Which of the following statements is necessarily true given that a random variable X satisfies P(X in 0,1) = 1?

The statement $P(X \in \{0,1\}) = 1$ means that the random variable $X$ can only take on the values 0 or 1. The probability of $X$ taking any other value is 0.

Let $P(X=1) = p$. Since $X$ can only take values 0 or 1, it must be that $P(X=0) = 1-p$.
The expected value of $X$ is $E[X] = 0 \cdot P(X=0) + 1 \cdot P(X=1) = 0 \cdot (1-p) + 1 \cdot p = p$.
Since $p$ is a probability, $0 \le p \le 1$.
If $p=0$, then $E[X]=0$. If $p=1$, then $E[X]=1$.
The interval $(0,1)$ excludes 0 and 1. Therefore, $E[X]$ is not necessarily in $(0,1)$ because it could be 0 or 1.
So, statement A is not necessarily true.

A random variable $X$ is defined as a Bernoulli random variable if it takes exactly two values, typically 0 and 1, with probabilities $P(X=1)=p$ and $P(X=0)=1-p$ for some $p \in [0,1]$.
The given condition $P(X \in \{0,1\}) = 1$ means that $X$ can only take values 0 or 1. This is precisely the definition of a Bernoulli random variable.
So, statement B is necessarily true.

For a Bernoulli random variable, the variance is given by $Var(X) = p(1-p)$.
To find the maximum value of $p(1-p)$ for $p \in [0,1]$, we can consider the function $f