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f(x) = x - 1x and g(x) = xx - 1. We have f(g(a)) = a. What is the value of a?

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f(x) = x - 1x and g(x) = xx - 1. We have f(g(a)) = a. What is the value of a?

by Max@Math Revolution » Wed Jul 15, 2020 11:32 pm

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[GMAT math practice question]

f(x) = x - 1/x and g(x) = x/x - 1. We have f(g(a)) = a. What is the value of a?

A. -2
B. -1
C. 1
D. 2
E. 3
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Re: f(x) = x - 1x and g(x) = xx - 1. We have f(g(a)) = a. What is the value of a?

by deloitte247 » Sat Jul 18, 2020 11:32 am
$$Given that:\ f(g(a))=a.What\ is\ the\ value\ of\ a?$$
$$Expres\sin g\ g(a)\ in\ terms\ of\ g(x)$$
$$g\left(x\right)=\frac{x}{x-1}$$
$$g\left(a\right)=\frac{a}{a-1}$$
$$f\left(g\left(a\right)\right)=f\left(\frac{a}{a-1}\right)$$
$$=>f\left(\frac{a}{a-1}\right)=\frac{\left(\frac{a}{a-1}\right)-\left(\frac{1}{1}\right)}{\frac{a}{a-1}}$$
$$=\frac{\frac{a-a+1}{a-1}}{\frac{a}{a-1}}$$
$$=\frac{1}{a-1}\div\frac{a}{a-1}$$
$$f\left(\frac{a}{a-1}\right)=\frac{1}{a-1}\cdot\frac{a-1}{a}=\frac{1}{a}$$
$$f\left(a\right)=\frac{1}{a}$$
$$so\ \frac{1}{a}=a$$
$$\sqrt{1}=\sqrt{a^2}$$
$$a=+1$$
$$if\ a=+1$$
$$g\left(a\right)=g\left(1\right)=\frac{1}{1-1}=\frac{1}{0}$$
$$This\ gives\ a\ zero\ division\ error\ so\ a\ cannot\ =+1\ because\ g\left(1\right)is\ undefined$$
$$if\ a=-1$$
$$g\left(a\right)=g\left(-1\right)=\frac{-1}{-1-1}=\frac{+1}{+2}=\frac{1}{2}=0.5$$
$$Hence\ a=-1$$
$$Answer\ =\ B$$
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Re: f(x) = x - 1x and g(x) = xx - 1. We have f(g(a)) = a. What is the value of a?

by Max@Math Revolution » Sun Jul 19, 2020 6:04 pm
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7.16ps(a).png
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