फलन (f(x)=\frac{1}{\sqrt{16-x-2}}) का परिसर क्या है?
What is the range of (f(x)=\frac{1}{\sqrt{16-x-2}})?
Explanation opens after your attempt
Correct Answer
A. \( \left[\frac{1}{4},\infty\right\) )
Concept
The maximum of \(\sqrt{16-x^2}\) is (4), so the minimum fraction is \(\frac{1}{4}\). Near the endpoints, the denominator tends to (0).
Why this answer is correct
The correct answer is A. \( \left[\frac{1}{4},\infty\right\) ). The maximum of \(\sqrt{16-x^2}\) is (4), so the minimum fraction is \(\frac{1}{4}\). Near the endpoints, the denominator tends to (0).
Exam Tip
\(\sqrt{16-x^2}\) का अधिकतम (4) है, इसलिए भिन्न का न्यूनतम \(\frac{1}{4}\) है। किनारों के पास हर (0) की ओर जाता है।
Login to save your score, XP, coins and progress.