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Choose the range of (f(x)=\sqrt{x}+\sqrt{1-x}).फलन (f(x)=\sqrt{x}+\sqrt{1-x}) का परिसर चुनिए।
Answer and explanationउत्तर और व्याख्या
Correct answer:सही उत्तर: ([1,\sqrt{2}])([1,\sqrt{2}])
On ([0,1]), the minimum at endpoints is (1) and the maximum at (x=\frac{1}{2}) is (\sqrt{2}). Hence the range is ([1,\sqrt{2}]).प्रांत ([0,1]) में न्यूनतम मान किनारों पर (1) और अधिकतम (x=\frac{1}{2}) पर (\sqrt{2}) है। इसलिए परिसर ([1,\sqrt{2}]) है।
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