# AQA questionsWatch

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#1
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.
0
6 years ago
#2
(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
0
#3
Oh of course, f(x) and f-1(x) are reflections in y=x. Ok done that now 12 (a) (ii)
0
6 years ago
#4
(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
0
#5
Wait a sec in the formula book it says the integral of tanx is Ln(secx)
0
6 years ago
#6
(Original post by thers)
Wait a sec in the formula book it says the integral of tanx is Ln(secx)
ok

I tend to use -ln|cosx|

either way
0
#7
(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?
0
6 years ago
#8
(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

Do you know a rule that will turn tan^2x into sec^2x
0
#9
Oh right tan^2x + 1 = sec^2x

Got it now. ty
0
6 years ago
#10
(Original post by thers)
Oh right tan^2x + 1 = sec^2x

Got it now. ty

so obvious when I looked again
0
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