Consider the three statements -p: forall n in | vedclass

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(C) $1$. Analyze statement $p$: For $n=1$,$10(1)-3 = 7$ (prime). For $n=2$,$10(2)-3 = 17$ (prime). For $n=4$,$10(4)-3 = 37$ (prime). For $n=5$,$10(5)-

Vedclass · Accepted Answer

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

Vedclass · Answer

(C) $1$. Analyze statement $p$: For $n=1$,$10(1)-3 = 7$ (prime). For $n=2$,$10(2)-3 = 17$ (prime). For $n=4$,$10(4)-3 = 37$ (prime). For $n=5$,$10(5)-3 = 47$ (prime). For $n=7$,$10(7)-3 = 67$ (prime). For $n=8$,$10(8)-3 = 77 = 7 	imes 11$ (not prime). Thus,$p$ is False $(F)$.$2$. Analyze statement $q$: The sum of squares of direction cosines must be $1$. Here,$(\frac{2}{\sqrt{3}})^2 + (\frac{-2}{\sqrt{3}})^2 + (\frac{-1}{\sqrt{3}})^2 = \frac{4}{3} + \frac{4}{3} + \frac{1}{3} = \frac{9}{3} = 3 
eq 1$. Thus,$q$ is False $(F)$.$3$. Analyze statement $r$: $\sin x$ is strictly increasing on $[-\frac{\pi}{2}, \frac{\pi}{2}]$. Thus,$r$ is True $(T)$.$4$. Evaluate options:$A: (F \wedge F) \leftrightarrow T$ $\Rightarrow F \leftrightarrow T$ $\Rightarrow F$$B: (F$ $ightarrow F)$ $ightarrow \sim T$ $\Rightarrow T$ $ightarrow F$ $\Rightarrow F$$C: (\sim F \vee F) \wedge T$ $\Rightarrow (T \vee F) \wedge T$ $\Rightarrow T \wedge T$ $\Rightarrow T$$D: (\sim F \wedge \sim F) \leftrightarrow \sim T$ $\Rightarrow (T \wedge T) \leftrightarrow F$ $\Rightarrow T \leftrightarrow F$ $\Rightarrow F$Therefore,the correct option is $C$.

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$(p \wedge \sim q) \wedge (\sim p \vee q)$ is :-

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