यदि `(1-ix)/(1+ix) =a -ib` , a का मान है

Options

A. `x^(2)`
B. `(x^(2) + 1)/(x^(2)-1)`
C. `(1-x^(2))/(1+x^(2))`
D. इनमे से कोई नहीं

Answer

C

Explanation

`(1-ix)/(1+ix) =a -ib`
`rArr (1-ix)/((1-ix)) xx ((1-ix))/((1-ix)) =a -ib`
`rArr (1-x^(2)-2ix)/(1+x^(2)) =a -ib`
यहाँ, `(1-x^(2))/(1+x^(2)) =a`
एवं ` (2x)/(1+x^(2)) =b`
`therefore (1-ix)/(1+ix) = (a-ib)/(1)`
`rArr (1-ix + 1 +ix)/(1-ix-1-ix) = (a-ib+1)/(a-ib-1)` [योगान्तरानुपात नियम से]
`rArr (2)/(-2ix) = (a-ib +1)/(a-ib-1)`
`rArr -ix = (a-ib +1)/(a-ib +1)`
`rArr (x)/(i) = (a-ib-1)/(a-ib + 1)`
`therefore x = (ai +b-i)/(a-ib +1)`
` = (ai +b-i)/((1+a) = ib) xx ((1+a) + ib)/((1+a) + ib)`
`(ai + a^(2)i -ab +b +ab +ib^(2) -i -ai +b)/((1+a)^(2) + b^(2))`
`x = ((a^(2) +b^(2))i + 2b)/((1+a)^(2) +b^(2))`
`x = (2b)/((1+a)^(2) +b^(2)) + (i(a^(2) + b^(2)))/((1+a^(2)) + b^(2))`
`because x` वास्तविक संख्या है ।
`therefore x = (2b)/((1+a)^(2) +b^(2))`