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by vivek.kapoor83 » Sat Oct 11, 2008 9:30 am
IMO E.
coz f(a+b) = f(a)+f(b)
we have to prove.,
then
putiing x=a+b in E
f(a+b) = -3(a+b) = -3a-3b
put x = a
f(a) = -3a
put x =b
f(b) = -3b
f(a)+f(b) = -3a-3b

Hence,E.
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by cramya » Sat Oct 11, 2008 9:44 am
It should be E

This ha sbeen solved in this forum before but cant find the link

Pick numbers lets say a=2 b=3

f(5) = f(2) + f(3)

x=5, x=2,x=3 WEvaluate the choices and E will be true
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by smallsorrow » Sat Oct 11, 2008 10:44 am
e is correct. thanks for explaining!
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by elmiko » Sat Oct 11, 2008 3:45 pm
cramya wrote:It should be E

This ha sbeen solved in this forum before but cant find the link

Pick numbers lets say a=2 b=3

f(5) = f(2) + f(3)

x=5, x=2,x=3 WEvaluate the choices and E will be true
I don't understand how you plug in the choices into each of the functions in the answer choices. Can you explain further?
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by canuckclint » Sat Oct 11, 2008 5:58 pm
elmiko wrote:
cramya wrote:It should be E

This ha sbeen solved in this forum before but cant find the link

Pick numbers lets say a=2 b=3

f(5) = f(2) + f(3)

x=5, x=2,x=3 WEvaluate the choices and E will be true
I don't understand how you plug in the choices into each of the functions in the answer choices. Can you explain further?
You can plug for example
f(x) = x^2 take a = 2 b = 3
f(a+b) = f(5) = 5^2
not equal to
f(a) + f(b) = f(2) + f(3) = 2^2 + 3^2

Actually there is no need to plug if you got the right answer.
Just remember factoring from algebra.

k(a+b) = ka + kb for any k. just from algebra
Part E is just sucah a case with k=-3

Cheers,
Clinton
Math Minor
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