MHT CET · Maths · Mathematical Reasoning
Dual of \(\left(x^{\prime} \vee y^{\prime}\right)=x \wedge y\) is
- A \(\left(x^{\prime} \vee y^{\prime}\right)=x \vee y_{1}\)
- B \(\left(x^{\prime} \wedge y^{\prime}\right)^{\prime}=x \vee y\)
- C \(\left(x^{\prime} \wedge y^{\prime}\right)^{\prime}=x \wedge y\)
- D None of the above
Answer & Solution
Correct Answer
(B) \(\left(x^{\prime} \wedge y^{\prime}\right)^{\prime}=x \vee y\)
Step-by-step Solution
Detailed explanation
Dual of \(\left(x^{\prime} \vee y^{\prime}\right)^{\prime}=x \wedge y\) is \(\left(x^{\prime} \wedge y^{\prime}\right)^{\gamma}=x \vee y\)
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