A) \[1/2\]
B) \[1\]
C) \[2\]
D) \[3\]
Correct Answer: B
Solution :
[b] \[\underset{x\to 0}{\mathop{\lim }}\,\frac{1-\cos x}{{{e}^{ax}}-bx-1}=2\] \[\Rightarrow \,\,\,\,\underset{x\to 0}{\mathop{\lim }}\,\frac{{{x}^{2}}}{1+\frac{ax}{1!}+\frac{{{a}^{2}}{{x}^{2}}}{2!}+....-bx-1}=2\] \[\Rightarrow \,\,\,\,\underset{x\to 0}{\mathop{\lim }}\,\frac{{{x}^{2}}}{(a-b)x+\frac{{{a}^{2}}{{x}^{2}}}{2!}+....}=2\] For limit to exist, \[a-b=0\] \[\therefore \,\,\,\,\,\,\,\frac{2}{{{a}^{2}}}=2\] \[\Rightarrow \,\,\,\,\,a=\pm 1\] If \[a=1\] then \[b=1\]. If \[a=-1\]then \[b=-1\]. \[\therefore \,\,\,\,\,a/b=1\]
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