Click to Chat

1800-2000-838

+91-120-4616500

CART 0

• 0

MY CART (5)

Use Coupon: CART20 and get 20% off on all online Study Material

ITEM
DETAILS
MRP
DISCOUNT
FINAL PRICE
Total Price: Rs.

There are no items in this cart.
Continue Shopping
`        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`
8 years ago

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
```
8 years ago
Think You Can Provide A Better Answer ?

## Other Related Questions on Integral Calculus

View all Questions »
• Complete JEE Main/Advanced Course and Test Series
• OFFERED PRICE: Rs. 15,900
• View Details