$(a)\;2\qquad(b)\;3\qquad(c)\;4\qquad(d)\;5$

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A atoms are at the corners.

$\therefore$ No of A atoms per unit cell = $8\times\large\frac{1}{8}$

$\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\; = 1$

B atoms are face centre

$\therefore$ No of B atoms per unit cell = $6\times\large\frac{1}{2}$

$\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;=3$

$\therefore AB_x = AB_3$

$\Rightarrow x = 3$

Hence answer is (b)

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