If the Rolle’s theorem is applicable to the function f defined by  $f\left(x\right)=\left\{\begin{array}{ll}a{x}^2+b,& \left|x\right|\le 1\\ 1,& \left|x\right|=1\\ \frac{c}{\left|x\right|},& \left|x\right|>1\end{array}\right.$  In the interval $\left[-3,3\right]$ then which of the following alternative(S) is/are correct?

As rolles theorem is applicable the function should be continuous and differentiable in $\left[-3,3\right]$

So at $x=1$ it is continuous

$\underset{x\to 1^{-}}{\lim }f\left(x\right)=\underset{x\to 1^{+}}{\lim }f\left(x\right)-f\left(1\right)$

$a+b=c=1$  ….(1)

Differentiable at   $x=1$

$f^1\left(1^{+}\right)=f^1\left(1^{-}\right)$

$\underset{h\to 0}{\lim }\frac{\frac{c}{1+h}-1}{h}=2a\Rightarrow 2a=\underset{h\to 0}{\lim }\frac{c-1-h}{h\left(h+1\right)}$ exists only when $c=1$

$2a=\underset{h\to 0}{\lim }\frac{-h}{h\left(h+1\right)}=\underset{h\to 0}{\lim }\frac{-1}{h+1}=-1$

$a=\frac{-1}{2}$  and    $c=1$  …(2)

From (1), (2) $b=\frac{3}{2}$

$a+b+c=2\Rightarrow \left|a\right|+\left|b\right|+\left|c\right|=3$

$2a+4b+3c=8$$\Rightarrow 4a^2+4b^2+5c^2=15$