Question

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

• A. $1$
• B. $2$
• C. $3$
• D. $0$

Answer 1

Answer: (B)

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