Question

If f: R⁺ —>R, f(x)=x/x+1 is ......,Select Proper option from the given options.
(a) one-one and onto
(b) one-one and not onto
(c) not one-one and not onto
(d) onto but not one-one

Answer 1

Answer:

The correct option from the given options is (b).

Step-by-step explanation:

In fact, $f$ is not onto because since $x$ is always positive, so will be $\frac{x}{x+1}$. Then given a negative number $y \in \mathbb{R}$, there is no $x \in \mathbb{R}^+$ such that $f(x) = y$.

To show $f$ is one-one, we have to conclude that $f(x) = f(y) \Rightarrow x = y$.

$f(x) = f(y) \Rightarrow \frac{x}{x+1} = \frac{y}{y+1} \Rightarrow xy+x = xy+y \Rightarrow x = y$

So $f$ is one-one and not onto.

Answer 2

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DEFINITION

INJECTIVE FUNCTION :

$\sf{A \: function \: f : A \to B \: is \: said \: to \: be \: injective \: if}$

$\sf{For \: \: x_1 \ne x_2 \: \: we \: have \: \: f(x_1) \ne f(x_2)}$

SURJECTIVE FUNCTION

$\sf{A \: function \: f : A \to B \: is \: said \: to \: be \: surjective}$

if for every element y in the co-domain B there exists a pre-image x in domain set A such that y = f(x)

BIJECTIVE FUNCTION

$\sf{A \: function \: f : A \to B \: is \: said \: to \: be \: bijective}$

if f is both injective and surjective

GIVEN

A function f: R⁺ —>R is defined by f(x)=x/x+1

TO CHOOSE THE CORRECT OPTION

(a) one-one and onto

(b) one-one and not onto

(c) not one-one and not onto

(d) onto but not one-one

CALCULATION

$\displaystyle \: \sf{ \:Here \: \: \: \: f(x) = \frac{x}{x + 1} }$

CHECKING FOR ONE TO ONE

$\displaystyle \: \sf{let \: \: x \: , \: y \in \mathbb{{R}^{ + } } \: \:such \: that \: \: \: f(x) = \:f( y) \: \: }$

$\displaystyle \: \sf{\: \: f(x) = \:f( y) \: \: } \: \: gives$

$\displaystyle \: \sf{ \: \frac{x}{x + 1} = \: \frac{y}{y + 1} \: }$

$\implies \: \displaystyle \: \sf{ xy + x = xy + y }$

$\implies \: \displaystyle \: \sf{ x = y }$

$\therefore \displaystyle \: \sf{\: \: f(x) = \:f( y) \: \: } \: \: gives \: \: x = y$

Hence f is one to one

CHECKING FOR ONTO

$\displaystyle \: \sf{clearly \: \: \: 1 \in \mathbb{{R}^{ } } \: \: ( \: codomain \: set \: ) \: \: }$

But there done not exists any x in the domain set R⁺ such that.

$\sf{ \: f(x) = 1 \: }$

Hence f is not onto

RESULT

The function is

(b) one-one and not onto

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LEARN MORE FROM BRAINLY

f: R—>R,f(x)=x²+2x+3

Select Proper option from the given options.

https://brainly.in/question/5596321