The function $f(x) = \sqrt{1 - \sqrt{1 - x^2}}$

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

Consider the following three statements for the function $f : (0, \infty) \rightarrow \mathbb{R}$ defined by $f(x) = |\log_{e} x| - |x - 1|$:
$(I)$ $f$ is differentiable at all $x > 0$.
$(II)$ $f$ is increasing in $(0, 1)$.
$(III)$ $f$ is decreasing in $(1, \infty)$.
Then:

$A$ function $f(x) = \begin{cases} 1 + x, & x \le 2 \\ 5 - x, & x > 2 \end{cases}$ 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:

Let $f(x) = \begin{cases} \frac{1}{|x|}, & |x| \geqslant 1 \\ ax^2 + b, & |x| < 1 \end{cases}$ be continuous and differentiable everywhere. Then $a$ and $b$ are