JEE Advanced 2021 — Paper 1 — Question 29

Given equation: $3x^2 - 4|x^2 - 1| + x - 1 = 0$. Case I: $|x| \geq 1$ (so $x^2 \geq 1$), then $|x^2 - 1| = x^2 - 1$. Substitute: $3x^2 - 4(x^2 - 1) + x - 1 = 0$. Simplify: $3x^2 - 4x^2 + 4 + x - 1 = 0$. $-x^2 + x + 3 = 0$ or $x^2 - x - 3 = 0$. Discriminant: $\Delta = 1 + 12 = 13 > 0$. Roots: $x = \frac{1 \pm \sqrt{13}}{2}$. Check condition: $|x| \geq 1$ for both roots. $\frac{1 + \sqrt{13}}{2} \approx 2.3$ (valid), $\frac{1 - \sqrt{13}}{2} \approx -1.3$ (valid). Case II: $|x| < 1$ (so $x^2 < 1$), then $|x^2 - 1| = 1 - x^2$. Substitute: $3x^2 - 4(1 - x^2) + x - 1 = 0$. Simplify: $3x^2 - 4 + 4x^2 + x - 1 = 0$. $7x^2 + x - 5 = 0$. Discriminant: $\Delta = 1 + 140 = 141 > 0$. Roots: $x = \frac{-1 \pm \sqrt{141}}{14}$. Check condition: $|x| < 1$ for both roots. $\frac{-1 + \sqrt{141}}{14} \approx 0.85$ (valid), $\frac{-1 - \sqrt{141}}{14} \approx -0.99$ (valid). Total real roots: 4.