Let $f(x) = \begin{cases} |x|, & -\infty < x < 2 \\ |2x-4|, & 2 \leq x \leq 20 \end{cases}$. If $x=a$ is a point where $f(x)$ is continuous but not differentiable and $x=b$ is a point where $f(x)$ is not differentiable $(a \neq b)$,then $a+b=$

If a function $f(x)$ defined by $f(x) = \begin{cases} ax^2 + bx + c, & x \leq -1 \\ 2x^2 + 4x + 1, & -1 < x < 1 \\ cx^2 + bx + a, & x \geq 1 \end{cases}$ is continuous on $\mathbb{R}$,and $\lim_{x \rightarrow \frac{3}{2}} f(x) = 14$,then find $\lim_{x \rightarrow -2} f(x)$.

Let $f(x) = \begin{cases} x^3+8; x < 0 \\ x^2-4; x \ge 0 \end{cases}$ and $g(x) = \begin{cases} (x-8)^{1/3}; x < 0 \\ (x+4)^{1/2}; x \ge 0 \end{cases}$. Then the number of points, where the function $g \circ f$ is discontinuous, is ————

Let $f(x) = \begin{cases} \frac{x^4-5x^2+4}{|(x-1)(x-2)|} & , x \neq 1,2 \\ 6 & , x=1 \\ 12 & , x=2 \end{cases}$. Then $f(x)$ is continuous on the set:

If the function $f$ defined on $\left(-\frac{1}{3}, \frac{1}{3}\right)$ by $f(x) = \begin{cases} \frac{1}{x} \log_{e}\left(\frac{1+3x}{1-2x}\right) & x \neq 0 \\ k & x = 0 \end{cases}$ is continuous,then $k$ is equal to

If the function defined by $f(x) = \begin{cases} (x^2 + e^{\frac{1}{2-x}})^{-1}, & x \neq 2 \\ k, & x = 2 \end{cases}$ is right continuous at $x = 2$, then $k =$