Question

Number of solutions of the equation $ta{n}^{-1}\left(\frac{1}{2x+1}\right)+ta{n}^{-1}\left(\frac{1}{4x+1}\right)=ta{n}^{-1}\left(\frac{2}{x^2}\right)$ is

• A. $1$
• B. $2$
• C. $3$
• D. $4$

Answer 1

Answer: (B)

We have,
$ta{n}^{-1}\left(\frac{1}{2x+1}\right)+ta{n}^{-1}\left(\frac{1}{4x+1}\right)=ta{n}^{-1}\left(\frac{2}{x^2}\right)$
$\Rightarrow ta{n}^{-1}\left\{\frac{\frac{1}{2x+1}+\frac{1}{4x+1}}{1-\left(\frac{1}{2x+1}\right)\left(\frac{1}{4x+1}\right)}\right\}=ta{n}^{-1}\left(\frac{2}{x^2}\right)$
$\Rightarrow ta{n}^{-1}\left(\frac{6x+2}{8x^2+6x}\right)=ta{n}^{-1}\left(\frac{2}{x^2}\right)$
$\Rightarrow tan\left\{ta{n}^{-1}\left(\frac{2\left(3x+1\right)}{2x\left(4x+3\right)}\right)\right\}=\frac{2}{x^2}$
$\Rightarrow \frac{3x+1}{x\left(4x+3\right)}=\frac{2}{x^2}$
$\Rightarrow 3x^2-7x-6=0$
$\Rightarrow \left(x-3\right)\left(3x+2\right)=0$
$\Rightarrow x=3$ or $x=\frac{-2}{3}$
$\therefore$ The given equation has $2$ solutions.