In order that the function $f(x) = (x + 1)^{\cot x}$ is continuous at $x = 0$,$f(0)$ must be defined as

Let $a$ be a positive real number. If a real valued function $f(x) = \begin{cases} \frac{6^x-3^x-2^x+1}{1-\cos \left(\frac{x}{a}\right)} & \text{if } x \neq 0 \\ \log 3 \log 4 & \text{if } x=0 \end{cases}$ is continuous at $x=0$,then $a=$

If function $f$ is continuous at point $x = \pi$ and $f(x) = \begin{cases} kx+1; & x \leq \pi \\ \cos x; & x > \pi \end{cases}$ then the value of $k$ is $\dots \dots \dots$

If $f(x)$ is continuous at $x = 3$ where $f(x) = \begin{cases} ax + 1, & \text{for } x \leq 3 \\ bx + 3, & \text{for } x > 3 \end{cases}$,then

Let $[x]$ denote the greatest integer function and $f(x) = \max\{1+x+[x], 2+x, x+2[x]\}$ for $0 \leq x \leq 2$. Let $m$ be the number of points in $[0, 2]$ where $f$ is not continuous,and $n$ be the number of points in $(0, 2)$ where $f$ is not differentiable. Then $(m+n)^2+2$ is equal to:

Let $f(x)$ be a real-valued function. If $f^{\prime}(x)$ is a constant for all $x \in R$,$f(0)=2$,and $f^{\prime}(0)=1$,then