Let $f(x) = \begin{cases} 3-x & \text{if } x < -3 \\ 6 & \text{if } -3 \leq x \leq 3 \\ 3+x & \text{if } x > 3 \end{cases}$. Let $\alpha$ be the number of points of discontinuity of $f$ and $\beta$ be the number of points where $f$ is not differentiable. Then $\alpha+\beta=$

$f(x) = \begin{cases} \frac{\sin(x-[x])}{x-[x]} & , x \in (-2, -1) \\ \max \{2x, 3[|x|]\} & , |x| < 1 \\ 1 & , \text{otherwise} \end{cases}$ where $[t]$ denotes the greatest integer $\leq t$. If $m$ is the number of points where $f$ is not continuous and $n$ is the number of points where $f$ is not differentiable,then the ordered pair $(m, n)$ is

$f(x)$ is a differentiable function and given $f^{\prime}(2)=6$ and $f^{\prime}(1)=4$, then $L=\lim _{h \rightarrow 0} \frac{f\left(2+2 h+h^2\right)-f(2)}{f\left(1+h-h^2\right)-f(1)}$

If $y = \frac{\tan x \cos^{-1} x}{\sqrt{1-x^2}}$, then the value of $\frac{dy}{dx}$ when $x = 0$ is:

$(i)$ $f(x)$ is continuous and defined for all real numbers.
$(ii)$ $f'(-5) = 0$; $f'(2)$ is not defined and $f'(4) = 0$.
$(iii)$ $(-5, 12)$ is a point which lies on the graph of $f(x)$.
$(iv)$ $f''(2)$ is undefined,but $f''(x)$ is negative everywhere else.
$(v)$ The signs of $f'(x)$ are given below:
$f'(x)$ sign chart:
- For $x < -5$,$f'(x) > 0$
- For $-5 < x < 2$,$f'(x) < 0$
- For $2 < x < 4$,$f'(x) > 0$
- For $x > 4$,$f'(x) < 0$
From the possible graph of $y = f(x)$,we can say that:

Let $h$ be a twice continuously differentiable positive function on an open interval $J.$ Let $g(x) = \ln(h(x))$ for each $x \in J$. Suppose $(h'(x))^2 > h''(x) h(x)$ for each $x \in J$. Then