Question

Absolute value of the parameter 'a' for which the inverse of
f(x) = 1 + 2ax, a = 0 is itself, is
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Answer 1

Given : inverse of  f(x) = 1 + 2ax, a = 0 is itself,

To Find : Absolute value of the parameter 'a'

Solution:

f(x) =   1  + 2ax

=> 2ax  = f(x)  - 1

=>  x  =( f(x) - 1)/2a

=> f⁻¹(x) = (x - 1)/2a

f(x) =   1  + 2ax

f⁻¹(x) = ( x - 1)/2a

f(x)  =  f⁻¹(x)

=> 1  + 2ax   = ( x - 1)/2a

=> 2a + 4a²x = x - 1

=> x(4a² - 1)  + 2a + 1 = 0

=> x(2a + 1)(2a - 1) + 2a + 1 =  0

=> (2a + 1)(x(2a - 1) + 1) = 0

=> 2a + 1 = 0

=> a = -1/2

or we can do by equating coefficient's

1  + 2ax   = ( x - 1)/2a

=> 1 + 2ax  = x/2a  - 1/2a

=> 2a = 1/2a  and  1 = -1/2a

=> a² = 1/4   and   a = -1/2

=> a  =± 1/2   and   a = -1/2

combining both a = -1/2

f(x) =   1  - x

Absolute value of the parameter 'a' for which the inverse of

f(x) = 1 + 2ax, a = 0 is itself,   is  -1/2

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