यदि \(U={x:x \in \mathbb{Z},-10\le x\le 10}\) और \(A={x:x \in U,x^2\le 16}\), तो (n(A')) क्या है?

If \(U={x:x \in \mathbb{Z},-10\le x\le 10}\) and \(A={x:x \in U,x^2\le 16}\), what is (n(A'))?

Explanation opens after your attempt
Correct Answer

A. (12)

Step 1

Concept

\(x^2\le 16\) gives \(-4\le x\le 4\), so (A) has (9) elements. Since (U) has (21) elements, (n(A')=12).

Step 2

Why this answer is correct

The correct answer is A. (12). \(x^2\le 16\) gives \(-4\le x\le 4\), so (A) has (9) elements. Since (U) has (21) elements, (n(A')=12).

Step 3

Exam Tip

\(x^2\le 16\) से \(-4\le x\le 4\) मिलता है, इसलिए (A) में (9) सदस्य हैं। (U) में (21) सदस्य हैं, अतः (n(A')=12) है।

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Mathematics Answer, Explanation and Revision Hints

यदि \(U={x:x \in \mathbb{Z},-10\le x\le 10}\) और \(A={x:x \in U,x^2\le 16}\), तो (n(A')) क्या है? / If \(U={x:x \in \mathbb{Z},-10\le x\le 10}\) and \(A={x:x \in U,x^2\le 16}\), what is (n(A'))?

Correct Answer: A. (12). Explanation: \(x^2\le 16\) से \(-4\le x\le 4\) मिलता है, इसलिए (A) में (9) सदस्य हैं। (U) में (21) सदस्य हैं, अतः (n(A')=12) है। / \(x^2\le 16\) gives \(-4\le x\le 4\), so (A) has (9) elements. Since (U) has (21) elements, (n(A')=12).

Which concept should I revise for this Mathematics MCQ?

\(x^2\le 16\) gives \(-4\le x\le 4\), so (A) has (9) elements. Since (U) has (21) elements, (n(A')=12).

What exam hint can help solve this Mathematics question?

\(x^2\le 16\) से \(-4\le x\le 4\) मिलता है, इसलिए (A) में (9) सदस्य हैं। (U) में (21) सदस्य हैं, अतः (n(A')=12) है।