CBSE Class 12 Math 2022 Term I Solved Paper

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Question : 3
Total: 50
A function f:R→R is defined as f(x)=x3+1. Then the function has
Explanation: Given, f(x)=x3+1
∴‌f′(x)‌=3x2‌ and ‌f′′(x)=6x
‌ Put ‌‌f′(x)‌=0
⇒3x2‌=0⇒x=0
‌‌ At ‌‌x‌=0,f′′(x)=0
Thus, f(x) has neither maximum value nor minimum value.
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