A) -1
B) 1
C) 0
D) 2
Correct Answer: A
Solution :
[a] \[f(x)=\left\{ \begin{matrix} ax-2 & -2<x<-1 \\ -1 & -1\le x\le 1 \\ a+2{{(x-1)}^{2}} & 1<x<2 \\ \end{matrix} \right.\] If f(x) is continuous at x = -1 then, \[\underset{x\to -1}{\mathop{\lim }}\,(ax-2)=\underset{x\to -1}{\mathop{\lim }}\,(-1)\] \[\Rightarrow a(-1)-2=-1\Rightarrow a=-1\] If f(x) is continuous at x = 1 then, \[\underset{x\to 1}{\mathop{\lim }}\,a+2{{(x-1)}^{2}}=\underset{x\to 1}{\mathop{\lim }}\,-1\] \[\Rightarrow a+2{{(1-1)}^{2}}=-1\Rightarrow a=-1\]
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