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Question

Question: If 1\< x \<\(\sqrt{2}\), then number of solutions of the equation tan<sup>–1</sup>(x – 1) + tan<sup...

If 1< x <2\sqrt{2}, then number of solutions of the equation

tan–1(x – 1) + tan–1 x + tan–1(x + 1) = tan–1 3x, is/are

A

0

B

1

C

2

D

3

Answer

0

Explanation

Solution

Given equation can be written as

tan–1 (x – 1) + tan–1 (x + 1) = tan–1 3x – tan–1 x

̃ tan–1 x1+x+11(x1)(x+1)\frac{x - 1 + x + 1}{1 - (x - 1)(x + 1)} = tan–1 3xx1+3x2\frac{3x - x}{1 + 3x^{2}}

̃ 2x2x2=2x1+3x2\frac{2x}{2 - x^{2}} = \frac{2x}{1 + 3x^{2}}

̃ x + 3x3 = 2x – x3

̃ 4x3 – x = 0

̃ x(4x2 – 1) = 0

̃ x = 0, x = ± ½

none of which satisfies 1 < x < 2\sqrt{2}