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Why is (f:\mathbb{R}\to[0,\infty)), (f(x)=\sqrt{x^2+1}), not onto?फलन (f:\mathbb{R}\to[0,\infty)), (f(x)=\sqrt{x^2+1}), सर्वाच्छादक क्यों नहीं है?
Answer and explanationउत्तर और व्याख्या
Correct answer:सही उत्तर: Because (\frac{1}{2}) is not an imageक्योंकि (\frac{1}{2}) छवि नहीं बनता
Step 1: Since (x^2+1\ge1), (\sqrt{x^2+1}\ge1). Step 2: The codomain ([0,\infty)) contains (\frac{1}{2}), but it is not an image. Step 3: If the codomain is larger than the actual range, the function is not onto.चरण 1: (x^2+1\ge1), इसलिए (\sqrt{x^2+1}\ge1)। चरण 2: सहप्रांत ([0,\infty)) में (\frac{1}{2}) है, पर यह छवि नहीं बनता। चरण 3: सहप्रांत यदि वास्तविक परास से बड़ा हो जाए, तो फलन सर्वाच्छादक नहीं रहता।
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