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फलन \(f:\mathbb{R}\to[0,\infty\)), (f(x)=\(e^x-1\)2), सर्वाच्छादक है या नहीं?

Is \(f:\mathbb{R}\to[0,\infty\)), (f(x)=\(e^x-1\)2), onto or not?

Explanation opens after your attempt
Correct Answer

A. सर्वाच्छादक हैIt is onto

Step 1

Concept

(\(e^x-1\)2\ge0), and at (x=0), the value is (0).

Step 2

Why this answer is correct

For any \(y\ge0\), set \(e^x-1=\sqrt{y}\), giving (x=\ln\(1+\sqrt{y}\)), which is real.

Step 3

Exam Tip

For exponential expressions, logarithms help construct preimages. चरण 1: (\(e^x-1\)2\ge0) और (x=0) पर (0) मिलता है। चरण 2: किसी भी \(y\ge0\) के लिए \(e^x-1=\sqrt{y}\) लेने पर (x=\ln\(1+\sqrt{y}\)) वास्तविक मिलता है। चरण 3: घातीय फलन में धनात्मक पूर्वप्रतिबिंब बनाने के लिए लघुगणक का प्रयोग करें।

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