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What is the maximum value of \(f(x)=\sqrt{x+2}+\sqrt{6-x}\)?\(f(x)=\sqrt{x+2}+\sqrt{6-x}\) का अधिकतम मान क्या है?
Answer and explanationउत्तर और व्याख्या
Correct answer:सही उत्तर: \(4\)\(4\)
For square roots, \(x+2\ge 0\) and \(6-x\ge 0\Rightarrow -2\le x\le 6\). The expression is symmetric: \(x+2\) and \(6-x\) add to \(8\). The maximum occurs when the two radicands are equal: \(x+2=6-x\Rightarrow 2x=4\Rightarrow x=2\). Then \(f(2)=\sqrt{2+2}+\sqrt{6-2}=\sqrt{4}+\sqrt{4}=2+2=4\).वर्गमूल के लिए \(x+2\ge 0\) और \(6-x\ge 0\Rightarrow -2\le x\le 6\) होना चाहिए। यह अभिव्यक्ति सममित है क्योंकि \(x+2\) और \(6-x\) का योग \(8\) है। अधिकतम तब होता है जब दोनों अंतर्गत राशियाँ बराबर हों: \(x+2=6-x\Rightarrow 2x=4\Rightarrow x=2\). तब \(f(2)=\sqrt{2+2}+\sqrt{6-2}=\sqrt{4}+\sqrt{4}=2+2=4\).
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