Question

Let $f:$ $R\to R$ be a differentiable function at $x=1$ such that $f\left(1\right)=4$ and $f^{\prime }\left(1\right)=2$ and $\alpha ={lim}_{x\to 1}{\int }_4^{f\left(x\right)}\frac{2t}{x-1}dt$, then $\alpha$ is

• A. 8
• B. 16
• C. 5
• D. 2

Answer 1

Answer: (B)

16

$\underset{x\to 1}{lim}{\int }_4^{f\left(x\right)}\frac{2t}{x-1}dt=\underset{x\to 1}{lim}\frac{1}{x-1}{\int }_4^{f\left(x\right)}2tdt$

$=\underset{x\to 1}{lim}\frac{1}{x-1}{\left[t^2\right]}_4^{f\left(x\right)}=\underset{x\to 1}{lim}\frac{(f\left(x\right)-4)(f\left(x\right)+4)}{x-1}$

$=\underset{x\to 1}{lim}\frac{(f\left(x\right)-f\left(1\right))(f\left(x\right)+f\left(1\right))}{x-1}=2f\left(1\right).f^{\prime }\left(1\right)=2\times 4\times 2=16$
Hence, option 'B' is correct.