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If (f:\mathbb{R}\to\mathbb{R}) is given by (f(x)=\begin{cases}x+3,&x\le1\x^2+1,&x\ge1\end{cases}), why is it not a function?यदि (f:\mathbb{R}\to\mathbb{R}) को (f(x)=\begin{cases}x+3,&x\le1\x^2+1,&x\ge1\end{cases}) से दिया गया है, तो यह फलन क्यों नहीं है?
Answer and explanationउत्तर और व्याख्या
Correct answer:सही उत्तर: Because at (x=1), two different values (4) and (2) occurक्योंकि (x=1) पर दो अलग मान (4) और (2) मिलते हैं
The input (x=1) belongs to both parts and gives different values (4) and (2). Two different outputs for one input do not define a function.(x=1) दोनों भागों में आता है और मान (4) तथा (2) अलग हैं। एक ही इनपुट पर दो अलग आउटपुट फलन नहीं बनाते।
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