Let $f, g: R \rightarrow R$ be two real-valued functions defined as $f(x)=\begin{cases} -|x+3| & , x < 0 \\ e^{x} & , x \geq 0 \end{cases}$ and $g(x)=\begin{cases} x^{2}+k_{1} x & , x < 0 \\ 4 x+k_{2} & , x \geq 0 \end{cases}$,where $k_{1}$ and $k_{2}$ are real constants. If $(g \circ f)$ is differentiable at $x=0$,then $(g \circ f)(-4)+(g \circ f)(4)$ is equal to

Which of the following statements is true for the function $f(x) = \begin{cases} \sqrt{x} & x \ge 1 \\ x^3 & 0 \le x < 1 \\ \frac{x^3}{3} - 4x & x < 0 \end{cases}$

Give the correct order of initials $T$ or $F$ for the following statements. Use $T$ if the statement is true and $F$ if it is false.
Statement-$1$: If $f: R \rightarrow R$ and $c \in R$ is such that $f$ is increasing in $(c - \delta, c)$ and $f$ is decreasing in $(c, c + \delta)$,then $f$ has a local maximum at $c$. Where $\delta$ is a sufficiently small positive quantity.
Statement-$2$: Let $f: (a, b) \rightarrow R, c \in (a, b)$. Then $f$ cannot have both a local maximum and a point of inflection at $x = c$.
Statement-$3$: The function $f(x) = x^2 |x|$ is twice differentiable at $x = 0$.
Statement-$4$: Let $f: [c - 1, c + 1] \rightarrow [a, b]$ be a bijective map such that $f$ is differentiable at $c$ and $f'(c) \neq 0$,then $f^{-1}$ is also differentiable at $f(c)$.

Let $f:\left[-\frac{1}{2}, 2\right] \rightarrow R$ and $g:\left[-\frac{1}{2}, 2\right] \rightarrow R$ be functions defined by $f(x)=\left[x^2-3\right]$ and $g(x)=|x| f(x)+|4 x-7| f(x)$,where $[y]$ denotes the greatest integer less than or equal to $y$ for $y \in R$. Then
$(A)$ $f$ is discontinuous exactly at three points in $\left[-\frac{1}{2}, 2\right]$
$(B)$ $f$ is discontinuous exactly at four points in $\left[-\frac{1}{2}, 2\right]$
$(C)$ $g$ is $NOT$ differentiable exactly at four points in $\left(-\frac{1}{2}, 2\right)$
$(D)$ $g$ is $NOT$ differentiable exactly at five points in $\left(-\frac{1}{2}, 2\right)$

Let $a_1, a_2, \ldots, a_{100}$ be non-zero real numbers such that $a_1+a_2+\ldots+a_{100}=0$. Then,

If $f(x) = \cos x \cos 2x \cos 4x \cos 8x \cos 16x$,then $f'\left( \frac{\pi}{4} \right)$ is