Continuity practice · Easy

Option D ($e$) is correct because $f(x) = \frac{e(e^{x-1}-1)}{x-1}$, so $f(1) = e$.

Option D (None of these) is correct because the left-hand limit at x=0 does not equal the right-hand limit, regardless of the value of k.

The correct answer is B ($f\left( x \right) + g\left( x \right)$ must be discontinuous) because the sum of a continuous function and a discontinuous function is always discontinuous.

Option B (-5) is correct because for the function to be continuous at x=0, the limit as x approaches 0 must equal f(0), leading to α = -5.

Option A ($e(-e+1)$) is correct because, for continuity at x=2, the left-hand limit, right-hand limit, and the function's value at x=2 must be equal, leading to $\lambda = -e^2$ and $\mu = e$.

Option A (1) is correct because the limit of the given function as x approaches 0 must equal α for the function to be continuous at x = 0, which evaluates to 1.

The correct answer is B (Discontinuous) because the left-hand limit and right-hand limit at $x = \frac{3}{2}$ are not equal.

Option D (None of these) is correct because the limit as x approaches 0 of the given function is 7/64, which is not any of the other options.

Option D ($\log 4$) is correct because the limit of the function as x approaches 0, calculated using L'Hopital's rule, is equal to $\log 4$, which is the value of f(0) for continuity.

Option B ($\frac{1}{4}$) is correct because the right-hand limit of the function as x approaches 2 is equal to $\frac{1}{4}$, and for the function to be continuous from the right, f(2) must equal this limit.

Option D ($\frac{1}{4}$) is correct because it is the limit of the given function as x approaches $\frac{\pi}{4}$, ensuring continuity.

The correct answer is 'If $f$ is discontinuous at exactly one point, then $a + b + c \ne 1$' because continuity conditions lead to constraints on a, b, and c, and the function cannot be continuous at all points.

Option A ($\frac{\pi}{5}$) is correct because for f(x) to be continuous at x=0, the limit of f(x) as x approaches 0 must equal f(0), which leads to k = $\frac{\pi}{5}$.

Option C is correct because the limit of the function as x approaches 0 is equal to the function's value at x=0, satisfying the condition for continuity.

Option D ($\frac{n(n+1)}{2}$) is correct because by using the squeeze theorem and properties of the greatest integer function, the limit of $f(x)$ as $x$ approaches 0 is found to be $\frac{n(n+1)}{2}$, which must equal $k$ for continuity.

Option C ($3a^2$) is correct because for the function to be continuous at x=a, the limit of the function as x approaches a must equal b.

Option A (2:1) is correct because the length of the subtangent is twice the abscissa for the given parabola.

Option D (80) is correct because the limit of the function as x approaches 2 is 80, and for continuity, f(2) must equal this limit.

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.

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 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 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.