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
Interesting GMATFix Problem-27
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- arora007
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is B the answer/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
- arora007
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cool...B is the answer... but the steps?
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- selango
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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
=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
--Anand--
diebeatsthegmat wrote:is B the answer/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
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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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)
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)