A) \[1\]
B) \[1/2\]
C) \[-1\]
D) \[-1/2\]
Correct Answer: C
Solution :
[c] \[\underset{x\to 0}{\mathop{\lim }}\,\frac{{{(4x-1)}^{\frac{1}{3}}}+a+bx}{x}=\frac{1}{3}\] \[\Rightarrow \,\,\,\,\,\,\,\,\,\,\underset{x\to 0}{\mathop{\lim }}\,\frac{-\left[ 1-\frac{4}{3}x \right]+a+bx}{x}=\frac{1}{3}\] \[\Rightarrow \,\,\,\,\,\,\,\,\,\,\underset{x\to 0}{\mathop{\lim }}\,\frac{(a-1)+\left( \frac{4}{3}+b \right)x}{x}=\frac{1}{3}\] For limit to exist, \[a=1\] \[\Rightarrow \,\,\,\,\,\,\,\,\,\,\underset{x\to 0}{\mathop{\lim }}\,\frac{\left( \frac{4}{3}+b \right)x}{x}=\frac{1}{3}\] \[\Rightarrow \,\,\,\,\,\,\,\,\,\,b=-1\]
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