फलन `y = (10^(x) - 10^(-x))/(10^(x) + 10^(-x))` का प्रतिलोम है

Options

A. `log_(10) (2-x)`
B. `(1)/(2) "log"_(10) (1 + x)/(1-x)`
C. `(1)/(2) log_(10) (2x - 1)`
D. `(1)/(4) log ((2x)/(2-x))`

Answer

B

Explanation

`y = (10^(x) - 10^(-x))/(10^(x) + 10^(-x))`

`=(10^ -x ( (10^ x)/( 10^ (-x) -1))/((10^ -x ( (10^ x)/( 10^ (-x) +1))`

`(10 ^ (2x) - 1))/((10 ^ (2x) + 1))`

`y(10 ^ (2x) + 1) ``= 10 ^ (2x) - 1`

`y * 10 ^ (2x) + y ``= 10 ^ (2x)-1`

` y * 10 ^ (2x) - 10 ^ (2x)`` = - y - 1`

`10 ^ (2x) * (y - 1) ``= - (y + 1)`

`10^ (2x)* (1-y) ``= y+1 `
[दोनो पक्षो को - 1 से गुणा करने पर]
`10 ^ (2x) ``= (y + 1)/(1 - y)`
दोनो पक्षो में `log_10` लेने पर

`log_10 10 ^ (2 ^ x)`` = log_10 ((y + 1)/(1 - y))`

`2x log_(10) 10=log_10 ( y+1 )/(1-y )`

`2x=log _10 (y+1 )/(1-y )`

`x = 1/2 * log_10 ((y + 1)/(1 - y))`

`f ^ - 1 * (x) = 1/2 * log_10 ((x + 1)/(1 - x))`