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If \(x\in\mathbb{Z}\) and \(-5<x\le 1\), how many integer values of \(x\) are possible?

Answer and explanation

Correct answer: 6

In \(-5<x\le 1\), the value \(-5\) is excluded, while \(1\) is included. Thus, the possible integers are \(-4,-3,-2,-1,0,1\), giving a total of 6 values. A common mistake is to count \(-5\) as well, but it is not allowed because the inequality is strict at that endpoint. Exam tip: an open inequality sign excludes the boundary value, whereas \(\le\) or \(\ge\) includes it.

Related tags

Linear InequalitiesInteger SolutionsNumber LineCountingRepresentation On Number LineMathematicsClass 11 Mcq

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