List of top Mathematics Questions on Continuity asked in CBSE CLASS XII

If \[ f(x) = \begin{cases} \frac{\sin^2(ax)}{x^2}, & x \neq 0 \\ 1, & x = 0 \end{cases} \] is continuous at \( x = 0 \), then the value of \( a \) is:
Assertion (A): $f(x) = \begin{cases} x \sin \frac{1}{x}, & x \neq 0 \\ 0, & x = 0 \end{cases}$ is continuous at $x = 0$.
Reason (R): When $x \to 0$, $\sin \frac{1}{x}$ is a finite value between -1 and 1.
If $f(x) = \begin{cases} ax + b, & x \leq 3 \\ 7, & 5<x \end{cases}$ is continuous in $\mathbb{R}$, then the values of $a$ and $b$ are:
The function $f$ defined by \[ f(x) = \begin{cases} x, & \text{if } x \leq 1 \\ 5, & \text{if } x>1 \end{cases} \] is not continuous at :
If $f(x) = \left\{ \begin{array}{ll} \frac{1 - \sin^3 x}{3 \cos^2 x} & \text{for} \, x \neq \frac{\pi}{2}, \\ k & \text{for} \, x = \frac{\pi}{2}, \end{array} \right. $ is continuous at $x = \frac{\pi}{2}$, then the value of $k$ is:
Find $k$ so that \[ f(x) = \begin{cases} \frac{x^2 - 2x - 3}{x + 1}, & \text{if } x \neq -1 \\ k, & \text{if } x = -1 \end{cases} \] is continuous at $x = -1$.
The value of constant \( c \) that makes the function \( f \) defined by \(f(x) = \ x^2 - c^2\), \(\&\ \text{if } x<4\)\(cx + 20, \&\ \text{if } x \geq 4\)   continuous for all real numbers is: