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What is the range of (f(x)=\sqrt{x^2+4})?फलन (f(x)=\sqrt{x^2+4}) की रेंज क्या है?
Answer and explanationउत्तर और व्याख्या
Correct answer:सही उत्तर: ([2,\infty))([2,\infty))
The minimum value of (x^2+4) is (4), so (\sqrt{x^2+4}\ge 2). In exams first find the minimum value inside.(x^2+4) की न्यूनतम वैल्यू (4) है इसलिए (\sqrt{x^2+4}\ge 2)। परीक्षा में पहले अंदर की न्यूनतम वैल्यू निकालें।
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