Given the following two statements:

$\left(S_1\right):(q\vee p)\to (p\leftrightarrow \sim q)$ is a tautology.

$\left(S_2\right):\sim q\wedge (\sim p\leftrightarrow q)$ is a fallacy. Then:

$\left(p\vee q\right)\Rightarrow q$

= $\sim \left(p\vee q\right)\vee q$

= $\left(\sim p\wedge \sim q\right)\vee q$

= $\left(\sim p\vee q\right)\wedge \left(\sim q\vee q\right)$

= $\sim p\vee q$

now $\left(p\wedge q\right)\Delta \left(\sim p\vee q\right)istautolog\text{?}y$

pq $p\wedge q$ $\sim p\vee q$ $\left(p\wedge q\right)\Delta \left(\sim p\vee q\right)$

TTTTT

TFFFT

FTFTT

FFFTT

So,  $\Delta =\Rightarrow$

Let

P : Ramesh listens to music

Q : Ramesh is out of his village

R : It is Sunday

S : It is Saturday

Then the statement “Ramesh listens to music only if he is in his village and it is Sunday or Saturday can be expressed as

$P\equiv$ Ramesh listens to music

$\sim q\equiv$ He is in village.

$r\vee s\equiv$ Saturday or Sunday

$P\Rightarrow \left(\left(\sim q\right)\wedge \left(r\vee s\right)\right)$

${\int }_0^3?g\left(x\right)-f\left(x\right)={\int }_0^3?\left|\right|x-2|-2|dx-{\int }_0^3?|x-2|dx$

$=\left(\frac{1}{2}\times 2\times 2+1+\frac{1}{2}\times 1\times 1\right)-\left(\frac{1}{2}\times 2\times 2+\frac{1}{2}\times 1\times 1\right)$

$=\left(2+1+\frac{1}{2}\right)-\left(2+\frac{1}{2}\right)=1$

 $?\left(\left(\sim q\right)\wedge p\right)\vee \left(pv\left(\sim p\right)\right)$

$=\left(\sim q\wedge p\right)\vee t\left(tis\right)$

$\equiv t$

$\therefore$ option (C) is correct.