If `f(x)={((log(1+2ax)-log(1-bx))/(x)",", x ne 0),(k",", x =0):}` is continuous at x = 0, then value of k is

Options

A. ` b + a `
B. ` b - 2a `
C. `2a - b `
D. ` 2a + b `

Answer

D

Explanation

Given ,
` f ( x ) = {{:( (log ( 1 + 2ax ) - log ( 1 - bx )) /( x ) ,",",x ne 0 ) ,( k,",", x = 0 ):} ` is continuous at ` x = 0 `
` therefore f ( 0 ) = lim _ ( x to 0) ( log ( 1 + 2a x ) - log ( 1 - bx ) ) /( x ) `
` ( (0) /( 0 ) ` form )
` rArr k = lim _ ( x to 0) ( ( 1 )/( 2ax + 1 ) ( 2a ) - ( 1 ) /( 1 - bx ) ( - b ) ) /( + 1 ) `
` " " ` (by 'L' Hospital's rule )
` rArr k = ( 2a )/( 0 + 1 ) + ( b ) /( 1 - 0 ) = 2a + b `