Question

Find the number of solutions of the equation `tan^-1 (x - 1) + tan^-1x + tan^-1(x + 1) = tan^-1(3x)`

Answer 1

tan^–1(x – 1) + tan^–1 x + tan^–1(x + 1)

= tan^–1(x – 1) + tan^–1(x + 1) + tan^–1x

= `tan^-1 [(x - 1 + x + 1)/(1 - (x - 1)(x + 1))] + tan^-1x`

= `tan^-1 [(2x)/(1 - (x^2 - 1))] + tan^-1x`

= `tan^-1 [(2x)/(1 - x^2 + 1)] + tan^-1x`

= `tan^-1 [(2x)/(2 - x^2)] + tan^-1x`

= `tan^-1 [((2x)/(2 - x^2) + x)/(1 - (2x)/(2 - x^2) * x)]`

= `tan^-1 [((2x + 2x - x^3)/(2 - x^2))/((2 - x^2 - 2x^2)/(2 - x^2))]`

= `tan^-1 [(4x - x^3)/(2 - 3x^2)]`

Given L.H.S. = R.H.S

`tan^-1 [(4x - x^3)/(2 - 3x^2)] = tan^-1 3x`

`(4x - x^3)/(2 - 3x)` = 3x

4x – x³ = 6x – 9x³

8x³ = 2x

8x³ – 2x = 0

2x(x² – 1) = 0

x = 0, x² = 1

x = ±1

Number of solutions are three (0, 1 – 1)