Let $f (x) = |\begin{matrix} 1 + sinx & \cos (x + \frac{π}{6}) & \cos (x + \frac{π}{3}) \\ \sin 2 x & \cos (2 x + \frac{π}{6}) & \cos (2 x + \frac{π}{3}) \\ \sin 3 x & \cos (3 x + \frac{π}{6}) & \cos (3 x + \frac{π}{3}) \end{matrix}|$ and $\int f (x) dx = F (x) + C,$ where F is a differentiable function in R such that F(0) = 0 and C is the constant of integration, then which of the following statements is (are) correct?

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Range of F is $[- \frac{1}{2}, \frac{1}{2}]$

Maximum value of $f (x) + F (x)$ is $\frac{\sqrt{2} - 1}{2}$

$F (\frac{π}{3}) = - \frac{1}{4}$

Number of solutions of F(x) = x is 3.

answer is B, C.

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Detailed Solution

$f (x) = |\begin{matrix} sinx & cosx & 1 \\ \sin 2 x & \cos 2 x & 0 \\ \sin 3 x & \cos 3 x & 0 \end{matrix}| \times |\begin{matrix} 1 & - \sin \frac{π}{6} & - \sin \frac{π}{3} \\ 0 & \cos \frac{π}{6} & \cos \frac{π}{3} \\ 1 & 0 & 0 \end{matrix}|$
$= (\sin 2 x . \cos 3 x - \cos 2 xsin 3 x) \times (- \sin \frac{π}{6} \cos \frac{π}{3} + \cos \frac{π}{6} \sin \frac{π}{3})$
$= (- sinx) \times \sin \frac{π}{6} = - \frac{1}{2} sinx$