A) \[X=\overline{\overline{A}}\cdot \overline{\overline{B}}\]
B) \[X=\overline{A\cdot B}\]
C) \[X=A\cdot B\]
D) \[X=\overline{A+B}\]
Correct Answer: C
Solution :
\[X=\overline{\overline{AB}}=A\cdot B\](i.e., AND gate) If the output X of NAND gate is connected to the input of NOT gate (made from NAND gate by joining two inputs) from the given figure then we get back an AND gate.
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