Question

IF f(x)=log(1+x/1-x), then f(a) + f(b) is equal to:
A) f(a+b)
B) f(ab)
C) (a+b/1+ab)
D) f(a-b/1+ab)

Answer 1

Step-by-step explanation:

f ( x ) = log ( 1+x / 1–x)

For f (a) put x = a

f (a) = log ( 1+a / 1–a ) – ( 1 )

For f (b) put x = b

f (b) = log ( 1+b / 1–b ) – ( 2 )

Adding ( 1 ) and ( 2 )

f (a) + f (b) = log ( 1 + a / 1 – a) + log ( 1 + b / 1 – b )

= log { ( 1+a )(1 + b) / (1-a) (1-b) }

= log { ( 1 + a + b + ab) / ( 1 – a – b + ab) }

= log { [ 1 + ( a + b + ab) ] / [ 1 – ( a + b + ab) ]

= f ( a + b + ab )

Please recheck once I think this is the correct answer.