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Let A(r) be the area of the region bound...

Let `A_(r)` be the area of the region bounded between the curves `y^(2)=(e^(-kr))x("where "k gt0,r in N)" and the line "y=mx ("where "m ne 0)`, k and m are some constants
`A_(1),A_(2),A_(3),…` are in G.P. with common ratio

A

`e^(-k)`

B

`e^(-2k)`

C

`e^(-4k)`

D

None of these

Text Solution

Verified by Experts

The correct Answer is:
B
Solving the two equations, we get
`m^(2)x^(2)=(e^(-kr))x`
`x_(1)=0,x_(2)=(e^(-kr))/(m^(2)).`

`"So, "A_(r)=overset(x_(2))underset(0)int(e^((-kr)/(2))sqrt(x)-mx)dx`
`=(2)/(3)e^(-kr//2)x_(2)^(3//2)-m(x_(2)^(2))/(2)`
`=(2)/(3)e^(-kr//2)(e^(-3kr//2))/(m^(3))-(m)/(2)(e^(-2kr))/(m^(4))=(e^(-2kr))/(6m^(3)).`
`"Now, "(A_(r+1))/(A_(r))=(e^(-2k(r+1)))/(e^(-2kr))=e^(-2k)="constant".`
So, the sequence `A_(1),A_(2),A_(3),...` is in G.P.
`"Sum of n terms "=(e^(-2k))/(6m^(3))(e^(-2nk)-1)/(e^(-2k)-1)=(1)/(6m^(3))(e^(-2nk)-1)/(1-e^(2k))`
`"Sum to infinte terms "=A_(1)(1)/(1-e^(-2k))`
`=(e^(-2k))/(6m^(3))xx(e^(2k))/(e^(2k)-1)=(1)/(6m^(3)(e^(2k-1))).`
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