QJEEAdv21Ma49b9: JEE Advanced 2021 Mathematics - Class 11 JEEAdv21Ma49b9 Solution: For x R, the number of real roots of the equation is ________.

Given,

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


For $-$ 1 $\le$ x $\le$ 1 i.e., x $\in$ [ $-$ 1, 1]

From Eq. (i), we get

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

$\Rightarrow 3{x^2} + 4{x^2} - 4 + x - 1 = 0$

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

$\Rightarrow x = {{ - 1 \pm \sqrt {1 + 140} } \over {(2 \times 7)}}$

Here, both values of x are acceptable.

For | x | > | i.e. x $\in$ ( $-$ $\infty$ , $-$ 1) $\cup$ (1, $\infty$ )

From Eq. (i), we get

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

$\Rightarrow {x^2} - x - 3 = 0$

$\Rightarrow x = {{1 \pm \sqrt {1 + 12} } \over 2}$

Again here, both values of x are acceptable.

Hence, total number of solutions is 4.