Question

Please solve the above question ​

Answer 1

Answer:

option B is the answer

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Answer 2

Step-by-step explanation:

$\tt {tan}^{-1}(\frac{1}{2x+1}) + {tan}^{-1}(\frac{1}{4x+1}) = {tan}^{-1}(\frac{2}{{x}^{2}} )\\ \\ \tt \longrightarrow {tan}^{-1}(\frac{\frac{1}{2x+1} + \frac{1}{4x+1}}{1 - \frac{1}{2x+1} \frac{1}{4x+1}}) = {tan}^{-1}(\frac{2}{{x}^{2}} )\\ \\ \tt \longrightarrow {tan}^{-1}(\frac{2x+1+4x+1}{(2x+1)(4x+1)-1}) = {tan}^{-1}(\frac{2}{{x}^{2}}) \\ \\ \tt \longrightarrow {tan}^{-1}(\frac{6x+2}{8{x}^{2}+6x} = {tan}^{-1}(\frac{2}{{x}^{2}}) \\ \\ \tt \frac{6x+2}{8{x}^{2}+6x} = \frac{2}{{x}^{2}} \\ \\ \tt \longrightarrow 6{x}^{3} + 2{x}^{2} = 16{x}^{2} + 12x \\ \\ \tt \longrightarrow 6{x}^{3} - 14{x}^{2} -12x = 0 \\ \\ \tt \longrightarrow 3{x}^{3} - 7{x}^{2} - 6x = 0 \\ \\ \tt \longrightarrow x(3{x}^{2} - 7x - 6) = 0 \\ \\ \tt \longrightarrow x(3{x}^{2} - 9x + 2x - 6 ) = 0 \\ \\ \tt \longrightarrow x(3x+2)(x-3) =0 \\ \\ \tt x = 0 , 3 , \frac{3}{2} \\ \\ \boxed{\bold{\underline{\red{\tt So\:number\:of\:solutions\:are\:3}}}}$