Question

If the derivative of the function $f\left(x\right)=\{\begin{array}{cc}b{x}^2+ax+4;x\ge -1& \\ a{x}^2+b;x<-1^{\prime }& \end{array}$ is continuous everywhere. 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, $\{\begin{array}{cc}a{x}^2+b,x<-1& \\ b{x}^2+ax+4,x\ge -1& \end{array}$
$f^{\prime }\left(x\right)=\{\begin{array}{cc}2ax,x<-1& \\ 2bx+a,x\ge -1& \end{array}$
Since, f(x) is differentiable at x=-1, therefore it is continuous at x = -1 and hence,
${lim}_{x\to -1^{-}}f\left(x\right)={lim}_{x\to -1^{+}}f\left(x\right)\Rightarrow a+b=b-a+4\Rightarrow a=2$
For differentiability at x = -1,
${lim}_{x\to -1^{-}}{f}^{\prime }\left(x\right)={lim}_{x\to -1^{+}}{f}^{\prime }\left(x\right)\Rightarrow -2a=-2b+a\Rightarrow 3a=2b\Rightarrow b=3$
a =2, b = 3