यदि एक समुच्चय A में 3 अवयव हैं और दूसरे समुच्चय B में 6 अवयव हैं, तो (A∪B) में न्यूनतम कितने अवयव हो सकते हैं?

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NDA (Held On: 17 Nov 2019) Maths Previous Year paper
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  1. 3
  2. 6
  3. 8
  4. 9

Answer (Detailed Solution Below)

Option 2 : 6
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अवधारणा:

यदि A और B दो समुच्चय  हैं तो, n (A ∪ B) = n (A) + n (B) - n (A ∩ B)

गणना:

दिया गया है: n (A) = 3 और n (B) = 6

जैसा कि हम जानते हैं कि, यदि A और B दो समुच्चय हैं तो, n (A ∪ B) = n (A) + n (B) - n (A ∩ B)

⇒ n (A ∪ B) = 3 + 6 - n (A ∩ B)

 n (A ∪ B) को कम करने के लिए हमें n (A ∩ B)  को अधिकतम करना होगा।

यदि A, B का उप-समुच्चय है, तो A ∩ B = A ⇒ n (A ∩ B) = n (A) = 3

n (A ∪ B) = 3 + 6 - 3 = 6

इसलिए, घटकों की न्यूनतम संख्या जो (A∪B) में हो सकती है,  वह 6 है। 

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