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Q.limx01cosxcos2xcos3xsin25x\lim_{x \to 0} \dfrac{1 - \cos x \cos 2x \cos 3x}{\sin^2 5x} equals

a
150\dfrac{1}{50}
b
125\dfrac{1}{25}
c
225\dfrac{2}{25}
d
110\dfrac{1}{10}

Correct Answer: Option B

The correct solution involves applying the fundamental concept to derive the final value step by step...

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