Question

If $f\left(x\right)$ be an identity function in $R$ and $g\left(x\right)={\sum }_{k=1}^3{\left(f\right(x)-(2016+k\left)\right)}^{-1},$ then

• A. $g\left(x\right)$ is strictly increasing in $(2018,2019)$
• B. $g\left(x\right)$ has two distinct real roots
• C. slope of tangent to the curve $g\left(x\right)$ at $x=f\left(2016\right)$ is $\frac{-49}{36}.$
• D. $\underset{x\to -\infty }{lim}g\left(x\right)=0$

Answer 1

Answer: (B), (C), (D)

$\underset{x\to -\infty }{lim}g\left(x\right)=0$

$f\left(x\right)=x,g\left(x\right)=\frac{1}{x-2017}+\frac{1}{x-2018}+\frac{1}{x-2019};xϵR-\{2017,2018,2019\}$
$\underset{x\to -\infty }{lim}g\left(x\right)=0$
$\underset{x\to \infty }{lim}g\left(x\right)=0$
$\underset{x\to -{2017}^{-}}{lim}g\left(x\right)=-\infty$,$\underset{x\to -{2017}^{+}}{lim}g\left(x\right)=\infty$


$g^{\prime }\left(x\right)=-\frac{1}{(x-2017{)}^2}-\frac{1}{(x-2018{)}^2}-\frac{1}{(x-2019{)}^2};xϵR-\{2017,2018,2019\}$
$g^{\prime }\left(2016\right)=-(1+\frac{1}{4}+\frac{1}{9})=\frac{-49}{36}$