Q6 of 137 Page 19

#Mark the correct alternative in each of the following

The minimum value of f(x) = x4 – x2 – 2x + 6 is


f(x)=x4-x2-2x+6


f’(x)=4x3-2x-2


so here put f’(x)=0


4x3-2x-2=0


2(x-1)(2x2+2x+1)=0


x=1 and other roots are complex.


so we will consider x=1


Hence by second derivative test


f’’(x)>0 so it’s a point of minimum.


f”(x)=12x2-1


f” (1) =12(1)-2


=10>0


At x=1 minimum


So, f (1) =1-1-2(1) +6


=4


Option(B)

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