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AQA questions

Question 6 (d)(i) I set f(x) equal to f-1(x) but can't get it into the form required.

Question 12 (a) (ii) did part (i) and got xtanx -Ln(secx) + C
Don't know what to do next.
AQA-XMCA2-W-QP-Jun09.pdf90.7KB
Reply 1
Avatar for TenOfThem
TenOfThem
18
Original post by thers
Question 6 (d)(i) I set f(x) equal to f-1(x) but can't get it into the form required.


It is easier than that

Those graphs will meet when y=x

Use that
Reply 2
Avatar for thers
thers
OP
5
Oh of course, f(x) and f-1(x) are reflections in y=x. Ok done that now 12 (a) (ii)
Reply 3
Avatar for TenOfThem
TenOfThem
18
Original post by thers


Question 12 (a) (ii) did part (i) and got xtanx -Ln(secx) + C
Don't know what to do next.


OMG thick me
(edited 11 years ago)
Reply 4
Avatar for thers
thers
OP
5
Wait a sec in the formula book it says the integral of tanx is Ln(secx) :confused:
Reply 5
Avatar for TenOfThem
TenOfThem
18
Original post by thers
Wait a sec in the formula book it says the integral of tanx is Ln(secx) :confused:


ok

I tend to use -ln|cosx|

either way :smile:
Reply 6
Avatar for thers
thers
OP
5
Original post by TenOfThem
Then use the fact that

xtan^2x = (xsec^x)sin^2x and do parts again [I think]


Ok I've tried this but its got way too complex. Its only 2 marks, is there an easy way?
Reply 7
Avatar for TenOfThem
TenOfThem
18
Original post by thers
Ok I've tried this but its got way too complex. Its only 2 marks, is there an easy way?


Yes

Me being dumb :frown:

Do you know a rule that will turn tan^2x into sec^2x
Reply 8
Avatar for thers
thers
OP
5
Oh right tan^2x + 1 = sec^2x

Got it now. ty
Reply 9
Avatar for TenOfThem
TenOfThem
18
Original post by thers
Oh right tan^2x + 1 = sec^2x

Got it now. ty


:biggrin:

so obvious when I looked again

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