If tan^(-1)1/(2x+1)+tan^(-1)1/(4x+1)=cot^(-1)(x^2)/2, then the number of all possible values of x is/are

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Correct answer: (B) 2. tan^(-1)1/(2x+1)+1/(4x+1)1-1/(2x+1)⋅1/(4x+1)=tan^(-1)2/(x^2) tan^(-1)(6x+2)/(8x^2+6x)=tan^(-1)2/(x^2) Therefore, (3x+1)/(4x^2+3x)=2/(x^2) x^23x+1=24x^2+3x 3x^3-7x^2-6x=0 x3x^2-7x-6=0 xx-33x+2=0 We take x=3,0 because when x=-2/3, LHS of the given equation will be negative whereas tan^(-1)2/(x^2) is positive. Hence, the number of values =2

If $t a n^{- 1} \frac{1}{2 x + 1} + t a n^{- 1} \frac{1}{4 x + 1} = c o t^{- 1} (\frac{x^{2}}{2})$ , then the number of all possible values of $x$ is/are

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More Inverse Trigonometric Functions Questions

If tan-112x+1+tan-114x+1=cot-1x22, then the number of all possible values of x is/are

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More Inverse Trigonometric Functions Questions

If $t a n^{- 1} \frac{1}{2 x + 1} + t a n^{- 1} \frac{1}{4 x + 1} = c o t^{- 1} (\frac{x^{2}}{2})$ , then the number of all possible values of $x$ is/are