Maxima and MinimaMediumBloom L3
Question
The set of values of p for which the points of extremum of the function, f (x) = x3 -3px2 + 3(p2 -1)x +1 lie in the interval (-2, 4) is -
Options
A.(-3, 5)
B.(-3, 3)
C.(-1, 3)
D.(-1, 5)
Solution
f(x) = x3 -3px2 + 3(p2 -1)x +1
f′(x) = 3{x2 - 2px + (p -1) (p +1)}
f′(x) = 3{x - (p -1)}{x - (p +1)}
-2 < p -1< 4 and -2 < p +1 < 4
p ∈ (-1, 3)
f′(x) = 3{x2 - 2px + (p -1) (p +1)}
f′(x) = 3{x - (p -1)}{x - (p +1)}
-2 < p -1< 4 and -2 < p +1 < 4
p ∈ (-1, 3)
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