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Interesting GMATFix Problem-27

by arora007 » Tue Sep 21, 2010 1:03 pm
If f(x)=x^2 -x -16 and g(x)=1/(x+10) , which of the following CANNOT be the value of g(f(x)) ?
A)1
B)0
C)-2
D)-3
E)-10
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by diebeatsthegmat » Tue Sep 21, 2010 11:16 pm
arora007 wrote:If f(x)=x^2 -x -16 and g(x)=1/(x+10) , which of the following CANNOT be the value of g(f(x)) ?
A)1
B)0
C)-2
D)-3
E)-10
is B the answer/
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by arora007 » Wed Sep 22, 2010 2:45 am
cool...B is the answer... but the steps?
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by selango » Wed Sep 22, 2010 4:05 am
g(f(x))=1/(x^2 -x -16+10)

=1/(x^2-x-6)

=1/(x-3)(x+2)

In this equation 0 is not possible as the numerator is not equal to 0.Even if we have to move the fraction from denominator to numerator,the fraction 1/0 is not possible.

Other values can be obtained by sub x as fraction.

Pick B
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by pzazz12 » Wed Sep 22, 2010 5:05 am
diebeatsthegmat wrote:
arora007 wrote:If f(x)=x^2 -x -16 and g(x)=1/(x+10) , which of the following CANNOT be the value of g(f(x)) ?
A)1
B)0
C)-2
D)-3
E)-10
is B the answer/



g(f(x)) means we have to place f(x) in the place of x
g(f(x))=1/(x^2-x-16+10)
=1/(x^2-x-6)
=1/(x-3)(x+2)
so g(f(x)) should not be the 0.
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by kaushiksin » Wed Sep 22, 2010 5:16 am
If f(x)=x^2 -x -16 and g(x)=1/(x+10)

g(f(x) = 1/(x^2-x-16+10) = 1/(x-3)(x+2)

x-3 # 0 or
x+2 #0

g(f(x)) # 0

Hence, Answer is (B)
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