Option : (B)
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