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# P2 question watch

1. show that:

3x-1-cos2x=0

has only one real root.

How would you do that without drawing it?
2. (Original post by Freddy C)
show that:

3x-1-cos2x=0

has only one real root.

How would you do that without drawing it?
Let f(x) = 3x - 1 - cos2x

Then f'(x) = 3 + 2sin2x

Note that f'(x) >0 and so the function is always increasing

In particular it can have at most one root [it still could have no roots]

But note that f(0) = -2 and f(1) = 3-1-cos2 >0

and so the function has a root between 0 and 1.

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