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MGMAT - Functions

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by camitava » Fri Sep 21, 2007 3:29 am
f(x)=(x-1)^2, What is the value of f(x-1)?

A. x^2+2x+1
B. x^2-4x+2
C. x^2-4x+4
D. x^2+4x+4
Just put (x - 1) in place of x.
So f(x -1) = (x - 1 - 1)^2
= x^2 - 4x + 4 Using the formula - (a - b)^2 = a^2 - 2ab + b^2.

Got it?
Correct me If I am wrong


Regards,

Amitava
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by cveluswamy » Fri Sep 21, 2007 3:56 am
Perfect. Thanks
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by Suyog » Fri Sep 21, 2007 4:08 am
Use Formula (a-b)^2 = a^2 - 2ab + b^2

f(x) = (x-1)^2
Therefore,

f(x-1) = [(x-1)^2 - 1]^2
= [(x-1)^2 - [2(x-1)(1)] + (1)^2] ; Using the formula
= [(x-1)^2 - 2x + 3]
= [(x^2 - 2x +1) - 2x +3] ; Using the formula
= x^2 -2x +1 -2x+3
= x^2 - 4x +4

Ans C.
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