Question

Let $f\left(x\right)$ be a differentiable function and $f\left(1\right)=2$.
If $\underset{x\to 1}{lim}{\int }_2^{f\left(x\right)}\frac{2t}{x-1}dt=4$, then the value of $f^{\prime }\left(1\right)$, is

• A. $1$
• B. $2$
• C. $4$
• D. none of these

Answer 1

Answer: (A)

$1$

$\underset{x\to 1}{lim}{\int }_2^{f\left(x\right)}\frac{2t}{x-1}dt=4$

$\Rightarrow \underset{x\to 1}{lim}\frac{{\int }_2^{f\left(x\right)}2tdt}{x-1}=4$
Using leibnitz theorem and L'Hospital's rule,
$\underset{x\to 1}{lim}(f^{\prime }\left(x\right)2f\left(x\right)-0)=4$
Substituting, $f\left(1\right)=2$
We get
$f^{{}^{\prime }}\left(1\right)2f\left(1\right)=4\Rightarrow f^{{}^{\prime }}\left(1\right)=1$

$\because f\left(1\right)=2$
Hence, option 'A' is correct.