If `y = tan^(-1).(1)/(2) + tan^(-1) b, (0 b lt 1) and 0 lt y le (pi)/(4)`, then the maximum value of b is

Options

A. `1//2`
B. `1//3`
C. `1//4`
D. `2//3`

Answer

B

Explanation

`y = tan^(-1).(1)/(2) + tan^(-1) b, (0 lt b lt 1) and 0 lt y le (pi)/(4) tan^(-1) b = y - tan^(-1).(1)/(2)`
Thus, maximum value of b occurs when y is maximum
`:. tan^(-1) b = (pi)/(4) - tan^(-1).(1)/(2)`
`rArr tan^(-1) b = tan^(-1) 1 - tan^(-1).(1)/(2)`
`rArr tan^(-1) b = tan^(-1).(1 -(1)/(2))/(1 + (1)/(2))`
`rArr tan^(-1) b = tan^(-1).(1)/(3)`
`rArr b_("max") = (1)/(3)`