यदि \(f:\mathbb{R}\to\mathbb{R}\) को (f(x)=\frac{x-2+1}{x-2+2}) से दिया गया है तो परिसर क्या है?

If \(f:\mathbb{R}\to\mathbb{R}\) is given by (f(x)=\frac{x-2+1}{x-2+2}), what is the range?

Explanation opens after your attempt
Correct Answer

A. \(\left[\frac{1}{2},1\right\))

Step 1

Concept

The value is \(1-\frac{1}{x^2+2}\), so the minimum is \(\frac{1}{2}\) and (1) is never attained. The range is \(\left[\frac{1}{2},1\right\)).

Step 2

Why this answer is correct

The correct answer is A. \(\left[\frac{1}{2},1\right\)). The value is \(1-\frac{1}{x^2+2}\), so the minimum is \(\frac{1}{2}\) and (1) is never attained. The range is \(\left[\frac{1}{2},1\right\)).

Step 3

Exam Tip

मान \(1-\frac{1}{x^2+2}\) है, इसलिए न्यूनतम \(\frac{1}{2}\) और (1) कभी नहीं मिलता। परिसर \(\left[\frac{1}{2},1\right\)) है।

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Mathematics Answer, Explanation and Revision Hints

यदि \(f:\mathbb{R}\to\mathbb{R}\) को (f(x)=\frac{x-2+1}{x-2+2}) से दिया गया है तो परिसर क्या है? / If \(f:\mathbb{R}\to\mathbb{R}\) is given by (f(x)=\frac{x-2+1}{x-2+2}), what is the range?

Correct Answer: A. \(\left[\frac{1}{2},1\right\)). Explanation: मान \(1-\frac{1}{x^2+2}\) है, इसलिए न्यूनतम \(\frac{1}{2}\) और (1) कभी नहीं मिलता। परिसर \(\left[\frac{1}{2},1\right\)) है। / The value is \(1-\frac{1}{x^2+2}\), so the minimum is \(\frac{1}{2}\) and (1) is never attained. The range is \(\left[\frac{1}{2},1\right\)).

Which concept should I revise for this Mathematics MCQ?

The value is \(1-\frac{1}{x^2+2}\), so the minimum is \(\frac{1}{2}\) and (1) is never attained. The range is \(\left[\frac{1}{2},1\right\)).

What exam hint can help solve this Mathematics question?

मान \(1-\frac{1}{x^2+2}\) है, इसलिए न्यूनतम \(\frac{1}{2}\) और (1) कभी नहीं मिलता। परिसर \(\left[\frac{1}{2},1\right\)) है।