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If (f:\mathbb{R}\to\mathbb{R}) is given by (f(x)=\begin{cases}x+4,&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+4,&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), two different values (6) and (3) occurक्योंकि (x=2) पर दो अलग मान (6) और (3) मिलते हैं
The input (x=2) belongs to both parts and gives different values (6) and (3). Two different outputs for one input do not define a function.(x=2) दोनों भागों में आता है और मान (6) तथा (3) अलग हैं। एक ही इनपुट पर दो अलग आउटपुट फलन नहीं बनाते।
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