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Question

For $$x\epsilon R - \left \{0, 1\right \}$$, let $$f_{1}(x) = \dfrac {1}{x}, f_{2}(x) = 1 - x$$ and $$f_{3}(x) = \dfrac {1}{1 - x}$$ be three given functions. If a function, $$J(x)$$ satisfies $$(f_{2}^{\circ} J^{\circ}f_{1})(x) = f_{3}(x)$$ then $$J(x)$$ is equal to


A
f3(x)
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B
f1(x)
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C
f2(x)
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D
1xf3(x)
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Solution

The correct option is A $$f_{3}(x)$$
Given $$f_{1}(x) = \dfrac {1}{x}, f_{2}(x) = 1 - x$$ and $$f_{3}(x) = \dfrac {1}{1 - x}$$
$$(f_{2}\cdot J\cdot f_{1})(x) = f_{3}(x)$$
$$f_{2}\cdot (J(f_{1}(x))) = f_{3}(x)$$
$$f_{2}\cdot \left (J\left (\dfrac {1}{x}\right )\right ) = \dfrac {1}{1 - x}$$
$$1 - J \left (\dfrac {1}{x}\right ) = \dfrac {1}{1 - x}$$
$$J \left (\dfrac {1}{x}\right ) = 1 - \dfrac {1}{1 - x} = \dfrac {-x}{1 - x} = \dfrac {x}{x - 1}$$
Now $$x \rightarrow \dfrac {1}{x}$$
$$J(x) = \dfrac {\dfrac {1}{x}}{\dfrac {1}{x} - 1} = \dfrac {1}{1 - x} = f_{3}(x)$$.

Mathematics

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