Explanation:

- Option 1: Continuous at \( x = \frac{1}{2} \)

- At \( x = \frac{1}{2} \), \( |x| = \frac{1}{2} \) and \([x] = 0\).

- \( f(x) = \cos(p(\frac{1}{2} + 0)) = \cos(\frac{p}{2}) \), which is continuous.

- Option 2: Continuous at \( x = 0 \)

- At \( x = 0 \), \( |x| = 0 \) and \([x] = 0\).

- \( f(x) = \cos(p(0 + 0)) = \cos(0) = 1 \).

- Check limits: \( x \to 0^- \) and \( x \to 0^+ \) give different integer values, hence it is not continuous.

- Option 3: Differentiable in \((-1,0)\)

- \([x]\) is constant in \((-1,0)\), hence \( f(x) \) is cos applied to linear, making it differentiable.

- Option 4: Differentiable in \((0,1)\)

- \([x]\) is constant in \((0,1)\), meaning \( f(x) \) is cos applied to linear, therefore differentiable.

Correct Answer: Option 2: Continuous at \( x = 0 \) is incorrect.

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