The Student Room Group

OCR MEI C3 Maths June 2015

Scroll to see replies

Everyone got 5*(1/100pi) for the rate of change question right? It went to 0.0159 or something?
Original post by conorw6
Everyone got 5*(1/100pi) for the rate of change question right? It went to 0.0159 or something?


YES! :wink:
Original post by conorw6
Everyone got 5*(1/100pi) for the rate of change question right? It went to 0.0159 or something?



I simplified to 120π\frac{1}{20\pi}
What were the limits for the integral on question 8??? Was it 4 and 1?? Some ppl are telling me we had to use x coordinate of P?
Original post by corleone12
What were the limits for the integral on question 8??? Was it 4 and 1?? Some ppl are telling me we had to use x coordinate of P?


It is 1 and 4, then you subtract it from the area of rectangle 3.
Original post by RemainSilent
same her lol i just did something like sin x = cos x then tan came out as pi/4 and it didnt wokr out


Wait my I got tan(x) = 1
So my x was pi/4


Posted from TSR Mobile
Original post by Primus2x
It is 1 and 4, then you subtract it from the area of rectangle 3.


What was the final area ?
For 9iii) I was correct up until substituting in my limits and where I got e^2x, I just substituted in ln(3) as when it should have been 2ln(3) causing me to get the wrong answer. How many marks do you think I will get out of 5?

3?
Original post by corleone12
What was the final area ?


I don't remember, sorry. If someone tell me what the function was, I could do it.
(edited 8 years ago)
Original post by Primus2x
I don't remember, sorry.


Somehow i got 3 for the integral and 4 for rectangle over all i got area of 1 :frown:
Original post by corleone12
YES! :wink:

yessss!
Original post by Primus2x
I simplified to 120π\frac{1}{20\pi}


I'm pretty sure they accept it in exact form or with suitable rounding!
Please vote on this poll for A boundary predictions:
http://www.thestudentroom.co.uk/showthread.php?t=3399719
Wait, I remember now.
(Someone seriously needs to fix latex on TSR)
Copy the code to this site http://www.codecogs.com/latex/eqneditor.php
Unparseable latex formula:

[br]f(x)=\frac{(x-2)^2}{x}=x-4+\frac{4}{x}[br]\\[br]\int_{1}^{4} f(x) dx=\left [ \frac{x^2}{2}-4x+4\ln{x} \right ]_{1}^{4}=[8-16+8\ln2]-\left [\frac{1}{2}-4+0 \right ][br]\\[br]=\frac{-9}{2}+8\ln{2}[br]\\[br]Area=3-\left [\frac{-9}{2}+8\ln{2}[br] \right ]=\frac{15}{2}-8\ln{2}[br]

Reply 493
Avatar for ETRC
ETRC
17
Original post by Connorbwfc
For arcsin(x)=arcos(x)

Would you get any marks for just putting the answer as Root(2) / 2 ?????


it was 2 marks so i think you will get 1. this was probably the trickiest question on the paper to get 2 marks in i think- mei always have an a* question in c3/c4 papers so this is probably it.

i used arccos x + arcsin x= pi/2 and the equation to get root2 /2

i think you might have to say because they are the smae values at pi/4 or something or solve it properly using an identity
If you can't read the equations that were parsed into that mess:

The answer to the area bounded by f(x) and y=1 should be 152−8ln⁡2\frac{15}{2}-8\ln{2}

And the integral of g(x) with respect to x between 0 and 3 is −152+8ln⁡2-\frac{15}{2}+8\ln{2} because if you translate y=1, x=1, x=4 by the same vector that transformed f(x) into g(x), they become y=0, x=0, and x=3 and that is equivalent to integrating g(x) between 0 and 3.
Reply 495
Avatar for ETRC
ETRC
17
Original post by Primus2x
If you can't read the equations that were parsed into that mess:

The answer to the area bounded by f(x) and y=1 should be 152−8ln⁡2\frac{15}{2}-8\ln{2}

And the integral of g(x) with respect to x between 0 and 3 is −152+8ln⁡2-\frac{15}{2}+8\ln{2} because if you translate y=1, x=1, x=4 by the same vector that transformed f(x) into g(x), they become y=0, x=0, and x=3 and that is equivalent to integrating g(x) between 0 and 3.


you can also say area is the same but under x-axis instead of above it
or you can say modulus area is the same
I think I have got between 59 and 64 on this paper. Is it still possible to get an A*?
53 = B grade boundary?
Original post by ETRC
it was 2 marks so i think you will get 1. this was probably the trickiest question on the paper to get 2 marks in i think- mei always have an a* question in c3/c4 papers so this is probably it.

i used arccos x + arcsin x= pi/2 and the equation to get root2 /2

i think you might have to say because they are the smae values at pi/4 or something or solve it properly using an identity


I did it like this
Unparseable latex formula:

[br]\arcsin{x}=\arccos{x}[br]\\[br]x=\sin\left ( \arccos{x} \right )[br]\\[br]x=\sqrt{1-\left ( \cos\left ( \arccos{x} \right ) \right )^2} \because \sin^2 \theta+\cos^2 \theta =1 \Rightarrow \sin \theta=\sqrt{1-\cos^2 \theta}[br]\\x=\sqrt{1-x^2}[br]x^2=1-x^2[br]\\[br]2x^2=1[br]\\[br]x^2=\frac{1}{2}[br]\\[br]x=\sqrt{\frac{1}{2}}[br]\\[br]\therefore x=\frac{\sqrt{2}}{2} \because \text{rationalise denominator}[br]

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=e2xcos⁡x[br]dydx=e2x(−sin⁡x)+2e2xcos⁡x[br]dydx=e2x(2cos⁡x−sin⁡x)  [2][br]−solveformax:[br]0=e2x(2cos⁡x−sin⁡x)[br]2cos⁡x=sin⁡x[br]tan⁡x=2  [2][br]x=1.1 to2s.f.[br]−subintoy=e2xcos⁡x[br]y=4.1 to2s.f.[br]P(1.1,4.1)  [2]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.  ∫2x−13 dx[br]−solvebyinspectionofsubstitution.Substitutionwouldbeu=2x−1[br]=38(2x−1)43+c  [4]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.  ∫12x3ln⁡x dx[br]−integratebyparts:[br]dvdx=x3,u=lnx[br]v=x44,dudx=1x  [2][br]−followthrough[2],and[br]=4ln⁡2−1516  [1]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=13πr2h, dVdt=5[br]−drawatriangle:[br]tan⁡45=rh[br]r=h[br]V=13πh3  [2][br]dVdh=πh2[br]−chainrule:[br]dhdt=dVdt×dhdV=dVdtdVdh[br]dhdt=5πh2  [2][br]h=10, dhdt=5100π=120π  [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.  y2+2xln⁡y=x2[br]−justput(1,1)in−I′mnotwritingthatoutforyou...[1][br]−differentiateimplicitly:[br]2ydydx+2x1ydydx+2ln⁡y=2x[br]dydx2(y+xy)=2(x−ln⁡y)[br]dydx=x−ln⁡yy+xy  [3][br]−substitutein(1,1)[br]dydx=1−01+1=12 [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.  6sin⁡−1x−π=0[br]sin⁡−1x=π6[br]x=sin⁡π6=12  [2][br]method:[br]sin⁡−1x=cos⁡−1x[br]θ=sin⁡−1x, θ=cos⁡−1x[br]x=sin⁡θ, x=cos⁡θ[br]−dividethrough(solvethemsimultaneously)[br]1=tan⁡θ[br]θ=π4[br]x=sin⁡π4=12(=22)  [2]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.  −for the first part you just substitute f(x) in wherever there is an x in f(x) [2][br]−ifyoureallywanttoseeitreplytothis[br]−forthesecondpart,Ididn′tknowthisasanactualrule,but:[br]f−1(x)=f(x)  [1][br]−sub(−x)intog(x),provethatg(x)=g(−x),easy.[2][br]−g(x)hasalineofsymmetryinthey−axis/isreflectedinthey−axisowtte[1]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)=(x−2)2x=x2−4x+4x[br]f(x)=x−4+4x−1[br]−noquotientruleforme,thanks[br]f′(x)=1−4x−2=1−4x2  [2][br]f′′(x)=8x−3=8x3  [2][br]−solveforQ:[br]0=1−4x2[br]4x2=1[br]4=x2[br]x=±2[br]x=+2 issolutionforP[br]x=−2 forQ[br]f(−2)=−8[br]Q(−2,−8)  [2][br]f′′(−2)=−1<0 soQisamaximum  [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]

− verify is simple [2][br]−the next part can be done two ways...[br]−rectangle−integral:[br]A=1×3−∫14f(x) dx[br]−orasanintegral,topcurve−bottomcurve:[br]A=∫141−f(x) dx[br]−followeitherthrough[br]−theintegrationiseasyusingtheexpansionoff(x)[br]A=152−4ln⁡4  [4][br]−nextpart...[br]g(x)=f(x+1)−1 qed.  [3][br]∫03g(x) dx=4ln⁡4−152[br]=−[youranswerfrombefore]  [1][br]−I can′t really say what the words for this answer are  [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)=(ex−2)2−1[br]0=(ex−2)2−1[br]1=(ex−2)2[br]ex−2=±1[br]ex=1,3  (ln1 is 0 which is for O)[br]x=ln⁡3  [2][br]f′(x)=2ex(ex−2)[br]0=2ex(ex−2)[br]ex=2[br]x=ln⁡2  [3][br]f(ln⁡2)=−1  [1][br]Q(ln⁡2,−1)[br]−asIsaidearlier,itwantsthe′enclosed′area...[br]∫0ln⁡3(ex−2)2−1 dx[br]=∫0ln⁡3e2x−4ex+3 dx[br]−followthrough[4][br]A=3ln⁡3−4  [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]

− swap y and x and rearrange etc[br]f−1(x)=ln⁡(x+1+2)  [3][br]domain: x≥−1  [1][br]range: f−1(x)≥ln⁡2  [1][br]−graphisreflectioniny=x[2][br]−startsat(−1,ln2)interceptatln3,thencurve[br]−shouldcrosswithf(x)aty=x\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.
(edited 8 years ago)

Quick Reply

Related discussions

Latest

Trending

Trending

Articles for you