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Grade: 12

                        

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

11 years ago

Answers : (1)

Ramesh V
70 Points
							

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

11 years ago
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