Question

The inverse of the function y = `(16^x - 16^-x)/(16^x + 16^-x)` is

विकल्प

• log₁₆ (2 - x)

• `1/2 log_16  (1 + x)/(1 - x)`

• `1/2 log_16  (2x - 1)`

• `1/4 log_16  (2x)/(2 - x)`

Answer 1

`1/2 log_16  (1 + x)/(1 - x)`

Explanation:

Let y = f(x) = `(16^x - 16^-x)/(16^x + 16^-x)`

∴ y = `(16^(2x) - 1)/(16^(2x) + 1)`

`=> 16^(2x) = (1 + y)/(1 - y)`

`=> 2x = log_16  (1 + y)/(1 - y)`

`=> x = 1/2 log_16  (1 + y)/(1 - y)`

`=> "f"^-1 (y) = 1/2 log_16  (1 + y)/(1 - y)`

`=> "f"^-1 (x) = 1/2 log_16  (1 + x)/(1 - x)`