OCR MEI C3 Maths June 2015

isn't it reflection in y=x ?

Yeah that's right 😊

Did everyone remember to subtract their answer to the integration from 3 for the area between y=1 and x axis? ... ☺️

Rate one i got 0.023 or something like that


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That is what I meant lol

isnt it reflection in y=x?? andhow about inverse

yeah I put 0.016 to 3 s.f

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 )

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

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

$\displaystyle \int \sqrt [3]{2x - 1} dx$

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


$\displaystyle \int^2 _1 x^{3}\ln x$

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


$\displaystyle y^2 + 2x \ln y = x^2$

~ 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.


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

~ 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)

$\displaystyle g(x) = \frac {1 - x^2}{1 + x^2}$

~ 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.


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

~ 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:

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

~ 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 ~

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

~ 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:
$\displaystyle f^{-1}(x) = \ln {( \sqrt {x + 1} + 2)}$

~ 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:

$\displaystyle 6 \sin ^{-1} x - \pi = 0 , x = \frac {\pi}{6}$
$\displaystyle \sin ^{-1} x = \cos ^{-1} x ,x = \frac {1}{\sqrt 2}$

Thats 2 sig fig...

Think I got that. What was the radius equal to?

no it isn't

radius was equal to h

.016 is three decimal places but 2 sig fig as the 0 after the decimal place is not significant. Just as the 160 would be 2 sig fig not three.

Not that it really mattera that much.