A) \[(-3,\,-1)\]
B) \[(-3,\,\,1)\]
C) \[(3,\,\,1)\]
D) \[(3,\,-1)\]
Correct Answer: B
Solution :
Given \[f(x)\] is differentiable at \[x=0\]. Hence, \[f(x)\] will be continuous at \[x=0\]. \\[\underset{x\to {{0}^{-}}}{\mathop{\text{lim}}}\,({{e}^{x}}+ax)=\underset{x\to {{0}^{+}}}{\mathop{\text{lim}}}\,b{{(x-1)}^{2}}\] Þ \[{{e}^{0}}+a\times 0=b{{(0-1)}^{2}}\] Þ \[b=1\] ?.. (i) But \[f(x)\] is differentiable at \[x=0\], then \[L{f}'(x)=R{f}'(x)\] Þ \[\frac{d}{dx}({{e}^{x}}+ax)=\frac{d}{dx}b{{(x-1)}^{2}}\] Þ \[{{e}^{x}}+a=2b(x-1)\] At \[x=0,\] \[{{e}^{0}}+a=-2b\] Þ \[a+1=-2b\] Þ \[a=-3\] Þ \[(a,\,\,b)=(-3,\,\,1)\].
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