Question

f: R—>R,f(x)=x²+2x+3 is......,Select Proper option from the given options.
(a) a bijection
(b) one-one but not onto
(c) onto but not one-one
(d) many-one and not onto

Answer 1

Answer:

The proper option is (d) many-one and not onto.

Step-by-step explanation:

Not onto: We can rewrite the function as $f(x) = (x+1)^2+2$. Since it is the sum of two positive numbers, it is always positive. That is, given $y<0$, there is no $x \in \mathbb{R}$ such that $f(x) = y$. Then the function is not onto.

Many-one: We just need to give two numbers $x \neq y$ such that $f(x) = f(y)$. For example, taking x = 1 and y = -3, we have

$f(1) = 1+2+3 = 6$

and

$f(-3) = 9-6+3 = 6$

So $f(1) = f(-3)$

and the function is many-one.

Answer 2

$\displaystyle\huge\red{\underline{\underline{Solution}}}$

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 is defined as f(x)=x²+2x+3

TO CHOOSE THE CORRECT OPTION

f is

(a) a bijection

(b) one-one but not onto

(c) onto but not one-one

(d) many-one and not onto

CALCULATION

CHECKING FOR ONE TO ONE

$\sf{Here \: \: 0, - 2 \: \in \mathbb{ R}}$

Such that

$\sf{ f(0) = {0}^{2} + (2 \times0 ) + 3 = 3 \: \: }$

$\sf{ f( - 2) = {( - 2)}^{2} + (2 \times - 2 ) + 3 = 3 \: \: }$

$\sf{Hence \: \: 0 \ne 2 \: \: but \: \: f(0) = f(2)}$

Hence f is not one to one

CHECKING FOR ONTO

$\sf{Let \: y \in \mathbb{R} \: ( Co domain \: set) }$

If possible let there exists an element x in domain set such that

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

$\implies \sf{ \:y = {x}^{2} + 2x + 3 \: }$

$\implies \sf{ \ {x}^{2} + 2x + (3 - y) = 0 \: }$

$\sf{Then \: \: x \in \mathbb{R}} \: if \: discriminant \geqslant 0$

$\implies \sf{ {( 2)}^{2} - 4.1.(3 - y) \geqslant 0\: \: }$

$\implies \sf{ 4 - 12 + 4y \geqslant 0\: \: }$

$\implies \sf{ 4y \geqslant 8\: \: }$

$\implies \sf{ y \geqslant 2\: \: }$

So 1 in co-domain set has no pre-image in the domain set

Hence f is not onto

RESULT

f is many one & not onto function

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LEARN MORE FROM BRAINLY

Represent all possible one-one functions from the set A = {1, 2} to the set B = {3,4,5) using arrow diagram.

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