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If (f:[1,9]\to\mathbb{R}) is given by (f(x)=\sqrt{x}+\sqrt{9-x}), what is the maximum value of (f)?यदि (f:[1,9]\to\mathbb{R}) को (f(x)=\sqrt{x}+\sqrt{9-x}) से दिया गया है, तो (f) का अधिकतम मान क्या है?
Answer and explanationउत्तर और व्याख्या
Correct answer:सही उत्तर: (3\sqrt{2})(3\sqrt{2})
By symmetry the maximum occurs at (x=\frac{9}{2}), and the value is (2\sqrt{\frac{9}{2}}=3\sqrt{2}). For sums of square roots, check the balanced point.सममिति से अधिकतम (x=\frac{9}{2}) पर मिलता है और मान (2\sqrt{\frac{9}{2}}=3\sqrt{2}) है। वर्गमूल योग में संतुलित बिंदु जांचें।
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