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If (f:\mathbb{R}\to\mathbb{R}) is given by (f(x)=\begin{cases}x+1,&x\le2\x^2-1,&x\ge2\end{cases}), why is it not a function?यदि (f:\mathbb{R}\to\mathbb{R}) को (f(x)=\begin{cases}x+1,&x\le2\x^2-1,&x\ge2\end{cases}) से दिया गया है तो यह फलन क्यों नहीं है?
Answer and explanationउत्तर और व्याख्या
Correct answer:सही उत्तर: Because at (x=2), both rules apply and give different valuesक्योंकि (x=2) पर दोनों नियम लागू होकर अलग मान देते हैं
Both pieces give the same value (3) at (x=2), so the rule is a valid function. Overlap is allowed only when the assigned values agree.(x=2) पर पहला मान (3) और दूसरा मान (3) नहीं बल्कि (2^2-1=3) है, इसलिए यह वास्तव में फलन है / At (x=2), the first value is (3) and the second value is also (2^2-1=3), so it is actually a function
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