Question

If g(x) = 1 – x and \(h\left( x \right) = \frac{x}{{x - 1}},\;then\frac{{g\left( {h\left( x \right)} \right)}}{{h\left( {g\left( x \right)} \right)}}\;is:\)
1. \(\frac{{h\left( x \right)}}{{g\left( x \right)}}\)
2. -1/x
3. \(\frac{{g\left( x \right)}}{{h\left( x \right)}}\)
4. \(\frac{x}{{{{\left( {1 - x} \right)}^2}}}\)

Answer 1

Correct Answer - Option 1 : \(\frac{{h\left( x \right)}}{{g\left( x \right)}}\)

Data:

g(x) = 1 – x

\(h\left( x \right) = \frac{x}{{x - 1}}\) 

Calculation:

\(g\left( {h\left( x \right)} \right) = 1 - \;\frac{x}{{x - 1}} = \; - \frac{1}{{x - 1}}\)

\(h\left( {g\left( x \right)} \right) = \;\frac{{1 - x}}{{1 - x - 1}} = \frac{{1 - x}}{{ - x}}\;\)

\(\frac{{g\left( {h\left( x \right)} \right)}}{{h\left( {g\left( x \right)} \right)}} = \frac{{ - \frac{1}{{x - 1}}}}{{\frac{{\left( {1 - x} \right)}}{{ - x}}\;}} = \frac{{\frac{x}{{x - 1}}}}{{1 - x}}\; = \frac{{h\left( x \right)}}{{g\left( x \right)}}\)