Let A= { x∈R:x is not a positive integer} . Define a function f:A→R as fx=2x/(x-1) , then f is:

Correct answer: (A) Injective but not surjective. Given, function fx=2x/(x-1) and domain of fx is -∞, 0 Now f^(')x=(x-12-2x1-0x-1^2f^(′)x=-2)/(x-1^2)<0 ⇒f is always decreasing in the given domain. Clearly, given function is one-one or injective. Since, codomain -∞, ∞≠ Range 0, 2 , hence f is into.

Q48 / 90 · JEE Main 2019 (09 Jan Shift 2) · Mathematics

Let $A =$ { $x \in R : x$ is not a positive integer} $.$ Define a function $f : A \to R$ as $f (x) = \frac{2 x}{x - 1}$ , then $f$ is:

Correct answer
A
Injective but not surjective
B
Not injective
C
Surjective but not injective
D
Neither injective nor surjective

Solution

Given, function $f (x) = \frac{2 x}{x - 1}$ and domain of $f (x)$ is $(- \infty, 0]$

Now $f^{'} (x) = \frac{(x - 1) 2 - 2 x (1 - 0)}{{(x - 1)}^{2}}$

$f^{'} (x) = \frac{- 2}{{(x - 1)}^{2}} < 0$

$\Rightarrow f$ is always decreasing in the given domain.

Clearly, given function is one-one or injective.
Since, codomain $(- \infty, \infty) \neq$ Range $(0, 2)$ , hence $f$ is into.

Hence, the correct answer is option: 'A'.

Let A= { x∈R:x is not a positive integer} . Define a function f:A→R as fx=2xx-1 , then f is:

Solution

Relevant Questions

Let $A =$ { $x \in R : x$ is not a positive integer} $.$ Define a function $f : A \to R$ as $f (x) = \frac{2 x}{x - 1}$ , then $f$ is: