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The number of solutions of $\tan^{-1}(x-1) + \tan^{-1}(x) + \tan^{-1}(x+1) = \tan^{-1}(3x)$ is

  1. $1$
  2. $2$
  3. $3$
  4. $4$
in Numerical Ability by Boss (16.8k points)
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1 Answer

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The given equation can also be written as:

$\tan^{-1}(x-1) + \tan^{-1}(x+1) = \tan^{-1}(3x) - \tan^{-1}(x)\tag{1}$

We know that:
$\scriptsize{\mathrm{\tan^{-1} A + \tan^{-1} B = \tan^{-1}\bigg(\frac{A + B}{1-AB}\bigg ),tan^{-1}A - tan^{-1}B = tan^{-1}\bigg (\frac{A-B}{1+AB}\bigg)}}\tag{2}$

Let $\mathrm{A = x-1, B = x + 1}$

Now, On applying the above formula:

$\scriptsize {\begin{align} \tan^{-1}(x-1) + \tan^{-1}(x+1) &= \tan^{-1}\bigg(\frac{x-1+x+1}{1-(x-1)(x+1)} \bigg)\\ &=\tan^{-1}\frac{2x}{1-x^2+1}\\&=\tan^{-1}\frac{2x}{2-x^2}\end {align}}$

Similarly,

$\scriptsize{\begin{align} \tan^{-1}3x-\tan^{-1}x &= \tan^{-1}\bigg (\frac{3x-x}{1+3x(x)}\bigg )\\&= \tan^{-1}\bigg (\frac{2x}{1+3x^2}\bigg )\tag{3} \\&\text{From $(1)$, $(2)$ and $(3)$}\\&=\tan^{-1}\bigg (\frac{2x}{2-x^2}\bigg )\tag{4} \end{align}}$

Now, on comparing both the sides, in equations $(3)$ and $(4)$, we get:

$\begin{align} \frac{2x}{1+3x^2} = \frac{2x}{2-x^2} \Rightarrow 2-x^2 = 1 + 3x^2 \Rightarrow 1 = 4x^2 \Rightarrow x = \pm \sqrt{\frac{1}{2}}\end {align}$

$\therefore$ We have $2$ roots.

Hence, $\bf{B}$ is the correct option.
by Boss (13.8k points)
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