फलन (f(x)=\frac{1}{\sqrt{4-x-2}}) का परिसर क्या है?
What is the range of (f(x)=\frac{1}{\sqrt{4-x-2}})?
Explanation opens after your attempt
Correct Answer
A. \([\frac{1}{2},\infty\))
Concept
The maximum value of \(\sqrt{4-x^2}\) is (2), so the minimum value of (f) is \(\frac{1}{2}\). Near the endpoints, the value grows without bound.
Why this answer is correct
The correct answer is A. \([\frac{1}{2},\infty\)). The maximum value of \(\sqrt{4-x^2}\) is (2), so the minimum value of (f) is \(\frac{1}{2}\). Near the endpoints, the value grows without bound.
Exam Tip
\(\sqrt{4-x^2}\) का अधिकतम मान (2) है, इसलिए (f) का न्यूनतम मान \(\frac{1}{2}\) है। किनारों के पास मान असीमित बढ़ता है।
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