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Finding the range of 1 / (2x-1)

In the mark scheme , they are saying: x > 0 ....But I put down x>0 , x cannot equal 0.....Do you think i would be marked wrong::frown::frown::frown:
Reply 1
Are you given a domain? I can't decipher your unintelligible sprawl.
Reply 2
Original post by laurawoods
Finding the range of 1 / (2x-1)

In the mark scheme , they are saying: x > 0 ....But I put down x>0 , x cannot equal 0.....Do you think i would be marked wrong::frown::frown::frown:



isnt that the same thing -so youve written down the right answer
Reply 3
Original post by laurawoods
Finding the range of 1 / (2x-1)

In the mark scheme , they are saying: x > 0 ....But I put down x>0 , x cannot equal 0.....Do you think i would be marked wrong::frown::frown::frown:


x > 0 implies x =/= 0

So you were redundant, but not wrong. They may dock one mark, but only if they're being particularly harsh.
Reply 4
Original post by laurawoods
Finding the range of 1 / (2x-1)

In the mark scheme , they are saying: x > 0 ....But I put down x>0 , x cannot equal 0.....Do you think i would be marked wrong::frown::frown::frown:


The range would not be x anything
Reply 5
Original post by laurawoods
Finding the range of 1 / (2x-1)

In the mark scheme , they are saying: x > 0 ....But I put down x>0 , x cannot equal 0.....Do you think i would be marked wrong::frown::frown::frown:


Are you sure you are looking at the correct answer?

Reply 6
Original post by steve2005
Are you sure you are looking at the correct answer?



Presumably, the OP means y > 0, and the domain is restricted to x > 0.5
Reply 7
Original post by dendodge
Presumably, the OP means y > 0, and the domain is restricted to x > 0.5


So many possibilities :smile:

Probably this one :biggrin:

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