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Let $x_{1}, x_{2},…, x_{n} \in \mathbb{R}$ be distinct reals. Define the set

$$\text{A} = \left \{ \left ( f_{1}(t), f_{2} (t), \dots, f_{n}(t)\right ):t \in\mathbb{R} \right \},$$

where for $1 \leq  k \leq  n,$

$$f_{k}(t)=\left\{\begin{matrix}
1, & \text{if}  \;x_{k} \leq t\\ 
 0,& \text{otherwise}
\end{matrix}\right.$$

Then $\text{A}$ contains

  1. exactly $n$ distinct elements

  2. exactly $(n + 1)$ distinct elements

  3. exactly $2^{n}$ distinct elements

  4. infinitely many distinct elements

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The set $A$ contains ordered $n-tuple$ and each element of the ordered $n-tuple$ is either $0$ or $1$ based on the output of $f_k(t)$.

So, lets understand what this function $f_k(t)$ does – 

  • if $x_k$ is less than equal to $t$ then output is $1$.
  • if $x_k$ is greater than $t$ then output is $0$.

It is also given that $x_1, x_2, …, x_n \in \mathbb{R}$ are distinct.

$\implies$ For a given $t$, the $n-tuple$ encodes the relative position of $t$ with respect to $x_1, x_2, …, x_n$ on the number line.

Geometrically, $x_1, x_2, …, x_n$ divides the real number line into $R_1, R_2, …, R_n, R_{n+1}$ regions.

And $\forall t \in R_i \text{ } f_k(t)$ returns same value $\implies \forall t \in R_i$ we’ll have same ordered $n-tuple$.

$A$ is a set so no ordered $n-tuple$ repeats.

$\therefore A$ contains exactly $(n+1)$ distinct elements.

Answer :- B.

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