A) \[a=2,b=-3\]
B) \[a=3,b=2\]
C) \[a=-2,b=-3\]
D) \[a=-3,b=-2\]
E) \[a=-1,b=-2\]
Correct Answer: C
Solution :
\[f(x)=\left\{ \begin{matrix} b{{x}^{2}}+ax+4, & x\ge -1 \\ a{{x}^{2}}+b, & x<-1 \\ \end{matrix} \right.\] \[\therefore \]\[f(x)=\left\{ \begin{matrix} 2bx+a, & x\ge -1 \\ 2ax, & x<-1 \\ \end{matrix} \right.\] Given\[f(x)\]is continuous everywhere \[\therefore \]\[\underset{x\to -{{1}^{+}}}{\mathop{\lim }}\,f(x)=\underset{x\to -{{1}^{-1}}}{\mathop{\lim }}\,f(x)\] \[\Rightarrow \] \[-2b+a=-2a\]\[\Rightarrow \]\[-2b+3a=0\] \[\Rightarrow \] \[3a-2b=0\] \[\Rightarrow \]\[a=2,b=3\]or \[a=-2,b=-3\]
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