Let $\Delta \in \left\{\wedge ,\vee ,\Rightarrow ,\iff \right\}$ be such that $\left(p\wedge q\right)\Delta \left(\left(p\vee q\right)\Rightarrow q\right)$ is a tautology. Then $\Delta$ is equal to:

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 $TV\left(q\right)=F$

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

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

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

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

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

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

$\equiv t$

$\therefore$ option (C) is correct.