AQA A2 Mathematics MPC3 Core 3 - Wednesday 15th June 2016 [Official Thread]

Anyone know how to draw the graph for y=f(|x|), in comparison to y= |f(x)|

do the values not have to be satisfied by the ranges of both parent functions?
(i'm just asking in general not this particular question)

Anyone know how to draw the graph for y=f(|x|), in comparison to y= |f(x)|

Hi, can anyone help with question 1. I have differentiated and I know how to integrate for part b)ii) doing by parts but it says hence and the mark scheme gets straight to the answer after one step, is there a shorter way of integrating it?

http://www.mrbartonmaths.com/resources/A%20Level%20Past%20Papers/Core%203/2013%20-%20June/AQA-MPC3-QP-Jun13.pdf

question 6 . can anyone explain to me how the second graph works...? i thought it would be a reflection in the x axis and a movement of pi in the y direction. But in the mark scheme they've done what looks like just a reflection in the y axis.

Mark scheme
http://www.mrbartonmaths.com/resources/A%20Level%20Past%20Papers/Core%203/2013%20-%20June/AQA-MPC3-W-MS-June13.pdf

Are you familiar with integration by recognition/reverse chain rule?

The reflection in the x axis followed by a translation by pi in the y direction is, in this case, a reflection in the y axis.

I think so, I tried doing that so I put the 8 back in as the coefficient of x and then took 1/8 out of the integral to account for this, but then what happens to the 8x?

Way I think about it is that you must have had (x^2-6)^4 somewhere in order to differentiate and get back to it cubed, and if you differentiate that you get 4(x^2-6)^3 * 2x which gives you 8x(x^2-6)^3. You want what differentiates to give you just x(x^2-6)^3, so you want 1/8(x^2-6)^4 to get that. It's kinda hard to explain so https://www.youtube.com/watch?v=Fag7U3NtUmw might be a better bet if that doesn't make sense

lol i wouldn't know where to start if you gave me a substution then i could do it

Ah thank you so much! Yes that makes sense, the video is really useful too he goes through a lot of examples so thank you

One to one just means one value of x maps onto one value of y, and it depends on the function for example y=x^2 can have a domain for all real values but we know that f(x)>or equal to 0 as f(x) can't be negative, so just look at what they give and try a few numbers with your calculator both positive and negative and you will see its range.

Any ideas for 5B june 2014??

Try this integral.

My attempt, no idea if correct.

EDIT: +C of course