Let X = \{1, 2, 3, \ldots, 12\} and N be the number of pairs \{A, B\} such that A ⊆ X, B ⊆ X, A ≠ B and A ∩ B = \{2, 3, 5, 7, 8\}. Then the value of N is
Consider the graph of \(y = f(x)\) with key points \((-5,-1)\), \((-3,2)\), \((-1,1)\), \((0,3)\), \((2,3)\), \((4,2)\) (approaching \(y=2\)), \((5,-1)\). Find the number of solution(s) of \(x\) satisfying \(f(f(x)) = 2\).
Let $A=\{1,3,7,9,11\}$ and $B=\{2,4,5,7,8,10,12\}$. Then the total number of one-one maps $f:A\to B$, such that $f(1)+f(3)=14$, is:
\[f(x) = \begin{cases} x, & \text{if } x \text{ is rational} \\ 0, & \text{if } x \text{ is irrational} \end{cases}, \quad g(x) = \begin{cases} 0, & \text{if } x \text{ is rational} \\ x, & \text{if } x \text{ is irrational} \end{cases}\]\nThen, \(f \circ g\) is
Suppose that \( f: \mathbb{R} \to \mathbb{R} \) is a continuous function and satisfies the equation \( f(x)\, f(f(x)) = 1 \) for all \( x \in \mathbb{R} \). Further, if \( f(1000) = 999 \), then which of the following options are necessarily true?\( f(500) = \dfrac{1}{500} \)\( f(199) = \dfrac{1}{199} \)\( f(2000) = \dfrac{1}{2000} \)\( f(235) = \dfrac{1}{235} \)\( f(1099) = \dfrac{1}{1099} \)\( f(x) = \dfrac{1}{x} \; \forall x \in \mathbb{R} - \{0, 1000\} \)No such function existsEnter the product of the number of all correct options. For example, if correct options are 2 and 3, then enter 6.
Total number of functions = $3^5$. Since each of 1, 2, 3, 4, or 5 can correspond to any of $a$, $b$, or $c$. The number of functions that corresponds to only one element of $B$ is $^3C_1 imes 1^3$ and the number of functions that correspond to almost two elements of $B$ is $^3C_2 imes 2^5$. Total number of onto functions = $3^5 - ^3C_1 imes 1^3 - (^3C_2 imes 2^5)$ (using $^3C_1 imes 1^3$ repeated twice in $^3C_2 imes 2^5$). What is the result?