The Student Room Group

OCR MEI C3 Maths June 2015

Scroll to see replies

All the functions that appeared in the paper, as follows:

(I'll try and add questions as I remember them, though I have almost no idea of question numbers :P )

f(x)=e2xcos⁡x\displaystyle f(x) = e^{2x} \cos x

~ differentiate and find the coordinate of the maximum (1.1, 4.1)

∫2x−13dx\displaystyle \int \sqrt [3]{2x - 1} dx

~ Integrate for:
38(2x−1)43+c\displaystyle \frac {3}{8} (2x - 1)^{\frac {4}{3}} + c


∫12x3ln⁡x\displaystyle \int^2 _1 x^{3}\ln x

~ Integration by parts for something I can't quite remember, but someone will bother to work out or ask about...


y2+2xln⁡y=x2\displaystyle y^2 + 2x \ln y = x^2

~ verify point (1, 1) which means sub it in and prove it still equates, then differentiate implicitly to get the gradient at that point, with was 1/2.


f(x)=1−x1+x\displaystyle f(x) = \frac {1 - x}{1 + x}

~ prove f(f(x)) = x, which involves subbing f(x) in in place of each x, then rearranging.
~ hence write down the inverse of f(x), which is f(x) (I personally don't understand why this was a 'hence' question - I haven't seen a rule that would describe this before, although it's obvious there is one)

g(x)=1−x21+x2\displaystyle g(x) = \frac {1 - x^2}{1 + x^2}

~ show g(x) is even - sub (-x) for x, and show that is the same as g(x). This means that g(x) has a line of symmetry in the y-axis, or x = 0.


f(x)=(x−2)22\displaystyle f(x) = \frac {(x - 2)^2}{2}

~ you had to differentiate this twice. I found it easier to expand the top and divide through by x than use the quotient rule, but either should work and I think the answer was f'(x) = 1 - 4x^(-2), then differentiate again for f''(x) = 8x^-3 you should fin the point Q(-2, -8), then show it is a maximum because f''(x) < 0

~ I think this was the question where you had to integrate to find the area enclosed by the function and the line y = 1 (after proving the intersections were x = 1 and 4), which was easy enough - either use a rectangle, or take the integral of 1 - f(x). I think my answer was 15/2 - 4ln4

translate (-1, -1) for:

g(x)=x2−3xx+1\displaystyle g(x) = \frac {x^2 - 3x}{x + 1}

~ you had to show this, which just meant showing f(x + 1) - 1 = g(x).
~ then, it asked you to write down the integral from 0 to 3 of g(x) with no further calculation, which is the integral from the second part (I don't know if they would want you to indicate that it was this value but negative), and then describe why this was the case. I guess you just talk about the translation and stuff ~

f(x)=(ex−2)2−1\displaystyle f(x) = (e^x - 2)^2 - 1

~ I can't remember quite what the questions were here, but I think you had to differentiate it, find x-intercept at ln3, and minimum at (ln2, -1)

I got for the inverse:
f−1(x)=ln⁡(x+1+2)\displaystyle f^{-1}(x) = \ln {( \sqrt {x + 1} + 2)}

~ domain of inverse x>= -1 range f-1(x) >= ln2
~ then sketch the inverse by reflecting in y = x. The easy way to do this was plot the images of Q and P, and make sure you cross the line y=x in the same place, and continue the line without changing gradient much (it was pretty much a flat line past the intersection with y=x).


and somewhere in the paper, two parts to a question:

6sin⁡−1x−π=0,x=12\displaystyle 6 \sin ^{-1} x - \pi = 0 , x = \frac {1}{2}
sin⁡−1x=cos⁡−1x,x=12\displaystyle \sin ^{-1} x = \cos ^{-1} x ,x = \frac {1}{\sqrt 2}
(edited 8 years ago)
Original post by Duskstar
All the functions that appeared in the paper, as follows:

f(x)=e2xcos⁡x\displaystyle f(x) = e^{2x} \cos x
Unparseable latex formula:

\displaystyle \int \root {3}{2x - 1} dx


∫12x3ln⁡x\displaystyle \int^2 _1 x^{3}\ln x
y2+2xln⁡y=x2\displaystyle y^2 + 2x \ln y = x^2
verify point (1, 1)
f(x)=1−x1+x\displaystyle f(x) = \frac {1 - x}{1 + x}
g(x)=1−x21+x2\displaystyle g(x) = \frac {1 - x^2}{1 + x^2}
f(x)=(x−2)22\displaystyle f(x) = \frac {(x - 2)^2}{2}
translate (-1, -1) for:
g(x)=x2−3xx+1\displaystyle g(x) = \frac {x^2 - 3x}{x + 1}
f(x)=(ex−2)2−1\displaystyle f(x) = (e^x - 2)^2 - 1
I got for the inverse:
Unparseable latex formula:

\displaystyle f^{-1}(x) = \ln {\root {x + 1} + 2}



and somewhere in the paper, two parts to a question:
6sin⁡−1x−π=0(x=π6\displaystyle 6 \sin ^{-1} x - \pi = 0 (x = \frac {\pi}{6}
Unparseable latex formula:

\displaystyle \sin ^{-1} x = \cos ^{-1} x (x = \frac {1}{\root 2}



For arcsinX and arccosx I set them both equal to theta. Theta =pi/4 then x equals root2/2
Original post by chizz1889
Ahaha safe then 😆 I got 4-3ln3 as it has to be positive right 😊


They'll probably accept positive or negative, because conceptually you can think of it as a negative area. I solved for the negative area then wrote |Area| = 4-3ln3 as well just in case.
Implicit differentiation gradient = 0.5
Original post by chizz1889
For arcsinX and arccosx I set them both equal to theta. Theta =pi/4 then x equals root2/2


Yep, I just had some formatting trouble lol.
Original post by chizz1889
Implicit differentiation gradient = 0.5


I got the same ~
Reply 246
Avatar for _Caz_
_Caz_
10
Any idea of grade boundaries?

Was surprised to see no modulus questions.

AND NO DODGY PROOF. YES.


Posted from TSR Mobile
Original post by _Caz_
Any idea of grade boundaries?

Was surprised to see no modulus questions.

AND NO DODGY PROOF. YES.


Posted from TSR Mobile


I know thank f**k haha
Reply 248
Avatar for _Caz_
_Caz_
10
Original post by pierre225
Such a nice exam ahhh so annoyed at myself for losing a mark on integration! hopefully 71/72 will be enough?


Dude I'd happily take 71/72

Posted from TSR Mobile
Reply 249
Avatar for Buses
Buses
9
Original post by _Caz_
Any idea of grade boundaries?

Was surprised to see no modulus questions.

AND NO DODGY PROOF. YES.


Posted from TSR Mobile


definitely higher than last year, since this paper was freakishly easy
I got so many answers wrong :frown:

Will I lose only 1 mark for wrong answers each, as long as method was right?
Original post by _Caz_
Any idea of grade boundaries?

Was surprised to see no modulus questions.

AND NO DODGY PROOF. YES.


Posted from TSR Mobile


Grade boundaries will probably be pretty normal.

I personally found it very easy - it was very calculus heavy, function heavy, and you could check literally 90% of your answers with a graphical calculator.
Original post by chizz1889
(1.11,4.09)


How did you get that? I got lost at was it tanx=2 or something?
Original post by chizz1889
Implicit differentiation gradient = 0.5


Well that's one i know i got right now XD

Why was the paper so ****ing hard?!?!?!?!
Original post by JamesExtra
How did you get that? I got lost at was it tanx=2 or something?


Yeah that's right but you had to have the calculator in rad mode then do arctan 2
Original post by Duskstar
Grade boundaries will probably be pretty normal.

I personally found it very easy - it was very calculus heavy, function heavy, and you could check literally 90% of your answers with a graphical calculator.


Nah i reckon the grade boundaries will be a bit lower than normal, everyone i've spoken to found it pretty hard. Even my asian friend said so haha
Original post by chizz1889
Yeah that's right but you had to have the calculator in rad mode then do arctan 2


Ah ok I left it in degrees. I was wondering why I was getting a ridiculous answer for Y. Oh well hopefully have lost much seeing as I got up to there
Original post by _Caz_
Dude I'd happily take 71/72

Posted from TSR Mobile


yeah I would too I actually did quite bad I couldn't answer loads of the questions
Original post by Buses
definitely higher than last year, since this paper was freakishly easy


You must be crazy if you think that paper was easy lol
Ok not like a maths genius or anything and I found it fairly easy compared to other papers. Only ones I got stumped on we're Q1 after tanx=2 and the arcsin=arccos. Expect the grade boundaries to be high to be honest

Quick Reply

Related discussions

Latest

Trending

Trending

Articles for you