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Domain and Range of Basic Inverse Trigonometric Functions

If 1< x <√2...

Question

If $1 < x < \sqrt{2}$ , then number of solutions of the equation ${tan}^{- 1} (x - 1) + {tan}^{- 1} x + {tan}^{- 1} (x + 1) = {tan}^{- 1} 3 x,$ is

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Solution

The correct option is A $0$
Given equation can be written as

${tan}^{- 1} (x - 1) + {tan}^{- 1} (x + 1) = {tan}^{- 1} 3 x - {tan}^{- 1} x$

$\Rightarrow {tan}^{- 1} \frac{x - 1 + x + 1}{1 - (x - 1) (x + 1)} = {tan}^{- 1} \frac{3 x - x}{1 + 3 x^{2}}$

$\Rightarrow \frac{2 x}{2 - x^{2}} = \frac{2 x}{1 + 3 x^{2}}$

$\Rightarrow x + 3 x^{3} = 2 x - x^{3}$

$\Rightarrow 4 x^{3} - x = 0$

$\Rightarrow x = 0, x = \pm 1 / 2$

none of which satisfies $1 < x < \sqrt{2}$
Thus option A is correct.

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