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(D) $1$. Evaluate the truth value of each statement:- $p$: For matrices $A$ and $B$,$(A-B)(A+B) = A^2 + AB - BA - B^2$. Since $AB 
eq BA$,$A^2 - B^2

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$(s \vee \sim p) \leftrightarrow (\sim p \wedge \sim r)$

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(D) $1$. Evaluate the truth value of each statement:- $p$: For matrices $A$ and $B$,$(A-B)(A+B) = A^2 + AB - BA - B^2$. Since $AB 
eq BA$,$A^2 - B^2 
eq (A-B)(A+B)$. Thus,$p$ is False $(F)$.- $q$: $5 \leqslant 5$ is true. Thus,$q$ is True $(T)$.- $r$: We know $\sum_{k=0}^{n} { }^n C_k = 2^n$. Here,${ }^8 C_0 + { }^8 C_1 + \ldots + { }^8 C_8 = 2^8 = 256$. Since ${ }^8 C_0 = 1$,the sum ${ }^8 C_1 + \ldots + { }^8 C_8 = 256 - 1 = 255$. Thus,$r$ is False $(F)$.- $s$: The maximum value of ${ }^n C_r$ is ${ }^n C_{n/2}$ for even $n$. For $n=8$,it is ${ }^8 C_4 = \frac{8 	imes 7 	imes 6 	imes 5}{4 	imes 3 	imes 2 	imes 1} = 70$. Thus,$s$ is True $(T)$.$2$. Truth values: $p=F, q=T, r=F, s=T$.$3$. Check options:- $A: (F \wedge \sim F) \vee (\sim T \wedge \sim T) = (F \wedge T) \vee (F \wedge F) = F \vee F = F$.- $B: (F \vee \sim T) \leftrightarrow (\sim F$ $ightarrow T) = (F \vee F) \leftrightarrow (T$ $ightarrow T) = F \leftrightarrow T = F$.- $C: (F \leftrightarrow T) \wedge (\sim F \vee \sim T) = F \wedge (T \vee F) = F \wedge T = F$.- $D: (T \vee \sim F) \leftrightarrow (\sim F \wedge \sim F) = (T \vee T) \leftrightarrow (T \wedge T) = T \leftrightarrow T = T$.Therefore,option $D$ is true.

If $p, q, r, s$ are statements,where:
$p: A^2-B^2=(A-B)(A+B)$ where $A, B$ are matrices and $AB \neq BA$
$q: 5 \leqslant 5$
$r: { }^8 C_1+{ }^8 C_2+{ }^8 C_3+\ldots+{ }^8 C_8=256$
$s: \text{Maximum value of } { }^8 C_r \text{ is } 70$
Which of the following statements has a truth value of true?

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If $p, q, r, s$ are statements,where:$p: A^2-B^2=(A-B)(A+B)$ where $A, B$ are matrices and $AB \neq BA$$q: 5 \leqslant 5$$r: { }^8 C_1+{ }^8 C_2+{ }^8 C_3+\ldots+{ }^8 C_8=256$$s: \text{Maximum value of } { }^8 C_r \text{ is } 70$Which of the following statements has a truth value of true?