यदि \(f:\mathbb{R}\to\mathbb{R}\) और (f(x)=|x+5|+|x-2|) हो, तो परिसर क्या है?

If \(f:\mathbb{R}\to\mathbb{R}\) and (f(x)=|x+5|+|x-2|), what is the range?

Explanation opens after your attempt
Correct Answer

A. \([7,\infty\))

Step 1

Concept

For \(-5\le x\le2\), the value is (7), and outside this interval the value increases. Hence the minimum is (7) and the range is \([7,\infty\)).

Step 2

Why this answer is correct

The correct answer is A. \([7,\infty\)). For \(-5\le x\le2\), the value is (7), and outside this interval the value increases. Hence the minimum is (7) and the range is \([7,\infty\)).

Step 3

Exam Tip

\(-5\le x\le2\) पर मान (7) है और बाहर मान बढ़ता है। इसलिए न्यूनतम (7) और परिसर \([7,\infty\)) है।

Question me issue ya doubt hai?

Answer, explanation, typing mistake ya suggestion directly hamari team ko bhejein. 📱Helpline (Call / WhatsApp): +91 7272824365

Related Mathematics Questions

FAQs

Mathematics Answer, Explanation and Revision Hints

यदि \(f:\mathbb{R}\to\mathbb{R}\) और (f(x)=|x+5|+|x-2|) हो, तो परिसर क्या है? / If \(f:\mathbb{R}\to\mathbb{R}\) and (f(x)=|x+5|+|x-2|), what is the range?

Correct Answer: A. \([7,\infty\)). Explanation: \(-5\le x\le2\) पर मान (7) है और बाहर मान बढ़ता है। इसलिए न्यूनतम (7) और परिसर \([7,\infty\)) है। / For \(-5\le x\le2\), the value is (7), and outside this interval the value increases. Hence the minimum is (7) and the range is \([7,\infty\)).

Which concept should I revise for this Mathematics MCQ?

For \(-5\le x\le2\), the value is (7), and outside this interval the value increases. Hence the minimum is (7) and the range is \([7,\infty\)).

What exam hint can help solve this Mathematics question?

\(-5\le x\le2\) पर मान (7) है और बाहर मान बढ़ता है। इसलिए न्यूनतम (7) और परिसर \([7,\infty\)) है।