JEE MAIN - Mathematics (2019 - 11th January Evening Slot - No. 7)
Let a function f : (0, $$\infty $$) $$ \to $$ (0, $$\infty $$) be defined by f(x) = $$\left| {1 - {1 \over x}} \right|$$. Then f is :
not injective but it is surjective
neiter injective nor surjective
injective only
both injective as well as surjective
Explanation
$$f\left( x \right) = \left| {1 - {1 \over x}} \right| = {{\left| {x - 1} \right|} \over x} = \left\{ {\matrix{
{{{1 - x} \over x}} & {0 < x \le 1} \cr
{{{x - 1} \over x}} & {x \ge 1} \cr
} } \right.$$
_11th_January_Evening_Slot_en_7_1.png)
$$ \Rightarrow $$ f(x) is not injective
but range of function is $$\left[ {0,\infty } \right)$$
Remarks : If co-domain is $$\left[ {0,\infty } \right)$$, then f(x) will be surjective.
$$ \Rightarrow $$ f(x) is not injective
but range of function is $$\left[ {0,\infty } \right)$$
Remarks : If co-domain is $$\left[ {0,\infty } \right)$$, then f(x) will be surjective.
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