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# C3 ANSWERS (less discussion) watch

1. (Original post by Wooty)
How was the 6 mark question proof done? The one where it ended up 4k+sink4k-2=0

And also the last proof where you had to differentiate something to get sin2A..?
This is how i did it:

y = (2x-1)(tan(2x))

dy/dx = (2k-1).2Sec^2(2k) + tan(2k).2

= (2k-1)(2sec^2(2k) + 2tan(2k)

At S.P (Min point) dy/dx = 0

Therefore (2k-1)(2sec^2(2k) + 2tan(2k) = 0

So, (2k-1)(2/Cos^2(2k)) + 2(Sin(2k)/Cos(2k)) = 0

Mutliply through by Cos2k

(2k-1)(2/Cos(2k)) + 2(sin(2k)) = 0

(4k-2)/(Cos(2k)) + 2(Sin(2k)) = 0

Multiply through by Cos(2k)

(4k-2) + 2Sin(2k)Cos(2k) = 0

[Identity Sin2x = 2SinxCosx, so Sin4x = 2sin(2x)cos(2x)]

4k+ sin4k -2 =0
2. (Original post by puck_fella)
The domain i seem to remember was restricted. This confused me, i put f(x) > ln 0, which, on reflection is not right. But oh well, only one mark.
If i remember correctly, the domain was restricted to x> -k, because x=-k would be the asmyptote. This does not limit the range in any way. The graph of lnx has simply been shifted k units in the negative direction.

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