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VP
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functions [#permalink]

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New post 09 Feb 2009, 00:09
This topic is locked. If you want to discuss this question please re-post it in the respective forum.

Pls explain ur answer
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Senior Manager
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Re: functions [#permalink]

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New post 09 Feb 2009, 01:43
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) .

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Re: functions [#permalink]

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New post 09 Feb 2009, 06:11
ritula wrote:
Pls explain ur answer



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

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Re: functions [#permalink]

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New post 09 Feb 2009, 08:45
OA is E. Thanks for discussion

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Re: functions   [#permalink] 09 Feb 2009, 08:45
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