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If (f:\mathbb{R}\to [1,\infty)), (f(x)=\sqrt{x^2+1}), what type is (f)?यदि (f:\mathbb{R}\to [1,\infty)), (f(x)=\sqrt{x^2+1}), तो (f) कैसा है?
Answer and explanationउत्तर और व्याख्या
Correct answer:सही उत्तर: Ontoआच्छादक है
Step 1: For any (y\ge1), set (y=\sqrt{x^2+1}). Step 2: Then (x^2=y^2-1), so (x=\sqrt{y^2-1}) is real. Step 3: If you can construct (x) from the target value, onto is proved.चरण 1: किसी भी (y\ge1) के लिए (y=\sqrt{x^2+1}) मानें। चरण 2: इससे (x^2=y^2-1) और (x=\sqrt{y^2-1}) वास्तविक मिलता है। चरण 3: लक्ष्य मान से (x) बना सके तो आच्छादकता सिद्ध हो जाती है।
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