फलन (f(x)=\sqrt{x}+\sqrt{1-x}) का परिसर चुनिए।
Choose the range of (f(x)=\sqrt{x}+\sqrt{1-x}).
Explanation opens after your attempt
Correct Answer
A. \([1,\sqrt{2}]\)
Concept
On ([0,1]), the minimum at endpoints is (1) and the maximum at \(x=\frac{1}{2}\) is \(\sqrt{2}\). Hence the range is \([1,\sqrt{2}]\).
Why this answer is correct
The correct answer is A. \([1,\sqrt{2}]\). On ([0,1]), the minimum at endpoints is (1) and the maximum at \(x=\frac{1}{2}\) is \(\sqrt{2}\). Hence the range is \([1,\sqrt{2}]\).
Exam Tip
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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