DeterminantHard

Question

Let f(x) = , then

Options

A.
B.maximum value of f(x) is 4
C.
D.f′(0) = 0

Solution

f(x) =
f(x) = 8cos3x - 2cosx - 2cosx = 2cos3x + 2cosx
f′(x) = - 6sin3x - 2sinx = 0
∴ f′(0) = 0 (D)
f″(x) = - 18cos3x - 2cosx < 0, at x = 0
∴ f(0)max = 2 + 2 = 4
(2 cos 3x + 2 cos x) dx = 0

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