Question

If the derivative of the function $f\left(x\right)=b{x}^2+ax+4;x\ge -1a{x}^2+b;x<-1^{{}^{\prime }}$, 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

$Wehave,f\left(x\right)=\{\begin{array}{cc}a{x}^2+b,& x<-1\\ b{x}^2+ax+4,& x\ge -1\end{array}\therefore f^{{}^{\prime }}\left(x\right)=\{\begin{array}{cc}2ax,& <-1\\ 2bx+a,& x\ge -1\end{array}Since,f\left(x\right)isdifferentiableatx=-1,thereforeitiscontinuousatx=-1andhence,\underset{x\to -1-}{lim}f\left(x\right)=\underset{x\to -1+}{lim}f\left(x\right)\Rightarrow a+b=b-a+4\Rightarrow a=2andalso,\underset{x\to -1-}{lim}f\left(x\right)=\underset{x\to -1+}{lim}f\left(x\right)\Rightarrow -2a=-2b+a\Rightarrow 3a=2b\Rightarrow b=3(\because a=2)Hence,a=2,b=3$