If ⨍ x=19cos x1012cos x1012cos x, then d2⨍ dx2=

f(x)=19cos x1012cos x1012cos x =19cos x4 cos2x-1-1(2 cos x) f(x)=194 cos3x-3 cos x =19cos 3x f1x=19-3 sin 3x f11(x)=19-9 cos 3x =-cos 3x =cos π+3x

If ⨍ x=19cos x1012cos x1012cos x, then d2⨍ dx2=

f(x)=19cos x1012cos x1012cos x =19cos x4 cos2x-1-1(2 cos x) f(x)=194 cos3x-3 cos x =19cos 3x f1x=19-3 sin 3x f11(x)=19-9 cos 3x =-cos 3x =cos π+3x

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If $⨍ (x) = \frac{1}{9} |\begin{matrix} \cos x & 1 & 0 \\ 1 & 2 \cos x & 1 \\ 0 & 1 & 2 \cos x \end{matrix}|,$ then $\frac{d^{2} ⨍}{{dx}^{2}} =$

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$\cos 3 x$

$\cos (π + 3 x)$

$\sin 3 x$

$\sin (π + 3 x)$

answer is B.

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Detailed Solution

$f (x) = \frac{1}{9} |\begin{matrix} \cos x & 1 & 0 \\ 1 & 2 \cos x & 1 \\ 0 & 1 & 2 \cos x \end{matrix}| = \frac{1}{9} [\cos x (4 \cos^{2} x - 1) - 1 (2 \cos x)] f (x) = \frac{1}{9} [4 \cos^{3} x - 3 \cos x] = \frac{1}{9} (\cos 3 x) f^{1} (x) = \frac{1}{9} (- 3 \sin 3 x) f^{11} (x) = \frac{1}{9} (- 9 \cos 3 x) = - \cos 3 x = \cos (π + 3 x)$