• # question_answer If $f:R\to R$ is a differentiable function $f(1)=4.$ Then, the value of $\underset{x\to 1}{\mathop{\lim }}\,\int_{4}^{f(x)}{\frac{2t}{x-1}}dt$ A) $8f'(1)$ B) $4f'(1)$ C) $2f'(1)$             D) $f'(1)$

[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 $\underset{x\to 1}{\mathop{\lim }}\,\frac{2f(x)\cdot f'(x)}{1}$ $=2f(1)f'(1)$ $=2\cdot (4)\cdot f'(1)$   $[\because f(1)=4]$ $=8f'(1)$