A) (3, 1)
B) (3, 2)
C) \[\left( \frac{1}{3},2 \right)\]
D) (2, 1)
Correct Answer: A
Solution :
If the function is continuous at \[x=0.\] \[{{\lim }_{x\to 0}}f(x)\]exists and is equal to \[f(0)\] \[{{\lim }_{x\to 0}}f(x)\] \[={{\lim }_{x\to 0}}\frac{1}{x}-\frac{k-1}{{{e}^{2x}}-1}\] \[={{\lim }_{x\to 0}}\frac{{{e}^{2x}}-1-kx+x}{(x)({{e}^{2x}}-1)}\] Using Taylor?s expansion for \[{{e}^{2x}},\]we get \[={{\lim }_{x\to 0}}\frac{(1+2x+\frac{{{(2x)}^{2}}}{2!})+\frac{{{(2x)}^{2}}}{3!}+...)-1-kx+x}{(x)((1+2x+\frac{{{(2x)}^{2}}}{2!}+\frac{{{(2x)}^{3}}}{3!}+...)-1)}\] \[={{\lim }_{x\to 0}}\frac{(3-k)x+\frac{4{{x}^{2}}}{2!}+\frac{8{{x}^{3}}}{3!}+...}{(2{{x}^{2}}+\frac{4{{x}^{3}}}{2!}+\frac{8{{x}^{4}}}{3!}+...)}\] For the limit to exist, power of\[x\]in the numerator should be greator than or equal to the power of\[x\]in the denominator. Therefore, coefficient of in numerator is equal to zero \[\Rightarrow \]\[3-k=0\] \[\Rightarrow \]\[k=3\] So the limit reduces to \[{{\lim }_{x\to 0}}\frac{({{x}^{2}})(\frac{4}{2!}+\frac{8x}{3!}+...)}{({{x}^{2}})(2+\frac{4x}{2!}+\frac{8x2}{3!}+...)}\] \[={{\lim }_{x\to 0}}\frac{\frac{4}{2!}+\frac{8x}{3!}+...}{2+\frac{4x}{2!}+\frac{8x2}{3!}+...}\] =1 Hence, \[f(0)=1\] Therefore, answer is option A.
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