question 5 ? Specifically for finding the derivative?


Posted from TSR Mobile

set each side of the equation equal to theta (an angle)
rearrange to obtain x in terms of theta
divide through to obtain expression in terms of theta
work out theta
work out x using one of the 'x in terms of theta' equations
$\sin ^{-1} x = \cos ^{-1} x[br]\theta = \sin ^{-1} x, \, \theta = \cos ^{-1} x[br]x = \sin \theta, \, x = \cos \theta[br]\mathrm{- divide through (solve them simultaneously)}[br]1 = \tan \theta[br]\theta = \dfrac{\pi}{4}[br]x = \sin \dfrac{\pi}{4} = \dfrac{1}{\sqrt{2}} ( = \dfrac{\sqrt{2}}{2})$

I swear I didn't quote you...your working's fine .

Oh I thought you just wanted to know how they did it lol. And they didn't do it correctly anyway, but those steps are how you get the tan bit ~ thanks

You probably lost 2 marks for not doing the rectangle, and then 2 marks for the g(x) question, because the intention was to literally write down you answer before * -1. On the sketch question you probably lost 1 or 2 marks on the actual sketch (out of 2) depending what qualities it had of what it was supposed to be.

wasnt there an asymptote at x=3

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.

Didn't do this exam (I do edexcel), but I like snooping on other papers. Here's how I would do 6b)

Arcsin(x))=arccos(x)

Cos(arcsin(x))=x

Sin(90-arcsin(x))=x

90-arcsin(x)=arcsin(x)

Then solve for x like this

Neat