#Mark the correct alternative in each of the following
If x lies in the interval [0, 1], then the least value of x2 + x + 1 is
f(x)=![]()
f’(x)=![]()
f’(x)=0
⇒2x+1=0
⇒ x= - �
At extreme points,
f(0)=0
f(1)=1+1+1>0
so,x=1 is a least value.
Option(C)
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