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

Answer: 2

$|\mathrm{x}-4|+1$ 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 $(-8,4)$
$\mathrm{f}(\mathrm{x})=\frac{1}{\mathrm{x}+9}+\frac{1}{5-\mathrm{x}}$
$=\frac{(x+9{)}^2-(5-x{)}^2}{(x+9{)}^2(5-x{)}^2}=\frac{14(2x+4)}{(x+9{)}^2(5-x{)}^2}$
${\mathrm{f}}^{\prime }(\mathrm{x})=0\Rightarrow \mathrm{x}=-2\Rightarrow$ local minima
$\therefore$ local minimum value of $\mathrm{f}(\mathrm{x})$ is
$\mathrm{f}(-2)=\frac{1}{7}+\frac{1}{7}=\frac{2}{7}$
$\therefore \left[{\mathrm{x}}_0+14\lambda \right]=\left[-2+14\cdot \frac{2}{7}\right]=2]$