यदि \(f:\mathbb{R}\to\mathbb{R}\) और (f(x)=|x|+|x-2|) हो तो परिसर क्या है?

If \(f:\mathbb{R}\to\mathbb{R}\) and (f(x)=|x|+|x-2|), what is the range?

Explanation opens after your attempt
Correct Answer

A. \([2,\infty\))

Step 1

Concept

For \(0\le x\le2\), the value is (2), and it increases outside this interval. Since the minimum is (2), the range is \([2,\infty\)).

Step 2

Why this answer is correct

The correct answer is A. \([2,\infty\)). For \(0\le x\le2\), the value is (2), and it increases outside this interval. Since the minimum is (2), the range is \([2,\infty\)).

Step 3

Exam Tip

\(0\le x\le2\) पर मान (2) है और बाहर जाने पर मान बढ़ता है। न्यूनतम (2) होने से परिसर \([2,\infty\)) है।

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FAQs

Mathematics Answer, Explanation and Revision Hints

यदि \(f:\mathbb{R}\to\mathbb{R}\) और (f(x)=|x|+|x-2|) हो तो परिसर क्या है? / If \(f:\mathbb{R}\to\mathbb{R}\) and (f(x)=|x|+|x-2|), what is the range?

Correct Answer: A. \([2,\infty\)). Explanation: \(0\le x\le2\) पर मान (2) है और बाहर जाने पर मान बढ़ता है। न्यूनतम (2) होने से परिसर \([2,\infty\)) है। / For \(0\le x\le2\), the value is (2), and it increases outside this interval. Since the minimum is (2), the range is \([2,\infty\)).

Which concept should I revise for this Mathematics MCQ?

For \(0\le x\le2\), the value is (2), and it increases outside this interval. Since the minimum is (2), the range is \([2,\infty\)).

What exam hint can help solve this Mathematics question?

\(0\le x\le2\) पर मान (2) है और बाहर जाने पर मान बढ़ता है। न्यूनतम (2) होने से परिसर \([2,\infty\)) है।