Match the items given in List-$I$ with those of the items of List-$II$:

List-$I$	List-$II$
$a$. If $y=|x|+|x-2|$, then at $x=2, \frac{dy}{dx}=$	$i$. $2$
$b$. If $f(x)=|\cos 2x|$, then $f^{\prime}(\frac{\pi}{4}+)=$	$ii$. $0$
$c$. If $f(x)=\sin(\pi[x])$, where $[\cdot]$ denotes the greatest integer function, then $f^{\prime}(1-)=$	$iii$. $-2$
$d$. If $f(x)=\log|x-1|, x \neq 1$, then $f^{\prime}(\frac{1}{2})=$	$iv$. Does not exist

Let $f : (0, \pi) \rightarrow \mathbb{R}$ be a twice differentiable function such that $\lim _{t \rightarrow x} \frac{f(x) \sin t - f(t) \sin x}{t-x} = \sin^2 x$ for all $x \in (0, \pi)$. If $f \left(\frac{\pi}{6}\right) = -\frac{\pi}{12}$,then which of the following statement$(s)$ is (are) $TRUE$?
$(A) f \left(\frac{\pi}{4}\right) = \frac{\pi}{4 \sqrt{2}}$
$(B) f(x) < \frac{x^4}{6} - x^2$ for all $x \in (0, \pi)$
$(C)$ There exists $\alpha \in (0, \pi)$ such that $f^{\prime}(\alpha) = 0$
$(D) f^{\prime \prime}\left(\frac{\pi}{2}\right) + f\left(\frac{\pi}{2}\right) = 0$

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 $f(x) = \begin{cases} \frac{5 e^{1/x} + 2}{3 - e^{1/x}}, & x \neq 0 \\ 0, & x = 0 \end{cases}$. Then at $x = 0$,$x f(x)$ and $f(x)$ are respectively:

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

Let $f(x) = ax^2 - b|x|$,where $a$ and $b$ are constants. Then at $x = 0$,$f(x)$ has