Question

If the derivative of the functions $f\left(x\right)=\left\{\begin{array}{cc}b{x}^2+ax+4;& x\ge -1\\ a{x}^2+b;& x<-1\end{array}\right\}$ is everywhere continuous then

• A. $a=2,b=3$
• B. $a=3,b=2$
• C. $a=-2,b=-3$
• D. $a=-3,b=-2$

Answer 1

The correct option is A $a=2,b=3$

We have $f\left(x\right)=\{\begin{array}{cc}a{x}^2+b,& x<-1\\ b{x}^2+ax+4,& x\ge -1\end{array}$

$\therefore f\left(x\right)=\{\begin{array}{c}2ax,x,x<-1\\ 2bx+a,x\ge -1\end{array}$

Since, $f\left(x\right)$ is differentiable at $x=-1$ therfore it is continuous at $x=-1$ and hence,

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

$\Rightarrow a+b=b-a+4$

$\Rightarrow a=2$

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

$\Rightarrow -2a--2b+a$

$\Rightarrow 3a=2b$

$\Rightarrow b=3$ (Q $a=2$)

Hence, $a=2,b=3$.