If $f(x) = \begin{cases} ax+b, & \text{if } x \leq 1 \\ ax^2+c, & \text{if } 1 < x \leq 2 \\ \frac{dx^2+1}{x}, & \text{if } x > 2 \end{cases}$ is differentiable on $\mathbb{R}$, then $ad-bc = $

If $f(x) = \begin{cases} \int_{0}^{x} (5 + |1-t|) \, dt, & x > 2 \\ 5x + 1, & x \leq 2 \end{cases}$,then:

Using the fact that $\sin (A+B)=\sin A \cos B+\cos A \sin B$ and the differentiation,obtain the sum formula for cosines.

Let $f(x) = x|x|$,$g(x) = \sin x$ and $h(x) = (g \circ f)(x)$. Then

At the point $x=1$,the function $f(x) = \begin{cases} x^3-1, & 1 < x < \infty \\ x-1, & -\infty < x \leq 1 \end{cases}$ is

Let $f_1: R \rightarrow R, f_2:\left(-\frac{\pi}{2}, \frac{\pi}{2}\right) \rightarrow R, f_3:\left(-1, e^{\frac{\pi}{2}}-2\right) \rightarrow R$ and $f_4: R \rightarrow R$ be functions defined by:
$(i)$ $f_1(x)=\sin \left(\sqrt{1-e^{-x^2}}\right)$
$(ii)$ $f_2(x)=\begin{cases} \frac{|\sin x|}{\tan^{-1} x} & \text{if } x \neq 0 \\ 1 & \text{if } x=0 \end{cases}$,where the inverse trigonometric function $\tan^{-1} x$ assumes values in $\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$.
$(iii)$ $f_3(x)=\left[\sin \left(\log_e(x+2)\right)\right]$,where,for $t \in R, [t]$ denotes the greatest integer less than or equal to $t$.
$(iv)$ $f_4(x)=\begin{cases} x^2 \sin \left(\frac{1}{x}\right) & \text{if } x \neq 0 \\ 0 & \text{if } x=0 \end{cases}$

$LIST-I$	$LIST-II$
$P$. The function $f_1$ is	$1$. $NOT$ continuous at $x=0$
$Q$. The function $f_2$ is	$2$. Continuous at $x=0$ and $NOT$ differentiable at $x=0$
$R$. The function $f_3$ is	$3$. Differentiable at $x=0$ and its derivative is $NOT$ continuous at $x=0$
$S$. The function $f_4$ is	$4$. Differentiable at $x=0$ and its derivative is continuous at $x=0$

The correct option is: