यदि \(U=\{0,1,2,3,4,5\}\) और \(A={x:x^2=x}\), तो (n(\mathcal{P}(A'))) कितना है?
If \(U=\{0,1,2,3,4,5\}\) and \(A={x:x^2=x}\), what is (n(\mathcal{P}(A')))?
Explanation opens after your attempt
Correct Answer
A. (16)
Concept
\(x^2=x\) gives (x=0) or (x=1), so \(A=\{0,1\}\). (A') has (4) elements, hence \(2^4=16\).
Why this answer is correct
The correct answer is A. (16). \(x^2=x\) gives (x=0) or (x=1), so \(A=\{0,1\}\). (A') has (4) elements, hence \(2^4=16\).
Exam Tip
\(x^2=x\) से (x=0) या (x=1), इसलिए \(A=\{0,1\}\)। (A') में (4) तत्व हैं, अतः \(2^4=16\)।
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