यदि \(G={x\in \mathbb{Z}: x^2<2x+8}\), तो (G) का सही रोस्टर रूप कौन-सा है?
If \(G={x\in \mathbb{Z}: x^2<2x+8}\), which is the correct roster form of (G)?
Explanation opens after your attempt
Correct Answer
A. \(G=\{-1,0,1,2,3\}\)
Concept
Convert \(x^2<2x+8\) into \(x^2-2x-8<0\).
Why this answer is correct
From ((x-4)(x+2)<0), we get (-2<x<4), so the integer elements are (-1,0,1,2,3).
Exam Tip
In strict inequalities, boundary points are not included. चरण 1: \(x^2<2x+8\) को \(x^2-2x-8<0\) में बदलें। चरण 2: ((x-4)(x+2)<0) से (-2<x<4) मिलता है, इसलिए पूर्णांक (-1,0,1,2,3) हैं। चरण 3: सख्त असमानता में सीमा बिंदु शामिल नहीं किए जाते।
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