Math, asked by rajitdeb2018, 6 hours ago

Which of the following functions form Z into Z bijections? (a) f(x) = (b)) - 2 (c) f(x) = 2x + 1 malosole​

Answers

Answered by vashimurugan
1

Step-by-step explanation:

f(x)=x

3

one−one test:

Let x

1

and x

2

be the elements in the domain such that,

f(x

1

)=f(x

2

)

⇒ x

1

3

=x

2

3

⇒ x

1

=x

2

∴ f is one-one.

onto test :

Let y be an element in the co-domain (Z), such that,

f(x)=y

⇒ x

3

=y

Consider y=3

⇒ x

3

=3

⇒ x=

3

3

/

Z

∴ f is not onto.

∴ f is not bijection.

(B) f(x)=x+2

one−one test:

Let x

1

and x

2

be the elements in the domain such that,

f(x

1

)=f(x

2

)

⇒ x

1

+2=x

2

+2

⇒ X

1

=x

2

∴ f is one-one.

onto test :

Let y be an element in the co-domain (Z), such that,

f(x)=y

⇒ x+2=y

⇒ x=y−2∈Z(Domain)

∴ f is onto

∴ f is bijection.

(C) f(x)=2x+1

one−one test:

Let x

1

and x

2

be the elements in the domain such that,

f(x

1

)=f(x

2

)

⇒ 2x

1

+1=2x

2

+1

⇒ 2x

1

=2x

2

⇒ x

1

=x

2

∴ f is one-one.

onto test :

Let y be an element in the co-domain (Z), such that,

f(x)=y

⇒ 2x+1=y

Consider y=4

⇒ 2x+1=4

⇒ 2x=3

⇒ x=

2

3

/

Z

∴ f is not onto.

∴ f is not bijection.

(D) f(x)=x

2

+x

⇒ f(0)=(0)

2

+0=0

⇒ f(−1)=(−1)

2

+(−1)=1−1=0

We can see that 0 and −1 have the same image.

∴ f is not on-one.

∴ f is not bijection.

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