For x ∈ R , the number of real roots of the equation 3 x ^ 2 - 4| x...

\left|{x}^{2} - 1\right| = \left\{$$\begin{array}{ll}{x}^{2} - 1, {x}^{2} \ge 1 \\ - \left({x}^{2} - 1\right){x}^{2} < 1\end{array}$$\right

Case 1

${x}^{2} - 1 < 0 ⇒ x \in ( - 1, 1 )$

$3{x}^{2} + 4\left({x}^{2} - 1\right) + x - 1 = 0$

$7{x}^{2} + x - 5 = 0$

$x = \dfrac{- 1 \pm \sqrt{1 + 140}}{2 \times 7} = \dfrac{- 1 + \sqrt{141}}{14}, \dfrac{- 1 - \sqrt{141}}{14}$

We know, $\sqrt{141} \approx 11 . 5$

So, both roots lie in the interval \left(- 1, 1\right)$</p><p><strong>Case 2</strong></p><p>${x}^{2} - 1 \ge 0 ⇒ x \in ( - \infty , - 1 ] \cup [ 1, \infty )$</p><p>$3{x}^{2} - 4\left({x}^{2} - 1\right) + x - 1 = 0$</p><p>$- {x}^{2} + 4 + x - 1 = 0$</p><p>${x}^{2} - x - 3 = 0$</p><p>$x = \dfrac{1 \pm \sqrt{1 + 12}}{2} = \dfrac{1 + \sqrt{13}}{2}, \dfrac{1 - \sqrt{13}}{2}$</p><p>We know,$3 < \sqrt{13} < 4$</p><p>So, both roots lie in the interval$( - \infty , - 1 ] \cup [ 1, \infty )$</p><p>Hence,$4$ real roots.