यदि \(f:\mathbb{R}\to\mathbb{R}\) को (f(x)=\sqrt{x-2-10x+29}) से दिया गया है, तो परिसर क्या है?
If \(f:\mathbb{R}\to\mathbb{R}\) is given by (f(x)=\sqrt{x-2-10x+29}), what is the range?
Explanation opens after your attempt
Correct Answer
A. \([2,\infty\))
Concept
Inside the root, (x-2-10x+29=(x-5)2+4), so the minimum inside is (4). Hence the range is \([2,\infty\)).
Why this answer is correct
The correct answer is A. \([2,\infty\)). Inside the root, (x-2-10x+29=(x-5)2+4), so the minimum inside is (4). Hence the range is \([2,\infty\)).
Exam Tip
भीतर (x-2-10x+29=(x-5)2+4) है, इसलिए न्यूनतम भीतर (4) है। अतः परिसर \([2,\infty\)) है।
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