Question

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

• A. $0$
• B. $1$
• C. $2$
• D. $3$

Answer 1

Answer: (D)

$3$

$\begin{array}{c}Wehave\\ {\tan }^{-1}\left(\frac{1}{2x+1}\right)+{\tan }^{-1}\left(\frac{1}{4x+1}\right)={\tan }^{-1}\left(\frac{2}{x^2}\right)\\ {\tan }^{-1}\left(\frac{\left(\frac{1}{2x+1}\right)+\left(\frac{1}{4x+1}\right)}{1-\left(\frac{1}{2x+1}\right)\left(\frac{1}{4x+1}\right)}\right)={\tan }^{-1}\left(\frac{2}{x^2}\right)\left[u\sin gformula{\tan }^{-1}\left(a+b\right)\right]\\ {\tan }^{-1}\left(\frac{4x+1+2x+1}{\left(2x+1\right)\left(4x+1\right)-1}\right)={\tan }^{-1}\left(\frac{2}{x^2}\right)\\ \frac{6x+2}{8x^2+6x}=\frac{2}{x^2}\\ 6x^3+2x^2=16x^2+12x\\ 3x^3-14x^2-12x=0\\ x\left(3x^2-7x-6\right)=0\\ x\left(3x^2-9x+2x-6\right)=0\\ x\left(3x+2\right)\left(x-3\right)=0\\ So,\\ x=0,x=3,x=\frac{-3}{2}\end{array}$

So, there are $3$ solutions.

Hence, this is the answer.