Question

Let f:R $\to$ R is differentiable function & $f\left(1\right)=4$, then $g\left(x\right)=\underset{x\to 1}{lim}{\int }_4^{f\left(x\right)}\frac{2t}{x-1}dt$ equals

• A. $4f^{\prime }\left(1\right)$
• B. $2f^{\prime }\left(1\right)$
• C. $8f^{\prime }\left(1\right)$
• D. $f^{\prime }\left(1\right)$

Answer 1

Answer: (C)

$8f^{\prime }\left(1\right)$

$g\left(x\right)=\underset{x\to 1}{lim}{\int }_4^{f\left(x\right)}\frac{2t}{x-1}dt$ and $f\left(1\right)=4$
$g\left(x\right)=\underset{x\to 1}{lim}{\int }_4^{f\left(x\right)}\frac{2t}{x-1}dt=\underset{x\to 1}{lim}\frac{{\int }_4^{f\left(x\right)}2tdt}{x-1}$
applying L-Hospital rule
$\Rightarrow g\left(x\right)=\underset{x\to 1}{lim}2f\left(x\right)f^{\prime }\left(x\right)$
$\therefore g\left(x\right)=8f^{\prime }\left(1\right)$