question_answer

If the derivative of the function \[f(x)\] is everywhere continuous and is given by \[f(x)=\left\{ \begin{matrix}    b{{x}^{2}}+ax+4; & x\ge -l  \\    a{{x}^{2}}+b; & x<-l  \\ \end{matrix} \right.\] , then

A)  \[a=2,b=3\]      

B)  \[a=3,b=2\]

C)  \[a=-2,b=-3\]                   

D)  \[a=-3,b=-2\]

Correct Answer: A

Solution :

Here, \[f'(x)\,=\left\{ \begin{matrix}    2bx+a & x\ge -1  \\    2ax, & x<-1  \\ \end{matrix} \right.\]             Given, f?(x) is continuous everywhere,             \[\Rightarrow \,3a=2b\]  Also, \[\,\underset{x\to -{{1}^{+}}}{\mathop{Lim}}\,\,f(x)\,=\,\underset{x\to -{{1}^{-}}}{\mathop{Lim}}\,\,f(x)\] \[\Rightarrow \,b-a+4=a+b\Rightarrow \,2a=4\Rightarrow \,a=2\] Hence, b = 3[from equation 1]