Verified Solution • 2025–26 Session moderate

QJEEM19Ma395b: JEE Main 2019 Mathematics - Class 11 JEEM19Ma395b Solution: Let A = {x R : x is not a positive integer}. Define a function…

Let A = {x ∈\in R : x is not a positive integer}.

Define a function ff : A →\to   R   as   f(x)f(x) = 2xx−1{{2x} \over {x - 1}} ,

then ff is :

Multiple Choice Options

  • A not injective
  • B neither injective nor surjective
  • C surjective but not injective
  • D injective but not surjective

Step-by-Step Verified Solution

f(x) = 2xx−1{{2x} \over {x - 1}}

f(x) = 2 + 2x−1{2 \over {x - 1}}

f'(x) = −- 2(x−1)2{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 = 2xx−1{{2x} \over {x - 1}}

xy −- y = 2x

⇒\Rightarrow   x(y −- 2) = y

⇒\Rightarrow   x = yy−2{y \over {y - 2}}

given that x ∈\in R : x is not a +ve integer

∴\therefore    yy−2≠{y \over {y - 2}} \ne N    (N →\to  Natural number)

⇒\Rightarrow   y ≠\ne Ny −- 2N

⇒\Rightarrow   y ≠\ne 2NN−1{{2N} \over {N - 1}}

So range ∉\notin R (in to function)

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