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If (f:\mathbb{R}\to[1,\infty)), (f(x)=x^2+1), what preimage can be taken for (y\in[1,\infty))?यदि (f:\mathbb{R}\to[1,\infty)), (f(x)=x^2+1) है, तो (y\in[1,\infty)) का पूर्वप्रतिबिंब क्या लिया जा सकता है?
Answer and explanationउत्तर और व्याख्या
Correct answer:सही उत्तर: (x=\sqrt{y-1})(x=\sqrt{y-1})
Step 1: From (y=x^2+1), we get (x^2=y-1). Step 2: Since (y\ge1), (y-1\ge0), so (x=\sqrt{y-1}) is real. Step 3: To prove onto, solve for (x) in terms of (y) and check that (x) belongs to the domain.चरण 1: (y=x^2+1) से (x^2=y-1) मिलता है। चरण 2: (y\ge1) होने से (y-1\ge0), इसलिए (x=\sqrt{y-1}) वास्तविक है। चरण 3: आच्छादी सिद्ध करते समय (y) से (x) निकालकर जाँचें कि (x) प्रांत में है।
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