Let for all n belonging to real number,n-2>=2ln n,then the minimum value of the area bound by the curve |y|+1/n<=e^ -|x| is 2 sq. unit. True or false

From n-2 >= 2*ln(n) 

we can say n> 0  which also mean:  n^-1 > 0

Now from the curve
  |y|+1/n <= e^-|x|   or           |y| <= e^-|x| +1/n

Now the graph of e^-|x|  is symmetric about Y axis 

for 1/n > 0 ;

CASE 1: lets  take least value 1/n = 0

so,   |y| <= e^-|x|

For , |y| <= e^-|x|  the area bounded is  2*integral of e^-|x| with limits from 0 to infininty

so area is 2 sq. units

CASE 1: lets  take  value 1/n more than 0

So so |y| <= e^-|x| - 1/n

here e^-|x| - 1/n becomes negative as shown in graph

but |y| is always positive which can never be less than a negative no.

so case 2 is not possible

hence area bounded is 2 sq. units and its TRUE