Question

Consider, $f(x)=\frac{1}{|x-4|+1}+\frac{1}{|x+8|+1}$. If local minimum value of $f(x)$ is $\lambda$ and which occurs
$x=x_{0}$, then find the value of $\left[x_{0}+14 \lambda\right]$.
[Note: [k] denotes greatest integer function which is less than or equal to $\mathrm{k}$.]

Answer 1

$|\mathrm{x}-4|+1$ and $|\mathrm{x}+8|+1$ both are decreasing in $(-\infty, 8)$ and increasing in $(4, \infty)$ $\therefore \mathrm{f}(\mathrm{x})$ is increasing in $(-\infty,-8)$ and decreasing in $(4, \infty)$
Now, in $(-8,4)$
$\mathrm{f}(\mathrm{x})=\frac{1}{\mathrm{x}+9}+\frac{1}{5-\mathrm{x}}$
$=\frac{(x+9)^{2}-(5-x)^{2}}{(x+9)^{2}(5-x)^{2}}=\frac{14(2 x+4)}{(x+9)^{2}(5-x)^{2}}$
$\mathrm{f}^{\prime}(\mathrm{x})=0 \Rightarrow \mathrm{x}=-2 \Rightarrow$ local minima
$\therefore$ local minimum value of $\mathrm{f}(\mathrm{x})$ is
$ \mathrm{f}(-2)=\frac{1}{7}+\frac{1}{7}=\frac{2}{7} $
$\therefore {\left.\left[\mathrm{x}_{0}+14 \lambda\right]=\left[-2+14 \cdot \frac{2}{7}\right]=2\right] }$