Continuity practice · Mixed difficulty

Option A (continuous everywhere but not differentiable at $x=0$) is correct because the absolute value function introduces a sharp corner at $x=0$, making it non-differentiable there.

Option D (1) is correct because the limit of the function as x approaches 0 is 1, which is the value needed to make the function continuous.

Option B (Statement I is true but statement II is false) is correct because f(x) is differentiable for all real x, while g(x) is not thrice differentiable.

The correct answer is B ($f$ is not continuous at $x=1$) because the function has different limits as x approaches 1 from the left and right, and the value at x=1 is different from both limits.

Option C (Only three points) is correct because the function is discontinuous at x = 0, x = 1, and x = 3 due to jumps in the function's definition.

Option C ($\log a - \log b$) is correct because by applying L'Hopital's rule or using the definition of the derivative, the limit as x approaches 0 is $\log a - \log b$.

Option C ($a=2, b=-1$ and $f$ is not differentiable at $x=-1,1$) is correct because the function must be continuous at x = -1 and x = 1, and the derivatives do not match at these points.

Option A ($f$ is continuous at every point in $[0, \infty)$ and differentiable except at the integer points) is correct because the integral of the greatest integer function is continuous everywhere but not differentiable at integer points.

Option B ($f(x)$ is continuous, $\forall x \in R$) is correct because the given functional equation and continuity at x=0 imply continuity everywhere.

Option A (continuous on (-1,0)) is correct because the function f(x) simplifies to a constant value of 1 on the interval (-1,0), making it continuous.

Option A ($f^{\prime}$ and $f^{\prime \prime}$ are also periodic) is correct because the derivative of a periodic function is also periodic with the same period.

Option D ($f$ is continuous for all $x \neq -1$) is correct because by simplifying the expression for $f_n(x)$ and taking the limit as n approaches infinity, the resulting function $f(x)$ is continuous for all $x \neq -1$.