If p : the function f(x) = x is an even funct | vedclass

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(D) $1$. Analyze $p$: The function $f(x) = x$ is an odd function because $f(-x) = -x = -f(x)$. Thus,$p$ is false ($\sim p$ is true).$2$. Analyze $q$:

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$\sim p \vee (q \wedge r)$

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(D) $1$. Analyze $p$: The function $f(x) = x$ is an odd function because $f(-x) = -x = -f(x)$. Thus,$p$ is false ($\sim p$ is true).$2$. Analyze $q$: The binomial coefficient ${}^nC_m$ represents the number of ways to choose $m$ items from $n$,which is always a non-negative integer (whole number) for $0 \le m \le n$. Thus,$q$ is true.$3$. Analyze $r$: If $\overrightarrow{c} = \overrightarrow{a} + \lambda \overrightarrow{b}$,then $\overrightarrow{c} - \overrightarrow{a} - \lambda \overrightarrow{b} = 0$. Since there exists a non-zero linear combination equal to the zero vector,the vectors $\overrightarrow{a}, \overrightarrow{b}, \overrightarrow{c}$ are linearly dependent. Thus,$r$ is true.$4$. Evaluate the options:- $A: (p \wedge q) = (F \wedge T) = F$- $B: (p \vee q) \wedge \sim r = (F \vee T) \wedge F = T \wedge F = F$- $C: \sim (q \wedge r) \vee p = \sim (T \wedge T) \vee F = \sim T \vee F = F \vee F = F$- $D: \sim p \vee (q \wedge r) = \sim F \vee (T \wedge T) = T \vee T = T$Therefore,the statement in option $D$ is true.

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