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The number of positive solutions satisfying the equation $\tan^{- 1} (\frac{1}{2 x + 1}) + \tan^{- 1} (\frac{1}{4 x + 1}) = \tan^{- 1} (\frac{2}{x^{2}})$ , is

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Correct option is A

We have,tan−1⁡12x+1+tan−1⁡14x+1=tan−1⁡2x2⇒ tan−1⁡12x+1+14x+11−1(2x+1)(4x+1)=tan−1⁡2x2⇒ tan−1⁡3x+14x2+3x=tan−1⁡2x2⇒ 3x+14x2+3x=2x2⇒3x2−7x−6=0⇒x=−23,3

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The number of positive solutions satisfying the equation tan−1⁡12x+1+tan−1⁡14x+1=tan−1⁡2x2, is

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The number of positive solutions satisfying the equation $\tan^{- 1} (\frac{1}{2 x + 1}) + \tan^{- 1} (\frac{1}{4 x + 1}) = \tan^{- 1} (\frac{2}{x^{2}})$ , is