0 reads0 ratings0 helpful
If (f:\mathbb{R}\to\mathbb{R}), (f(x)=\frac{1}{1+x^2}), then (f) is not onto becauseयदि (f:\mathbb{R}\to\mathbb{R}), (f(x)=\frac{1}{1+x^2}) है, तो (f) आच्छादी नहीं है क्योंकि
Answer and explanationउत्तर और व्याख्या
Correct answer:सही उत्तर: Its range is ((0,1])इसका परास ((0,1]) है
Step 1: Since (1+x^2\ge 1), (\frac{1}{1+x^2}\le 1) and is always positive. Step 2: Its range is ((0,1]), while the codomain is (\mathbb{R}). So many real numbers such as (2) and (-1) are not obtained. Step 3: Always identify the exact range.चरण 1: (1+x^2\ge 1), इसलिए (\frac{1}{1+x^2}\le 1) और हमेशा धनात्मक है। चरण 2: इसका परास ((0,1]) है, जबकि सहप्रांत (\mathbb{R}) है। इसलिए कई वास्तविक संख्याएँ जैसे (2) और (-1) नहीं मिलतीं। चरण 3: हर बार परास को ठीक-ठीक पहचानें।
Student feedback
Was this question useful?
Was this question useful?
👍 0 Helpful 👎 0 Not helpful
Difficulty
Easy0%
Medium0%
Hard0%
Was the explanation clear?
Yes 0% No 0%
0 responsesStudent Reviews
No published reviews yet.
