Match List I with List II: |l|l| List I & List II \\ a. (x+y) & I. 1 \\ b. x+1 & II. x +y \\ c. (xy) & III. x · y \\ d. x+0 & IV. x \\ Choose the correct answer from the options given below.

(a) \((x+y)' = x' · y'\) (De Morgan) → III (b) \(x + 1 = 1\) → I (c) \((xy)' = x' + y'\) (De Morgan) → II (d) \(x + 0 = x\) → IV So A-III, B-I, C-II, D-IV. Answer: (D)

Match List I with List II: |l|l| List I &... - Solved

Question 65

Computer Awareness Boolean algebra Easy

Match List I with List II: \begin{array}{|l|l|} \hline \text{List I} & \text{List II} \\ \hline a. (x+y)^{\prime} & I. 1 \\ b. \mathrm{x}+1 & II. x^{\prime}+y^{\prime} \\ c. (\mathrm{xy})^{\prime} & III. x^{\prime} \cdot y^{\prime} \\ d. \mathrm{x}+0 & IV. x \\ \hline \end{array} Choose the correct answer from the options given below.

(A) \( \mathrm{A}-\mathrm{II} ; \mathrm{B}-\mathrm{I} ; \mathrm{C}-\mathrm{III} ; \mathrm{D}-\mathrm{IV} \)

(B) A-II; B-IV; C-I; D-III

(C) A-III; B-I; C-IV; D-II

(D) A-III; B-I; C-II; D - IV

✓ Correct Answer: Option D Verified Solution

📘 Step-by-Step Solution & Detailed Explanation:

(a) \((x+y)' = x' \cdot y'\) (De Morgan) → III
(b) \(x + 1 = 1\) → I
(c) \((xy)' = x' + y'\) (De Morgan) → II
(d) \(x + 0 = x\) → IV

So A-III, B-I, C-II, D-IV.

Answer: (D)

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Question 65

📘 Step-by-Step Solution & Detailed Explanation:

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