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A periodic signal x(t) has a trigonometric Fourier series expansion $$x\left( t \right) = {a_0} + \sum\limits_{n = 1}^\infty {\left( {{a_n}\cos n{\omega _0}t + {b_n}\sin n{\omega _0}t} \right)} $$
If $$x\left( t \right) = - x\left( { - t} \right) = - x\left( {t - \frac{\pi }{{{\omega _0}}}} \right),$$      we can conclude that

Answer & Solution
Correct Answer: Option A
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