Let f: R → R be defined as f(x) = 2x - 1 and g: R - \ 1\ → R . be defined as g(x) = (x - (1)/(2))/(x - 1) . Then the composition function f(g(x)) is:

Correct answer: B. one-one but not onto f(g(x)) = 2g(x) - 1 = 2((2x - 1)/(2(x - 1))) - 1 . = (x)/(x - 1) = 1 + (1)/(x - 1) Range of f(g(x)) = R - \ 1\ Range of f(g(x)) is not onto & f(g(x)) is one-one So f(g(x)) is one-one but not onto. one-one, into

JEE Main 202124 FebruaryMorning Shift

Exam: JEE Main
Year: 2021
Date: 24 February
Shift: Morning Shift
Subject: Mathematics
Chapter: Functions

JEE Main 2021 · 24 February · Morning Shift · Mathematics · Functions

Let f: R → R be defined as f(x) = 2x - 1 and g: R - \ 1\ → R . be...

Let

f: R \to R

be defined as

f\left(x\right) = 2x - 1

and

g: R - \left\{1\right\} \to R

. be defined as

g\left(x\right) = \dfrac{x - \dfrac{1}{2}}{x - 1}

. Then the composition function

f\left(g\left(x\right)\right)

is: $Let f: R → R be defined as f(x) = 2x - 1 and g: R - \ 1\ → R . be defined as g(x) = (x - (1)/(2))/(x - 1) . Then the$

Options

Aneither one-one nor onto
Bone-one but not onto
Conto but not one-one
Dboth one-one and onto

Correct answer

B. one-one but not onto

Step-by-step solution

f\left(g\left(x\right)\right) = 2g\left(x\right) - 1 = 2\left(\dfrac{2x - 1}{2\left(x - 1\right)}\right) - 1 .

$= \dfrac{x}{x - 1} = 1 + \dfrac{1}{x - 1}$

Range of $f\left(g\left(x\right)\right) = R - \left\{1\right\}$

Range of $f\left(g\left(x\right)\right)$ is not onto

& $f\left(g\left(x\right)\right)$ is one-one

So $f\left(g\left(x\right)\right)$ is one-one but not onto.

one-one, into

$Let f: R → R be defined as f(x) = 2x - 1 and g: R - \ 1\ → R . be defined as g(x) = (x - (1)/(2))/(x - 1) . Then the$

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