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?
If \(f:[1,9]\to\mathbb{R}\) is given by (f(x)=\sqrt{x}+\sqrt{9-x}), what is the maximum value of (f)?
Explanation opens after your attempt
Correct Answer
C. \(3\sqrt{2}\)
Concept
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.
Why this answer is correct
The correct answer is C. \(3\sqrt{2}\). 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.
Exam Tip
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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