Let f = {(2, 7), (3, 4), (7, 9), (-1, 6), (0, 2), (5,3)} be a function from A = {-1,0, 2, 3, 5, 7} to B = {2, 3, 4, 6, 7, 9}. Identify the type of function.

One-to-one

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Both a and b

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Onto

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Many-to-one

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Solution

The correct option is B Both a and b
Given: f = {(2, 7), (3, 4), (7, 9), (-1, 6), (0, 2), (5,3)}
All the elements in set A has unique images in set B. Similarly all the elements in set B has pre-images in set A. Hence the function is one-to one and onto.

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Similar questions

Q. Let f = { (2, 7), (3, 4), (7, 9), (-1, 6), (0, 2), (5, 3) } be a function from A = { -1, 0, 2, 3, 5, 7 } to B = { 2, 3, 4, 6, 7, 9 }. Is this (i) an one-one function (ii) an onto function (iii) both one-one and onto function?

If
A = {1, 2, 3, 4}

B = {3, 4, 5, 6, 7}

C = {6, 7, 8, 9}

D = {7, 8, 9, 10}

$\begin{matrix} C o l u m n A & C o l u m n B 1. A \cup B & a . {3, 4, 6, 7} 2. A \cap B & b . \emptyset 3. (B \cap C) \cup D & c . {3, 4} 4. B \cup C \cup D & d . {3, 4, 5, 6, 7, 8, 9, 10} 5. (A \cap B) \cap (B \cup C) & e . {1, 2, 3, 4, 5, 6, 7} f . {6, 7, 8, 9, 10} g . {7} \end{matrix}$

Q. Let set U = {0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10}, set A = {2, 4, 6, 8}, and set B = {1, 3, 5, 7, 9}. Prove that (A

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B)’ = A’

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Let f = {(2, 7), (3, 4), (7, 9), (-1, 6), (0, 2), (5,3)} be a function from A = {-1,0, 2, 3, 5, 7} to B = {2, 3, 4, 6, 7, 9}. Identify the type of function.

The correct option is B Both a and bGiven: f = {(2, 7), (3, 4), (7, 9), (-1, 6), (0, 2), (5,3)} All the elements in set A has unique images in set B. Similarly all the elements in set B has pre-images in set A. Hence the function is one-to one and onto.

The correct option is B Both a and b
Given: f = {(2, 7), (3, 4), (7, 9), (-1, 6), (0, 2), (5,3)}
All the elements in set A has unique images in set B. Similarly all the elements in set B has pre-images in set A. Hence the function is one-to one and onto.