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C3 Maths A2 AQA 2016 (unofficial mark scheme new)

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can anyone please explain how to prove:

secx - tanx = -5, is the same as secx + tanx = -0.2

I'm literally so confused :s-smilie:
Reply 81
Original post by Parhomus
You can do (sec(x)- tan(x)) x ((secx+tanx)/secx+tanx)=-5
Then you get (sec^2x-tan^2x)/secx+tanx =-5
sec^2x-tan^2x=1 so u get 1/(secx+tanx)=-5
So secx+tanx=-1/5

IMG_7125.jpg
This is exactly what I put in the exam, I'm sure about x=-21.3 but not the other one as desmos is only showing one solution
Attachment not found
Original post by mathewnutt
can anyone please explain how to prove:

secx - tanx = -5, is the same as secx + tanx = -0.2

I'm literally so confused :s-smilie:


(secx - tanx)*(secx + tanx) = sec2x -tan2x = 1 (using the trig identity)

you can then say that 1/(secx - tanx) = secx + tanx = 1/-5 = -0.2
Original post by jtebbbs
IMG_7125.jpg
This is exactly what I put in the exam, I'm sure about x=-21.3 but not the other one as desmos is only showing one solution
Attachment not found


Your answers are correct for the values of x, I messed up on this one and didn't include a value of x before i did the +70 then /2 so I only got the -88.7
Reply 84
Original post by Parhomus
Your answers are correct for the values of x, I messed up on this one and didn't include a value of x before i did the +70 then /2 so I only got the -88.7


Thing is when you put x=-88.7 into the original equation it doesn't work? It gives you -0.2 instead of -0.5 so I feel like there's something wrong but I just can't find it
Reply 86
Here's the answer to the volume of revolution question, 95% sure it's rightby parts.jpg
Original post by jtebbbs
Thing is when you put x=-88.7 into the original equation it doesn't work? It gives you -0.2 instead of -0.5 so I feel like there's something wrong but I just can't find it


-88.7 doesn't work, I'm confused as to why that's the case though. It gives you a different tan value...
Reply 88
Original post by -jordan-
-88.7 doesn't work, I'm confused as to why that's the case though. It gives you a different tan value...


Maybe it's outside the domain? Was it -90<x<90 or have I remembered it wrong? Cause -247.4, which -88.7 comes from, is definitely a solution of cosx=-5/13 between -250 and 110, which I got from applying 2x-70 to the original domain
Original post by Parhomus
fg(x) means f(g(x)) so when you got the equation; g(x) wouldn't be 1/x^2 it would 1/x because the x in the f(x) just means the input and so g(x) could only be 1/x. I understand why it would make sense for it to be 1/x^2.


ah i see, thank you
Original post by jtebbbs
Maybe it's outside the domain? Was it -90<x<90 or have I remembered it wrong? Cause -247.4, which -88.7 comes from, is definitely a solution of cosx=-5/13 between -250 and 110, which I got from applying 2x-70 to the original domain


Yep I got it too, it was definitely -90<x<90, strange.
For the translation question, wouldn't the vector be (-5/2,0)?
Reply 92
Original post by -jordan-
Yep I got it too, it was definitely -90<x<90, strange.


-88.7 is a solution to the equation sec(2x-70) PLUS tan(2x-70)=-0.2, maybe that's got something to do with it? Was adding the equations that you're given to find secx (i.e. cosx) wrong?
Original post by -jordan-
Yep I got it too, it was definitely -90<x<90, strange.


Dunno if it helps anyone but 1.3 degrees works which is what I put down, I too got -88.7 but tried it and saw it didn't work so played around with it and got 1.3 which worked.
Original post by jtebbbs
-88.7 is a solution to the equation sec(2x-70) PLUS tan(2x-70)=-0.2, maybe that's got something to do with it? Was adding the equations that you're given to find secx (i.e. cosx) wrong?


Nope it isn't wrong, they got you to find that before you even did the second part. Everyone I've spoken to did what we did. If someone can find a valid reason or explanation as to what the actual solution was I'd be grateful.
(edited 7 years ago)
Reply 95
why is 5)a) 8ln8 - 8? don't you differentiate put it equal to 0 and get f(x) < .5ln8
Original post by Jupers
why is 5)a) 8ln8 - 8? don't you differentiate put it equal to 0 and get f(x) < .5ln8


That is the x value of the stationary point. You have to put it into the curve and find y. The range is what you get out of the function.
Reply 97
Original post by freddy4321
Dunno if it helps anyone but 1.3 degrees works which is what I put down, I too got -88.7 but tried it and saw it didn't work so played around with it and got 1.3 which worked.


How did you get 1.3 mate?
Original post by jtebbbs
How did you get 1.3 mate?


Sorry just checking now and 1.3 doesn't get you the correct answer, it gets you 5 instead of -5. I think the only answer is -21.3
(edited 7 years ago)
Thanks for this!!
And question 2(e) you just do the integral of [5] with the limits minus the previous answer from the Simpsons rule I think

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