If $f(x) = \begin{cases} \frac{1 - \sin x}{\pi - 2x}, & x \neq \frac{\pi}{2} \\ \lambda, & x = \frac{\pi}{2} \end{cases}$ is continuous at $x = \frac{\pi}{2}$,then the value of $\lambda$ is:

Let $f(x) = \begin{cases} \frac{ax^{2}+2ax+3}{4x^{2}+4x-3}, & x \neq -\frac{3}{2}, \frac{1}{2} \\ b, & x = -\frac{3}{2}, \frac{1}{2} \end{cases}$ be continuous at $x=-\frac{3}{2}$. If $f(f(x)) = \frac{7}{5}$, then $x$ is equal to:

Let $f(x) = \begin{cases} \frac{x - 4}{|x - 4|} + a, & x < 4 \\ a + b, & x = 4 \\ \frac{x - 4}{|x - 4|} + b, & x > 4 \end{cases}$. Then $f(x)$ is continuous at $x = 4$ when

If a real-valued function $f(x) = \begin{cases} \frac{2x^2+(k+2)x+9}{3x^2-7x-6} & , \text{for } x \neq 3 \\ l & , \text{for } x=3 \end{cases}$ is continuous at $x=3$ and $l$ is a finite value,then $l-k=$

If $f(x) = \begin{cases} -x^3 + 1, & \text{if } -\infty < x \leq 1 \\ |x - 1| + \lambda, & \text{if } x > 1 \end{cases}$,then:

If $f(x)$ is continuous and $f\left( \frac{9}{2} \right) = \frac{2}{9}$, then $\lim_{x \to 0} f \left( \frac{1 - \cos 3x}{x^2} \right)$ is equal to: