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functions [#permalink] New post 08 Feb 2009, 23:09
Pls explain ur answer
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Re: functions [#permalink] New post 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) .
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Re: functions [#permalink] New post 09 Feb 2009, 05:11
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] New post 09 Feb 2009, 07:45
OA is E. Thanks for discussion
Re: functions   [#permalink] 09 Feb 2009, 07:45
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