Question

Let $a,b,c\in R$ such that $a+b+c=\pi$. If $f\left(x\right)=\{\begin{array}{cc}\frac{\sin (a{x}^2+bx+c)}{x^2-1},& \text{if}x<1\\ -1,& \text{if}x=1\\ a\text{sgn}(x+1)\cos (2x-2)+b{x}^2,& \text{if}1<x\le 2\end{array}$ is continuous at $x=1$, then the value of $(a^2+b^2)$ is
[Here, $\text{sgn}\left(k\right)$ denotes signum function of $k$]

Answer 1

$f\left(1^{-}\right)=\underset{x\to 1^{-}}{lim}\frac{\sin (a{x}^2+bx+c)}{(x^2-1)}$
Applying L'Hospital Rule
$f\left(1^{-}\right)=\underset{x\to 1^{-}}{lim}\frac{\cos (a{x}^2+bx+c)\times (2ax+b)}{2x}$
$=\frac{-(2a+b)}{2}$ as $a+b+c=\pi$

$f\left(1^{+}\right)=\underset{x\to 1^{+}}{lim}a\text{sgn}(x+1)\cos (2x-2)+b{x}^2=a+b$
And $f\left(1\right)=-1$

Since, $f\left(x\right)$ is continuous at $x=1,$
$\therefore f\left(1^{-}\right)=f\left(1^{+}\right)=f\left(1\right)$
$\Rightarrow \frac{-(2a+b)}{2}=a+b=-1$
$\Rightarrow a=3,b=-4$
$\therefore a^2+b^2=(3{)}^2+(-4{)}^2=25$