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# functions

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Joined: 18 May 2008
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functions [#permalink]  08 Feb 2009, 23:09
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Senior Manager
Joined: 08 Jan 2009
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Re: functions [#permalink]  09 Feb 2009, 00:43
Since a and b are positive integers i took a=b=1. so a+b=2.

only the last f(x) = -3x

f(2) = -6 and
f(1) = -3

therefore f(2) = f(1) + f(1) .
Manager
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Re: functions [#permalink]  09 Feb 2009, 05:11
ritula wrote:

f(a+b) = f(a) + f(b)

1) f(x) = $$x^2$$
f(a+b) = $$(a+b)^2$$ is not equal to f(a) = $$a^2$$ + f(b) = $$b^2$$

2) f(x) = x+1
f(a+b) = a+b+1 is not equal to f(a) = a+ f(b) = b

3) f(x) = $$(x)^1/2$$
f(a+b) = sqrt(a+b) is not equal to f(a) = $$a^1/2$$ + f(b) = $$b^1/2$$

4) f(x) = $$2/x$$
f(a+b) = $$2/(a+b)$$ is not equal to f(a) = $$2/(a)$$ + f(b) = $$2/(b)$$

5) f(x) = -3x
f(a+b) = -3(a+b) is equal to f(a) = -3a + f(b) = -3b

E
VP
Joined: 18 May 2008
Posts: 1300
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Re: functions [#permalink]  09 Feb 2009, 07:45
OA is E. Thanks for discussion
Re: functions   [#permalink] 09 Feb 2009, 07:45
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