If 0 leq x

Question

(B) Given equation: $\operatorname{Tan}^{-1}(2x-1) + \operatorname{Tan}^{-1}(2x+1) = \operatorname{Tan}^{-1}(4x) - \operatorname{Tan}^{-1}(2x)$.Using

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$1$

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(B) Given equation: $\operatorname{Tan}^{-1}(2x-1) + \operatorname{Tan}^{-1}(2x+1) = \operatorname{Tan}^{-1}(4x) - \operatorname{Tan}^{-1}(2x)$.Using the formula $\operatorname{Tan}^{-1} A + \operatorname{Tan}^{-1} B = \operatorname{Tan}^{-1} \left( \frac{A+B}{1-AB} \right)$:$\operatorname{Tan}^{-1} \left( \frac{(2x-1) + (2x+1)}{1 - (2x-1)(2x+1)} \right) = \operatorname{Tan}^{-1} \left( \frac{4x - 2x}{1 + (4x)(2x)} \right)$.$\operatorname{Tan}^{-1} \left( \frac{4x}{1 - (4x^2 - 1)} \right) = \operatorname{Tan}^{-1} \left( \frac{2x}{1 + 8x^2} \right)$.$\operatorname{Tan}^{-1} \left( \frac{4x}{2 - 4x^2} \right) = \operatorname{Tan}^{-1} \left( \frac{2x}{1 + 8x^2} \right)$.Equating the arguments: $\frac{4x}{2(1 - 2x^2)} = \frac{2x}{1 + 8x^2}$.$\frac{2x}{1 - 2x^2} = \frac{2x}{1 + 8x^2}$.This implies $2x = 0$ or $\frac{1}{1 - 2x^2} = \frac{1}{1 + 8x^2}$.Case $1$: $2x = 0 \implies x = 0$. Since $0 \leq 0 Case $2$: $1 - 2x^2 = 1 + 8x^2 \implies 10x^2 = 0 \implies x = 0$.Thus, the only solution is $x = 0$. The number of values is $1$.

If $0 \leq x < \frac{3}{4}$, then the number of values of $x$ satisfying the equation $\operatorname{Tan}^{-1}(2x-1) + \operatorname{Tan}^{-1}(2x) = \operatorname{Tan}^{-1}(4x) - \operatorname{Tan}^{-1}(2x+1)$ is:

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