Continuity practice · Mixed difficulty

Option C ($f(x)$ is continuous but not differentiable at $x = 2$) is correct because the function is continuous at x=2 but the left-hand derivative and right-hand derivative are not equal at that point.

Option D ((1,1)) is correct because for f(x) to be continuous at x=1, the left-hand limit must equal the right-hand limit and the function value at x=1. When a=1 and b=1, the left-hand limit is 0, which is not equal to 2, so the function is discontinuous.

Option A (Statement 1 is true; Statement 2 is false) is correct because f is discontinuous at x=0, 1, 2, and 3, but the provided piecewise function in Statement 2 is continuous at x=3.

Option B (4) is correct. By evaluating the limit using L'Hopital's rule or standard limit formulas and comparing it with the given value of f(0), we can find the value of p.

Option A ($c = 0, a = 2b$) is correct because it satisfies both the continuity and differentiability conditions at x = 1.

Option C ($f(x)$ is discontinuous at $x=0$) is correct because the right-hand limit at x=0 is infinity, which is not equal to the function's value at x=0.

Option C ($n = m\frac{\pi }{2}$) is correct because for the function to be continuous at $x = \frac{\pi}{2}$, the left-hand limit must equal the right-hand limit.

Option A is correct because the function is discontinuous at the roots of the denominator, which are 1, -1, and -3.

Option B (1) is correct because it is the limit of the given function as x approaches $\frac{\pi}{2}$, ensuring continuity.

Option D ($\ln \left(\frac{2}{\mathrm{e}}\right)$) is correct because it is the limit of the function as x approaches 0, making the function continuous.

The correct answer is C ($\left( {2n + 1} \right)\frac{\pi }{{2}},n \in I$) because the tangent function is undefined at odd multiples of $\frac{\pi}{2}$.

Option B ($\phi$) is correct because the function is discontinuous everywhere, as the limit does not exist at any point.

Option B (-$\frac{1}{2}$) is correct because it is the value of p that makes the function continuous at x=0.

Option A ($\left[ { - 1,1} \right]$) is correct because the function is continuous on the closed interval from -1 to 1.

Option D (1) is correct because for the function to be continuous at x=3, the left-hand limit must equal the function's value at x=3.

Option C ({1 / 2,1,2}) is correct because the composite function (g o f)(x) is discontinuous at x = 1/2, 1, and 2, considering the discontinuity of f(x) and the points where g(f(x)) is undefined.

Option D (1/4) is correct because the limit of the function as x approaches pi/4 is found using L'Hopital's rule, and for continuity, this limit must equal a.

Option B ($f$ is continuous at x=4) is correct because the left-hand limit, right-hand limit, and the function value at x=4 are all equal.

Option C (e) is correct because the limit of (x+1)^{(cot x)} as x approaches 0 is e, which is the value of f(0) for continuity.

Option C (1) is correct because for the function to be continuous at x=3, the left-hand limit must equal the function's value at x=3.

Option B (Not continuous $x = - 2$) is correct because the left-hand limit and right-hand limit of the function at $x = -2$ are not equal, indicating a discontinuity.

Option B is correct because the limit of the function as x approaches 0 is a², which is not equal to f(0) = 1 when a ≠ ±1, indicating a discontinuity.

Option B (f is not continuous at $x=0$ for any real $k$) is correct because the limit as $x$ approaches 0 does not exist.

Option B (-1) is correct because the continuity of the function at x=1 and x=-1 leads to the equation a+b=-1.

Option C ($f(x)$ is discontinuous at $x = 0$) is correct because the left-hand limit and right-hand limit are not equal at $x = 0$.

Option D (103) is correct because the greatest integer function is discontinuous at every integer value, and there are 103 integers within the specified interval.

Option C (Discontinuous at $x = 1/2$) is correct because the function has a jump discontinuity at that point, as evident from the piecewise definition.

Option B (1) is correct because for the function to be continuous at x=0, the limit as x approaches 0 must equal f(0), which means k=1.

Option B ($f(x)$ is discontinuous at $x=a$) is correct because the left-hand limit and right-hand limit are not equal at $x=a$.

Option B ($\phi(x_0)$) is correct because it directly results from the definition of the derivative and the continuity of $\phi(x)$ at $x_0$.