For $x \in \mathbb{R}$, the number of real roots of the equation $3 x^{2}-4\left|x^{2}-1\right|+x-1=

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For $x \in \mathbb{R}$, the number of real roots of the equation $3 x^{2}-4\left|x^{2}-1\right|+x-1=0$ is

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(4) : For $x^{2}<1$ i.e., $-1<x<1$ The equation $3 x^{2}-\left|4 x^{2}-4ight|+x-1=0$ $$ \begin{aligned} & 3 x^{2}+4\left(x^{2}-1ight)+x-1=0 \ \Rightarrow \quad & 7 x^{2}+x-5=0 \ & x=\frac{-1 \pm \sqrt{141}}{14} \end{aligned} $$ Both lie in $(-1,1)$.\ For $x^{2} \geq 1$ i.e., $x \geq 1$ or $x \leq-1$\ The equation becomes\ $3 x^{2}-4\left(x^{2}-1ight)+x-1=0$\ $\Rightarrow \quad-x^{2}+x+3=0$\ $\Rightarrow x^{2}-x-3=0 \Rightarrow x=\frac{1 \pm \sqrt{13}}{2}$\ Both lie in $(-\infty,-1] \cup[1, \infty)$\ So, 4 real solutions exist.\