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