Question

If $tan^{- 1}\frac{1}{2 x + 1}+tan^{- 1}\frac{1}{4 x + 1}=cot^{- 1}\left(\frac{x^{2}}{2}\right)$ , then the number of all possible values of $x$ is/are

• A. $1$
• B. $2$
• C. $3$
• D. $0$

Answer 1

Answer: (B)

$tan^{- 1}\left(\frac{\left\{\frac{1}{\left(2 x + 1\right)}\right\} + \left\{\frac{1}{\left(4 x + 1\right)}\right\}}{1 - \left\{\frac{1}{\left(2 x + 1\right)}\right\} \cdot \left\{\frac{1}{\left(4 x + 1\right)}\right\}}\right)=tan^{- 1}\frac{2}{x^{2}}$
$tan^{- 1}\left(\frac{6 x + 2}{8 x^{2} + 6 x}\right)=tan^{- 1}\frac{2}{x^{2}}$
Therefore, $\frac{3 x + 1}{4 x^{2} + 3 x}=\frac{2}{x^{2}}$
$x^{2}\left(3 x + 1\right)=2\left(4 x^{2} + 3 x\right)$
$3x^{3}-7x^{2}-6x=0$
$x\left(3 x^{2} - 7 x - 6\right)=0$
$x\left(x - 3\right)\left(3 x + 2\right)=0$
We take $x=3,0$ because when $x=-\frac{2}{3},$ $LHS$ of the given equation will be negative whereas $tan^{- 1}\left(\frac{2}{x^{2}}\right)$ is positive.
Hence, the number of values $=2$