(1, 2)
(1, 1)
(1, -1)
(1, -2)
Correct answer is C
y=4x3+kx2−6x+4
dydx=12x2+2kx−6
At P(1, m)
dydx=12+2k−6=0 (parallel to the x- axis)
6+2k=0⟹k=−3
\(P(1, m) \implies m = 4(1^{3}) - 3(1^{2}) - 6(1) + 4)
= -1
P = (1, -1)
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