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If b-c, bx-cy, bx^(2)-cy^(2) (b,c ne 0) ...

If `b-c`, `bx-cy`, `bx^(2)-cy^(2)` (`b,c ne 0`) are in `G.P`, then the value of `((bx+cy)/(b+c))((bx-cy)/(b-c))` is

A

`x^(2)`

B

`-x^(2)`

C

`2y^(2)`

D

`3y^(2)`

Text Solution

Verified by Experts

The correct Answer is:
A
`(a)` `(bx-cy)^(2)=(b-c)(bx^(2)-cy^(2))`
`b^(2)x^(2)+c^(2)y^(2)-2bxy=b^(2)x^(2)-bcy^(2)-bcx^(2)+c^(2)y^(2)`
`bcx^(2)+bcy^(2)-2bcxy=0`
`bc(x-y)^(2)=0`
`impliesx=y` (`:'b`, `c ne 0`)
`implies((bx+cy)/(b+c))((bx-cy)/(b-c))=x^(2)`
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