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Assume $P \to R$ is TRUE and $Q \to R$ is FALSE. Now when $P$ is false, $Q$ is true and $R$ is false, $X$ is FALSE but $Y$ is true. So $Y \to X$ is not a tautology, which rules out options B and D. C is in fact a tautology - can verify by Truth table.
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$\large X : \left ( P \vee Q \right ) \rightarrow R$

$Y : (P \rightarrow R) \vee (Q \rightarrow R)$

 

$Y : (\overline{P} \vee R) \vee (\overline{Q} \vee R)$

$Y : (\overline{P} \vee \overline{Q} \vee R)$

$Y : (\overline{P \wedge Q}) \vee R$

$\large Y : (P \wedge Q) \rightarrow R$

now Y can be false if and only if $P$ and $Q$ are true but $R$ is false but in such case $X$ is also false

From this we can conclude that there exist no case where $X \rightarrow Y$ results in False.

$X \rightarrow Y$ is a tautology

 

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