Question

Which one of the following function is a bijective
function on the set of all real numbers?
el xl
x3 + x² - x + 6
+3
x² + 3x + 2
+ 2
x3 - x2 + 5x + 4​

Answer 1

SOLUTION

TO CHOOSE THE CORRECT OPTION

Which one of the following function is a bijective

function on the set of all real numbers

$\sf{1. \: \: {e}^{ |x| } }$

$\sf{2. \: \: {x}^{3} + {x}^{2} - x + 6}$

$\sf{3. \: \: {x}^{2} + 3x + 2}$

$\sf{4. \: \: {x}^{3} - {x}^{2} + 5 x + 4}$

EVALUATION

CHECKING FOR OPTION : 1

Let the function is

$\sf{f(x) = {e}^{ |x| } }$

Then

$\sf{f(1) = {e}^{ |1| } } = e$

$\sf{f( - 1) = {e}^{ | - 1| } = e}$

$\sf{ \therefore \: \: - 1 \ne \:1 \: \: but \: \: f( - 1) = f(1)}$

So f is not one - one

So f is not bijective

CHECKING FOR OPTION : 2

Let the given function

$\sf{g(x) = {x}^{3} + {x}^{2} - x + 6}$

Then

$\sf{g(1) = {1}^{3} + {1}^{2} - 1 + 6 = 7}$

$\sf{g( - 1) = {( - 1)}^{3} + {( - 1)}^{2} - ( - 1) + 6 = 7}$

$\sf{ \therefore \: \: - 1 \ne \:1 \: \: but \: \: g( - 1) = g(1)}$

So g is not one - one

So g is not bijective

CHECKING FOR OPTION : 3

Let the given function is

$\sf{h(x) = {x}^{2} + 3x + 2}$

$\sf{h( - 1) = {( - 1)}^{2} + 3 \times ( - 1) + 2 = 0}$

$\sf{h( - 2) = {( - 2)}^{2} + 3 \times ( - 2) + 2 =0 }$

$\sf{ \therefore \: \: - 1 \ne \: - 2 \: \: but \: \: h( - 1) = h( - 2)}$

So h is not one - one

So h is not bijective

CHECKING FOR OPTION : 4

Let the given function is

$\sf{p(x) = {x}^{3} - {x}^{2} + 5 x + 4}$

It is clear that

For every real numbers a and b

$\sf{ \: a \ne \: b \: \: but \: \: p( a) \ne p( b)}$

So p is one - one

Again for every element in the Co-domain set R there is a pre-image in the domain set R

So p is onto

So p is bijective

FINAL ANSWER

Hence the correct option is

$\sf{4. \: \: {x}^{3} - {x}^{2} + 5 x + 4}$

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Answer 2

Answer:

Step-by-step explanation: