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Which of the Following Functions Form Z to Itself Are Bijections? (A) F ( X ) = X 3 (B) F ( X ) = X + 2 (C) F ( X ) = 2 X + 1 (D) F ( X ) = X 2 + X - Mathematics

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Question

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

 

 

 
 

Options

  • \[f\left( x \right) = x^3\]

  • \[f\left( x \right) = x + 2\]

  • \[f\left( x \right) = 2x + 1\]

  • \[f\left( x \right) = x^2 + x\]

MCQ
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Solution

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.
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Chapter 2: Functions - Exercise 2.6 [Page 76]

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RD Sharma Mathematics [English] Class 12
Chapter 2 Functions
Exercise 2.6 | Q 14 | Page 76

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