Let $f: R \rightarrow R$ be defined as $f(x)=2 x-1$ and $g: R-\{1\} \rightarrow R$ be defined as $g(x)=\frac{x-\frac{1}{2}}{x-1}$. Then the composition function $\mathrm{f}(\mathrm{g}(x))$ is :

JEE Main · 2021 · 24 Feb 2021 · Shift 1 · Sets Relations and Functions

Let

f : R \to R

be defined as

f (x) = 2 x - 1

and

g : R - {1} \to R

be defined as

g (x) = \frac{x - \frac{1}{2}}{x - 1}

. Then the composition function

f (g (x))

is :

Answer: (D)

First, compute the composition:

f (g (x)) = f (\frac{x - \frac{1}{2}}{x - 1}) = 2 (\frac{x - \frac{1}{2}}{x - 1}) - 1 = \frac{2 x - 1}{x - 1} - 1 = \frac{2 x - 1 - ( x - 1 )}{x - 1} = \frac{x}{x - 1}

Domain:

R - {1}

, Codomain:

R

For one-one: If

\frac{x _{1}}{x _{1} - 1} = \frac{x _{2}}{x _{2} - 1}

, then

x_{1} (x_{2} - 1) = x_{2} (x_{1} - 1)

, which gives

x_{1} x_{2} - x_{1} = x_{2} x_{1} - x_{2}

, so

x_{1} = x_{2}

. Hence one-one.

For onto: We need

\frac{x}{x - 1} = y

for all

y \in R

. Solving:

x = y (x - 1) \Rightarrow x = y x - y \Rightarrow x (1 - y) = - y \Rightarrow x = \frac{y}{y - 1}

. This is undefined when

y = 1

, so the range is

R - {1} \neq = R

. Not onto.

Therefore,

f (g (x))

is one-one but not onto.

JEE Main · 2021 · 24 Feb 2021 · Shift 1 · Single correct option