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A real value of x satisfies the equation...

A real value of x satisfies the equation `(3-4ix)/(3+4ix)=alpha-ibeta(alpha,beta in R)`, if `alpha^2+beta^2=`

A

` x = 2n +1`

B

`x = 4n`

C

`x = 2n`

D

`x = 4n + 1`

Text Solution

Verified by Experts

The correct Answer is:
A
Given equation, ` ((3 - 4 ix)/(3 + 4 ix )) = alpha - i beta ( alpha , beta in R ) `
`rArr [(3 - 4 ix)/(3 + 4 ix )] = alpha - i beta`
Now, ` (alpha -ibeta ) =((3 -4ix)(3 - 4ix))/((3 + 4ix)(3 - 4 ix) )=(9+ 26i^(2) x^(2) - 24ix)/(9 - 16i^(2)x^(2))`
`rArr (alpha -ibeta ) =(9+ 26i^(2) x^(2) - 24ix)/(9 - 16i^(2)x^(2))`
`rArr alpha - ibeta = (9- 16x^(2))/(9 + 16x^(2) - i24x) /(9 + 16x^(2)`
`rArr alpha - ibeta = (9 - 16x^(2))/(9 + 16x^(2)) - (i24x)/(9 + 16x^(2)) ...(i)`
`:." " alpha + ibeta = (9 - 16x^(2))/(9 + 16x^(2)) - (i24x)/(9 + 16x^(2)) ...(ii)`
So, `(alpha - ibeta)(alpha + i beta)= ((9 - 16x^(2))/(9 + 16x^(2))^(2)) - ((i24x)/(9 + 16x^(2)))^(2)`
`alpha^(2) + beta^(2) = (81+ 256x^(4) - 288x^(2) + 576x^(2))/((9 + 16x^(2))^(2))`
`(( 9 + 16x^(2))^(2))/(9 + 16x^(2) )^(2) = 1`
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