यदि \(f:\mathbb{R}\to\mathbb{R}\) को (f(x)=\sqrt{x-2-6x+10}) से दिया गया है तो परिसर क्या है?
If \(f:\mathbb{R}\to\mathbb{R}\) is given by (f(x)=\sqrt{x-2-6x+10}), what is the range?
Explanation opens after your attempt
Correct Answer
A. \([1,\infty\))
Concept
Inside the root, (x-2-6x+10=(x-3)2+1), so the minimum inside is (1). Thus the range is \([1,\infty\)).
Why this answer is correct
The correct answer is A. \([1,\infty\)). Inside the root, (x-2-6x+10=(x-3)2+1), so the minimum inside is (1). Thus the range is \([1,\infty\)).
Exam Tip
भीतर (x-2-6x+10=(x-3)2+1) है, इसलिए न्यूनतम भीतर (1) है। अतः परिसर \([1,\infty\)) है।
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