A) 0
B) \[-1\]
C) 1
D) 2
Correct Answer: B
Solution :
Since the function is continuous at \[x=-2,\]then \[f(-2)=\underset{x\to -2}{\mathop{\lim }}\,f(x)\] \[=\underset{x\to -2}{\mathop{\lim }}\,\frac{3{{x}^{2}}+ax+a+3}{{{x}^{2}}+x-2}\] This limit will exist when \[15-a=0\] or \[a=15.\] ?(i) \[\therefore \] \[f(-2)=\underset{x\to -2}{\mathop{\lim }}\,\frac{3{{x}^{2}}+15x+18}{{{x}^{2}}+x-2}\] \[=\underset{x\to -2}{\mathop{\lim }}\,\frac{3(x+2)(x+3)}{(x+2)(x-1)}\] \[=\underset{x\to -2}{\mathop{\lim }}\,\frac{3(x+3)}{(x-1)}\] \[=\frac{3}{-3}=-1\]
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