For $\mathrm{x} \in \mathrm{R}-[0,1]$, let $\mathrm{f}_{1}(\mathrm{x})=\frac{1}{\mathrm{x}}, \mathrm{f}_{2}(\mathrm{x})=1-\mathrm{x}$ and $\mathrm{f}_{3}(\mathrm{x})=\frac{1}{1-\mathrm{x}}$ be three given functions. If a function, $J(x)$ satisfies $\left(f_{2} O J o f_{1}\right)(x)=f_{3}(x)$ then $J(x)$ is equal to:

(D) $\mathrm{f}_{1}(\mathrm{x})$

JEE Main · 2019 · 09 Jan 2019 · Shift 1 · Sets Relations and Functions

For

x \in R - [0, 1]

, let

f_{1} (x) = \frac{1}{x}, f_{2} (x) = 1 - x

and

f_{3} (x) = \frac{1}{1 - x}

be three given functions. If a function,

J (x)

satisfies

(f_{2} O J o f_{1}) (x) = f_{3} (x)

then

J (x)

is equal to:

Answer: (A)

The condition

(f_{2} \circ J \circ f_{1}) (x) = f_{3} (x)

means

1 - J (\frac{1}{x}) = \frac{1}{1 - x}

, so

J (\frac{1}{x}) = 1 - \frac{1}{1 - x} = \frac{x}{x - 1}

. Replacing

\frac{1}{x}

u

, so that

x = \frac{1}{u}

, gives

J (u) = \frac{1/ u}{1/ u - 1} = \frac{1}{1 - u}

. Thus

J (x) = \frac{1}{1 - x}

, which is exactly

f_{3} (x)

JEE Main · 2019 · 09 Jan 2019 · Shift 1 · Single correct option