A) \[a=1,\text{ }b=-\,1\]
B) \[a=-1,\text{ }b=1\]
C) \[a=1,\text{ }b=1\]
D) \[a=-\,1,\text{ }b=-\,1\]
Correct Answer: C
Solution :
[c] For \[f(x)\] to be continuous at x=0, we have \[f({{0}^{-}})=f({{0}^{+}})\] or \[a(0)+b=1\] or b=1 \[f({{0}^{+}})=\underset{h\to 0}{\mathop{\lim }}\,\frac{f(h)-f(0)}{h}=\underset{h\to 0}{\mathop{\lim }}\,\frac{{{e}^{{{h}^{2}}+h}}-b}{h}\] \[=\underset{h\to 0}{\mathop{\lim }}\,\frac{{{e}^{{{h}^{2}}+h}}-1}{h}\] \[=\underset{h\to 0}{\mathop{\lim }}\,\frac{{{e}^{{{h}^{2}}+h}}-1}{h(h+1)}(h+1)=1\] \[\therefore f'({{0}^{-}})=\alpha \] Hence, a=1.
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