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The Function F : R → R Defined by F ( X ) = ( X − 1 ) ( X − 2 ) ( X − 3 ) (A) One-one but Not onto (B) onto but Not One-one (C) Both One and onto (D) Neither One-one Nor onto - Mathematics

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Question

The function

\[f : R \to R\] defined by\[f\left( x \right) = \left( x - 1 \right) \left( x - 2 \right) \left( x - 3 \right)\]

(a) one-one but not onto
(b) onto but not one-one
(c) both one and onto
(d) neither one-one nor onto

Options

  • one-one but not onto

  • onto but not one-one

  • both one and onto

  • neither one-one nor onto

MCQ
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Solution

\[f\left( x \right) = \left( x - 1 \right)\left( x - 2 \right)\left( x - 3 \right)\]
Injectivity:
\[f\left( 1 \right) = \left( 1 - 1 \right)\left( 1 - 2 \right)\left( 1 - 3 \right) = 0\] 
\[f\left( 2 \right) = \left( 2 - 1 \right)\left( 2 - 2 \right)\left( 2 - 3 \right) = 0\] 
\[f\left( 3 \right) = \left( 3 - 1 \right)\left( 3 - 2 \right)\left( 3 - 3 \right) = 0\] 
\[\Rightarrow f \left( 1 \right) = f\left( 2 \right) = f\left( 3 \right) = 0\]
So, is not one-one.
Surjectivity :
Let y be an element in the co domain R, such that
\[y = f\left( x \right)\] 
\[ \Rightarrow y = \left( x - 1 \right)\left( x - 2 \right)\left( x - 3 \right)\] 
\[\text{Sincey} \in R \text{ and }x \in R, \text{f is onto}.\]
So, the answer is (b).
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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 19 | Page 76

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