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What is the range of \(f(x)=\frac{3x}{x^2+9}\)?फलन \(f(x)=\frac{3x}{x^2+9}\) का परिसर क्या है?
Answer and explanationउत्तर और व्याख्या
Correct answer:सही उत्तर: \( [-\frac{1}{2},\frac{1}{2}] \)\( [-\frac{1}{2},\frac{1}{2}] \)
Since \(|3x|\le \frac{x^2+9}{2}\), \(\left|\frac{3x}{x^2+9}\right|\le \frac{1}{2}\). Equality occurs at \(x=\pm 3\).\(|3x|\le \frac{x^2+9}{2}\), इसलिए \(\left|\frac{3x}{x^2+9}\right|\le \frac{1}{2}\)। समानता \(x=\pm 3\) पर मिलती है।
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