JEE MAIN - Mathematics (2019 - 9th January Evening Slot - No. 4)
Let A = {x $$ \in $$ R : x is not a positive integer}.
Define a function $$f$$ : A $$ \to $$ R as $$f(x)$$ = $${{2x} \over {x - 1}}$$,
then $$f$$ is :
Define a function $$f$$ : A $$ \to $$ R as $$f(x)$$ = $${{2x} \over {x - 1}}$$,
then $$f$$ is :
not injective
neither injective nor surjective
surjective but not injective
injective but not surjective
Explanation
f(x) = $${{2x} \over {x - 1}}$$
f(x) = 2 + $${2 \over {x - 1}}$$
f'(x) = $$-$$ $${2 \over {{{\left( {x - 1} \right)}^2}}}$$ < 0 $$\forall $$ x $$ \in $$ R
Hence f(x) is strictly decreasing
So, f(x) is one-one
Range : Let y = $${{2x} \over {x - 1}}$$
xy $$-$$ y = 2x
$$ \Rightarrow $$ x(y $$-$$ 2) = y
$$ \Rightarrow $$ x = $${y \over {y - 2}}$$
given that x $$ \in $$ R : x is not a +ve integer
$$ \therefore $$ $${y \over {y - 2}} \ne $$ N (N $$ \to $$ Natural number)
$$ \Rightarrow $$ y $$ \ne $$ Ny $$-$$ 2N
$$ \Rightarrow $$ y $$ \ne $$ $${{2N} \over {N - 1}}$$
So range $$ \notin $$ R (in to function)
f(x) = 2 + $${2 \over {x - 1}}$$
f'(x) = $$-$$ $${2 \over {{{\left( {x - 1} \right)}^2}}}$$ < 0 $$\forall $$ x $$ \in $$ R
Hence f(x) is strictly decreasing
So, f(x) is one-one
Range : Let y = $${{2x} \over {x - 1}}$$
xy $$-$$ y = 2x
$$ \Rightarrow $$ x(y $$-$$ 2) = y
$$ \Rightarrow $$ x = $${y \over {y - 2}}$$
given that x $$ \in $$ R : x is not a +ve integer
$$ \therefore $$ $${y \over {y - 2}} \ne $$ N (N $$ \to $$ Natural number)
$$ \Rightarrow $$ y $$ \ne $$ Ny $$-$$ 2N
$$ \Rightarrow $$ y $$ \ne $$ $${{2N} \over {N - 1}}$$
So range $$ \notin $$ R (in to function)
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