फलन (f(x)=5-\sqrt{x-2+1}) का परिसर क्या है?
What is the range of (f(x)=5-\sqrt{x-2+1})?
Explanation opens after your attempt
Correct Answer
A. ( \(-\infty,4] \)
Concept
Since \(\sqrt{x^2+1}\ge 1\), \(5-\sqrt{x^2+1}\le 4\). As (x) grows, the value decreases without bound.
Why this answer is correct
The correct answer is A. ( \(-\infty,4] \). Since \(\sqrt{x^2+1}\ge 1\), \(5-\sqrt{x^2+1}\le 4\). As (x) grows, the value decreases without bound.
Exam Tip
\(\sqrt{x^2+1}\ge 1\), इसलिए \(5-\sqrt{x^2+1}\le 4\)। (x) बड़ा होने पर मान असीम रूप से घटता है।
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