JEE Main 2019 — Functions Question with Solution
From: JEE Main 2019 (Online) 11th January Evening Slot
Question
Let a function f : (0, ) (0, ) be defined by f(x) = . Then f is :
Choose an option
Show full solutionCorrect option: B
Correct answer
Bneiter injective nor surjective
Step-by-step explanation
f\left( x \right) = \left| {1 - {1 \over x}} \right| = {{\left| {x - 1} \right|} \over x} = \left\{ {\matrix{
{{{1 - x} \over x}} & {0 < x \le 1} \cr
{{{x - 1} \over x}} & {x \ge 1} \cr
} } \right.
f(x) is not injective
but range of function is
Remarks : If co-domain is , then f(x) will be surjective.
f(x) is not injective
but range of function is
Remarks : If co-domain is , then f(x) will be surjective.
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