Arithmetic mean of the non-zero solutions of the equation `tan^-1 (1/(2x + 1)) + tan^-1 (1/(4x + 1)) = tan^-1 (2/x^2)`

Options

A. 2
B. 3
C. 4
D. none of these

Answer

B

Explanation

`tan^(-1).(1)/(1 + 2x) + tan^(-1).(1)/(1 + 4x) = tan^(-1).(2)/(x^(2))`
or `tan^(-1) [((1)/(1 + 2x) + (1)/(1+ 4x))/(1 - (1)/(1 + 2x) (1)/(1 + 4x))] = tan^(-1).(2)/(x^(2))`
or `(2 + 6x)/(6x + 8x^(2)) = (2)/(x^(2))`
or `6x^(3) - 14x^(2) - 12x = 0`
or `x(x -3) (3x + 2) = 0`
or `x = 3 " or " x = -2//3`(as `x != 0`)
But for `x = -2//3`, L.H.S. `lt 0 and R.H.S. gt 0`
Hence, the only solution is `x = 3`