The number of solutions of the equations ${tan}^{- 1} 2 x + {tan}^{- 1} 3 x = \frac{π}{4}$ is

2

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3

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1

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None of these

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Solution

The correct option is A $2$
given, ${tan}^{- 1} 2 x + {tan}^{- 1} 3 x = \frac{π}{4}$

We know that ${tan}^{- 1} A + {tan}^{- 1} B = {tan}^{- 1} (\frac{A + B}{1 - A B})$

$\Rightarrow {tan}^{- 1} (\frac{2 x + 3 x}{1 - 6 x^{2}}) = \frac{π}{4}$

$\Rightarrow {tan}^{- 1} (\frac{5 x}{1 - 6 x^{2}}) = \frac{π}{4}$

$\Rightarrow \frac{5 x}{1 - 6 x^{2}} = tan \frac{π}{4}$

$\Rightarrow \frac{5 x}{1 - 6 x^{2}} = 1$

$\Rightarrow 5 x = 1 - 6 x^{2}$

$\Rightarrow 6 x^{2} + 5 x - 1 = 0$

$\Rightarrow 6 x^{2} + 6 x - x - 1 = 0$

$\Rightarrow 6 x (x + 1) - 1 (x + 1) = 0$

$\Rightarrow (x + 1) (6 x - 1) = 0$

$\Rightarrow x = - 1, \frac{1}{6}$

$∴$ Number of solutions $= 2$

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