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 $\frac{a^2+b^2}{5}$ is
( Here, $\text{sgn}\left(k\right)$ denotes signum function of $k$ )

• A. $3$
• B. $5$
• C. $12$
• D. $8$

Answer 1

The correct option is B $5$

$f\left(1^{+}\right)=a+b;f\left(1\right)=-1$
$f\left(1^{-}\right)=\underset{x\to 1^{-}}{lim}\frac{\sin (a{x}^2+bx+c)}{x^2-1}\left[\frac{0}{0}\right]\text{form}$
Applying L' Hopitals' rule, we get
$f\left(1^{-}\right)=\underset{x\to 1^{-}}{lim}\frac{(2ax+b)\cos (a{x}^2+bx+c)}{2x}$
$=\frac{(2a+b)\cos (a+b+c)}{2}$
$=-\left(\frac{2a+b}{2}\right)$

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