Question

Which of the following functions form Z to itself are bijections?

 

 

 

 

Options

• \[f\left( x \right) = x^3\]

• \[f\left( x \right) = x + 2\]

• \[f\left( x \right) = 2x + 1\]

• \[f\left( x \right) = x^2 + x\]

Answer 1

f is not onto because for y = 3∈Co-domain(Z), there is no value of x∈Domain(Z)
\[ x^3 = 3\] 
\[ \Rightarrow x = \sqrt[3]{3} \not\in Z\] 
⇒ f is not onto
So, f is not a bijection

(b) Injectivity:
Let x and y be two elements of the domain (Z), such that

\[x + 2 = y + 2\] 
\[ \Rightarrow x = y\]

So, f is one-one.
Surjectivity:
Let y be an element in the co-domain (Z), such that

\[y = f\left( x \right)\] 
\[ \Rightarrow y = x + 2\] 
\[ \Rightarrow x = y - 2 \in Z \left( Domain \right)\]

\[\Rightarrow\]

⇒ f is onto.
So, f is a bijection.

\[\left( c \right) f\left( x \right) = 2x + 1 \text {is not onto because if we take }4 \in Z\left( co domain \right), then4 = f\left( x \right)\] 
\[ \Rightarrow 4 = 2x + 1\] 
\[ \Rightarrow 2x = 3\] 
\[ \Rightarrow x = \frac{3}{2} \not\in Z\] 
So,f is not a bijection.
\[\] 
 \[\left( d \right) f\left( 0 \right) = 0^2 + 0 = 0\] 
\[and f\left( - 1 \right) = \left( - 1 \right)^2 + \left( - 1 \right) = 1 - 1 = 0\] 

⇒ 0 and -1 have the same image.

⇒ f is not one-one. 

So,fis not a bijection.