यदि \(f:\mathbb{R}\to\mathbb{R}\), (f(x)=\frac{1}{1+x-2}) है, तो (f) आच्छादी नहीं है क्योंकि
If \(f:\mathbb{R}\to\mathbb{R}\), (f(x)=\frac{1}{1+x-2}), then (f) is not onto because
Correct answer and explanation
A. इसका परास ((0,1]) हैIts range is ((0,1])
Concept
\(1+x^2\ge 1\), इसलिए \(\frac{1}{1+x^2}\le 1\) और हमेशा धनात्मक है। / Since \(1+x^2\ge 1\), \(\frac{1}{1+x^2}\le 1\) and is always positive.
Why this answer is correct
इसका परास ((0,1]) है, जबकि सहप्रांत \(\mathbb{R}\) है। इसलिए कई वास्तविक संख्याएँ जैसे (2) और (-1) नहीं मिलतीं। / Its range is ((0,1]), while the codomain is \(\mathbb{R}\). So many real numbers such as (2) and (-1) are not obtained.
Exam Tip
हर बार परास को ठीक-ठीक पहचानें। / Always identify the exact range.
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