Let f(x) = Ax^3 - Bx - tan x cdot text{sgn}(x | vedclass

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(C) For $f(x)$ to be an even function,$f(x) = f(-x)$ must hold for all $x$ in the domain.Given $f(x) = Ax^3 - Bx - 	an x \cdot 	ext{sgn}(x)$.Since $

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$2$

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(C) For $f(x)$ to be an even function,$f(x) = f(-x)$ must hold for all $x$ in the domain.Given $f(x) = Ax^3 - Bx - 	an x \cdot 	ext{sgn}(x)$.Since $	an(-x) = -	an x$ and $	ext{sgn}(-x) = -	ext{sgn}(x)$,we have $	an(-x) \cdot 	ext{sgn}(-x) = (-	an x) \cdot (-	ext{sgn}(x)) = 	an x \cdot 	ext{sgn}(x)$.Thus,$f(-x) = A(-x)^3 - B(-x) - 	an x \cdot 	ext{sgn}(x) = -Ax^3 + Bx - 	an x \cdot 	ext{sgn}(x)$.For $f(x) = f(-x)$,we require $Ax^3 - Bx - 	an x \cdot 	ext{sgn}(x) = -Ax^3 + Bx - 	an x \cdot 	ext{sgn}(x)$.This simplifies to $2Ax^3 - 2Bx = 0$ for all $x$,which implies $A = 0$ and $B = 0$.$A = \sin^2 \alpha - \sin \alpha + \frac{1}{4} = (\sin \alpha - \frac{1}{2})^2 = 0 \Rightarrow \sin \alpha = \frac{1}{2}$.$B = 	an^2 \alpha + \frac{2}{\sqrt{3}} 	an \alpha + \frac{1}{3} = (	an \alpha + \frac{1}{\sqrt{3}})^2 = 0$ $\Rightarrow 	an \alpha = -\frac{1}{\sqrt{3}}$.We need $\alpha \in [-\frac{3\pi}{2}, 2\pi]$ such that $\sin \alpha = \frac{1}{2}$ and $	an \alpha = -\frac{1}{\sqrt{3}}$.$\sin \alpha = \frac{1}{2} \Rightarrow \alpha = \frac{\pi}{6}, \frac{5\pi}{6}, -\frac{7\pi}{6}, -\frac{11\pi}{6}$.$	an \alpha = -\frac{1}{\sqrt{3}} \Rightarrow \alpha = \frac{5\pi}{6}, -\frac{\pi}{6}, -\frac{7\pi}{6}, \frac{11\pi}{6}$.The common values are $\alpha = \frac{5\pi}{6}$ and $\alpha = -\frac{7\pi}{6}$.Thus,there are $2$ such values.

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