Contrapositive of the statement:

If a function $f$ is differentiable at $a$ , then it is also continuous at $a$ ', is:

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

= $\sim \left(p\vee q\right)\vee q$

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

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

= $\sim p\vee q$

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

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

Let TV (r) denotes truth value of a statement $r$ .

$\text{Now, if}TV\left(p\right)=TV\left(q\right)=T$

$\Rightarrow \mathrm{T}\mathrm{V}\left({\mathrm{S}}_1\right)=\mathrm{F}$

Also, if $TV\left(p\right)=T$ and $TV\left(q\right)=F$

$\Rightarrow \mathrm{T}\mathrm{V}\left({\mathrm{S}}_2\right)=\mathrm{T}$

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

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

$\equiv t$

$\therefore$ option (C) is correct.