JEE PYQ: Functions Question 11
Question 11 - 2021 (26 Feb Shift 2)
Let $A = {1, 2, 3, \ldots, 10}$ and $f : A \to A$ be defined as $f(k) = \begin{cases} k+1 & \text{if } k \text{ is odd} \ k & \text{if } k \text{ is even} \end{cases}$. Then the number of possible functions $g : A \to A$ such that $g \circ f = f$ is:
(1) $10^5$
(2) ${}^{10}C_5$
(3) $5^5$
(4) $5!$
Type: MCQ
Show Answer
Answer: (1) $10^5$
Solution
$g(f(x)) = f(x)$
$\Rightarrow g(x) = x$, when $x$ is even
5 elements in $A$ can be mapped to any 10
So, $10^5 \times 1 = 10^5$
BETA
