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# C3 Maths A2 AQA 2016 (unofficial mark scheme new) watch

1. can anyone please explain how to prove:

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

I'm literally so confused
2. (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

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 550633550635
Attached Images

3. (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
(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
4. (Original post by jtebbbs)

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 550633550635
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
5. (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
join me everyone
7. Here's the answer to the volume of revolution question, 95% sure it's right
8. (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...
9. (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
10. (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
11. (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.
12. For the translation question, wouldn't the vector be (-5/2,0)?
13. (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?
14. (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.
15. (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.
16. why is 5)a) 8ln8 - 8? don't you differentiate put it equal to 0 and get f(x) < .5ln8
17. (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.
18. (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?
19. (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
20. 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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