A) \[8f'(1)\]
B) \[4f'(1)\]
C) \[2f'(1)\]
D) \[f'(1)\]
Correct Answer: A
Solution :
[a] \[\underset{x\to 1}{\mathop{\lim }}\,\int\limits_{0}^{f(x)}{\frac{2t}{x-1}dt}\] \[=\underset{x\to 1}{\mathop{\lim }}\,\frac{1}{x-1}\left[ {{t}^{2}} \right]_{0}^{f(x)}=\underset{x\to 1}{\mathop{\lim }}\,\frac{{{f}^{2}}(x)-{{f}^{2}}(0)}{x-1}\] \[=\underset{x\to 1}{\mathop{\lim }}\,\frac{2f(x)f'(x)}{1}\] (L? Hospital Rule) \[=2f(1)f'(1)=8f'(1)\]
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