Let $f: R \rightarrow R$ be defined as:
$f(x) = \begin{cases} \frac{\lambda|x^{2}-5x+6|}{\mu(5x-x^{2}-6)}, & x < 2 \\ \mu, & x = 2 \\ e^{\frac{\tan(x-2)}{x-[x]}}, & x > 2 \end{cases}$
Where $[x]$ is the greatest integer less than or equal to $x$. If $f$ is continuous at $x = 2$,then $\lambda + \mu$ is equal to:

Let $f(x) = \min \{1, 1 + x \sin x \}$ for $0 \leq x \leq 2\pi$. If $m$ is the number of points where $f$ is not differentiable and $n$ is the number of points where $f$ is not continuous,then the ordered pair $(m, n)$ is equal to

Let $f$ and $g$ be real-valued functions. If $\lim _{x \rightarrow 0} \frac{2 f(x)-g(x)}{[f(x)+7]^{2 / 3}}=\frac{7}{4}$,$\lim _{x \rightarrow 0} f(x)=1$ and $\lim _{x \rightarrow 0} g(x)=\alpha$,then $h(x)= \begin{cases} \sin (\alpha x), & 0 \leq x \leq \frac{\pi}{10} \\ \cos (2 \alpha x), & \frac{\pi}{10} < x \leq \frac{\pi}{5} \end{cases}$ is:

Find all points of discontinuity of $f$,where $f$ is defined by $f(x) = \begin{cases} x^{10} - 1, & \text{if } x \le 1 \\ x^2, & \text{if } x > 1 \end{cases}$.

If $f(x) = \begin{cases} x, & x \le 0 \\ 0, & x > 0 \end{cases}$ then the function $f(x)$ at $x = 0$ is:

If $[x]$ denotes the greatest integer not exceeding the number $x$,then $f(x)$ defined by $f(x) = \begin{cases} [x], & \text{if } x < 2 \\ [x]-1, & \text{if } x \geq 2 \end{cases}$ is continuous in the interval.