फलन (f(x)=2-\sqrt{x-1}) की रेंज क्या है?
What is the range of (f(x)=2-\sqrt{x-1})?
Explanation opens after your attempt
Correct Answer
A. ( \(-\infty,2]\)
Concept
\(\sqrt{x-1}\ge 0\), so \(2-\sqrt{x-1}\le 2\) and it can go down without bound. In exams a negative sign reverses the direction of the range.
Why this answer is correct
The correct answer is A. ( \(-\infty,2]\). \(\sqrt{x-1}\ge 0\), so \(2-\sqrt{x-1}\le 2\) and it can go down without bound. In exams a negative sign reverses the direction of the range.
Exam Tip
\(\sqrt{x-1}\ge 0\), इसलिए \(2-\sqrt{x-1}\le 2\) और नीचे अनंत तक जा सकता है। परीक्षा में ऋण चिह्न रेंज की दिशा बदलता है।
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