Let A-1-2-3-10 andf-A- A be defined as fk- k-1- k - if k is odd- if k is even - Then the number of possible functions g-A- A such that g o f-f is-

Explanation for the correct option: Simplify the given function:Given, g:A→ A.Therefore, g(f(x))=f(x).If xis even, then g(x)=x.If xis odd, then g(x+1)=x+1.There ...

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Standard X

Mathematics

Cartesian Product

Let A=1,2,3,....

Question

Let $A = {1, 2, 3, . . . . . ., 10}$ and $f : A \to A$ be defined as $f (k) = \{\begin{array}{l} k + 1 \\ k \end{array} \begin{array}{r} i f k i s o d d \\ i f k i s e v e n \end{array}\}$ . Then the number of possible functions $g : A \to A$ such that $g o f = f$ is:

$10^{5}$

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$C_{5}^{10}$

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$5^{5}$

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$5!$

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Solution

The correct option is A
$10^{5}$

Explanation for the correct option:
Simplify the given function:
Given, $g : A \to A$ .
Therefore, $g (f (x)) = f (x)$ .
If $x$ is even, then $g (x) = x$ .
If $x$ is odd, then $g (x + 1) = x + 1$ .
Therefore, it is clear from the above that, $g (x) = x$ if $x$ is even.
However, $g (x)$ can take any value of $A$ when $x$ is odd.
Therefore, 5 elements in the set $A$ can be mapped to any $10$ elements.
Hence, the number of possible functions $g = 10^{5} \times 1 = 10^{5}$ .
Therefore, Option(A) is the correct answer.

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Let A={1,2,3,......,10} andf:A→A be defined as f(k)=k+1kifkisoddifkiseven. Then the number of possible functions g:A→A such that gof=f is:

Let $A = {1, 2, 3, . . . . . ., 10}$ and $f : A \to A$ be defined as $f (k) = \{\begin{array}{l} k + 1 \\ k \end{array} \begin{array}{r} i f k i s o d d \\ i f k i s e v e n \end{array}\}$ . Then the number of possible functions $g : A \to A$ such that $g o f = f$ is: