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"All pentagon are polygon".

Let, Pentagon $(x):x$ is a pentagon, and Polygon $(x):x$ is a polygon

Domain of discourse: All the shapes

If $x$ is a pentagon, then $x$ is a polygon

$\forall x[\text{Pentagon}(x) \rightarrow \text{Polygon}(x)]$

Now, we can negate the above statement.

$\neg\forall x[\text{Pentagon}(x) \rightarrow \text{Polygon}(x)] \equiv \neg \forall x [\neg \text{Pentagon}(x) \vee \text{Polygon}(x)] \equiv \exists x [\text{Pentagon}(x) \wedge \neg\text{Polygon}(x)] \quad [\because \text{Using Demorgan's law:}\;\neg \forall x P(x) \equiv \exists x\neg P(x),\;\; \neg \exists x P(x) \equiv \forall x\neg P(x)]$

We can write in English language, There exist at least one pentagon, which is not a polygon. (or) there is some pentagon which is not polygon.

So, the correct answer is $(B).$
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