A) \[8f'(1)\]
B) \[4f'(1)\]
C) \[2f'(1)\]
D) \[f'(1)\]
Correct Answer: A
Solution :
[a]| We have, \[\underset{x\to 1}{\mathop{\lim }}\,\int_{4}^{f(x)}{\frac{2t}{x-1}dt}\] |
| \[\underset{x\to 1}{\mathop{\lim }}\,\frac{\int_{4}^{f(x)}{2t}}{x-1}dt\] |
| Apply Leibnitz's rule, we get \[\underset{x\to 1}{\mathop{\lim }}\,\frac{2f(x)\cdot f'(x)}{1}\] |
| \[=2f(1)f'(1)\] |
| \[=2\cdot (4)\cdot f'(1)\] \[[\because f(1)=4]\] |
| \[=8f'(1)\] |
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