The number of points,where the curve $f(x) = e^{8x} - e^{6x} - 3e^{4x} - e^{2x} + 1$,$x \in R$ cuts the $x$-axis,is equal to

The function $f(x) = |x|$ at $x = 0$ is

Let $f: ( -\infty, \infty ) \to ( -\infty, \infty )$ be defined by $f(x) = x^3 + 1$.
Statement $1$: The function $f$ has a local extremum at $x = 0$.
Statement $2$: The function $f$ is continuous and differentiable on $( -\infty, \infty )$ and $f'(0) = 0$.

Let $f(x)=x^2+a x+b$,where $a, b \in R$. If $f(x)=0$ has all its roots imaginary,then the roots of $f(x)+f^{\prime}(x)+f^{\prime \prime}(x)=0$ are

Let $f: R \rightarrow R$ be given by
$f(x) = \begin{cases} x^5+5x^4+10x^3+10x^2+3x+1, & x < 0 \\ x^2-x+1, & 0 \leq x < 1 \\ \frac{2}{3}x^3-4x^2+7x-\frac{8}{3}, & 1 \leq x < 3 \\ (x-2)\log_e(x-2)-x+\frac{10}{3}, & x \geq 3 \end{cases}$
Then which of the following options is/are correct?
$(1)$ $f^{\prime}$ has a local maximum at $x = 1$
$(2)$ $f$ is onto
$(3)$ $f$ is increasing on $(-\infty, 0)$
$(4)$ $f^{\prime}$ is $NOT$ differentiable at $x = 1$

Let $f: R \rightarrow R$ be a function defined as $f(x) = \begin{cases} 3(1 - \frac{|x|}{2}) & \text{if } |x| \leq 2 \\ 0 & \text{if } |x| > 2 \end{cases}$. Let $g: R \rightarrow R$ be given by $g(x) = f(x+2) - f(x-2)$. If $n$ and $m$ denote the number of points in $R$ where $g$ is not continuous and not differentiable,respectively,then $n+m$ is equal to $....$