Question

A non zero integral solution of the equation is $ta{n}^{-1}\frac{1}{2x+1}+ta{n}^{-1}\frac{1}{4x+1}=ta{n}^{-1}\frac{2}{x^2}$ is

• A. 1
• B. 2
• C. 3
• D. 4

Answer 1

Answer: (C)

3

Given, ${\tan }^{-1}\left(\frac{1}{2x+1}\right)+{\tan }^{-1}\left(\frac{1}{4x+1}\right)$
$={\tan }^{-1}\left(\frac{4x+1+2x+1}{(4x+1)(2x+1)-1}\right)$
$={\tan }^{-1}(\frac{6x+2}{8x^2+6x+1-1}$
$={\tan }^{-1}\left(\frac{2(3x+1)}{2x(4x+3)}\right)$
$={\tan }^{-1}\left(\frac{2}{x^2}\right)$
$\therefore \frac{3x+1}{x(4x+3)}=\frac{2}{x^2}$
$(3x+1)x^2=2x(4x+3)$
$3x^3+x^2=8x^2+6x$
$3x^3-7x^2-6x=0$
$x(3x^2-7x-6)=0$
$x=0$ is one root.
$3x^2-7x-6=0$
$3x^2-9x+2x-6=0$
$(3x+2)(x-3)=0$
$x=\frac{-2}{3}$ and $x=3$
Hence there is only one non-zero integral solution.