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?
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?
Explanation opens after your attempt
A. क्योंकि (x=2) पर दोनों नियम लागू होकर अलग मान देते हैंBecause at (x=2), both rules apply and give different values
Concept
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.
Why this answer is correct
The correct answer is A. क्योंकि (x=2) पर दोनों नियम लागू होकर अलग मान देते हैं / Because at (x=2), both rules apply and give different values. 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.
Exam Tip
(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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