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What is the range of \(f(x)=\sqrt{12-x-x^2}\)?फलन \(f(x)=\sqrt{12-x-x^2}\) का परिसर क्या है?
Answer and explanationउत्तर और व्याख्या
Correct answer:सही उत्तर: \( [0,\frac{7}{2}] \)\( [0,\frac{7}{2}] \)
Since \(12-x-x^2=\frac{49}{4}-\left(x+\frac{1}{2}\right)^2\), the inside maximum is \(\frac{49}{4}\). Taking square root gives maximum \(\frac{7}{2}\).\(12-x-x^2=\frac{49}{4}-\left(x+\frac{1}{2}\right)^2\), इसलिए अंदर का अधिकतम \(\frac{49}{4}\) है। वर्गमूल से अधिकतम \(\frac{7}{2}\) मिलता है।
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