JEE Advanced Mathematics Question (2021) — Solution

Step-by-step solution

x 2 - 1 = x 2 - 1 ,                     x 2 ≥ 1 - x 2 - 1           x 2 < 1 Case 1 x 2 - 1 < 0 ⇒ x ∈ ( - 1 , 1 ) 3 x 2 + 4 x 2 - 1 + x - 1 = 0 7 x 2 + x - 5 = 0 x = - 1 ± 1 + 140 2 × 7 = - 1 + 141 14 , - 1 - 141 14 We know,  141 ≈ 11 . 5 So, both roots lie in the interval  - 1 , 1 Case 2 x 2 - 1 ≥ 0 ⇒ x ∈ ( - ∞ , - 1 ] ∪ [ 1 , ∞ ) 3 x 2 - 4 x 2 - 1 + x - 1 = 0 - x 2 + 4 + x - 1 = 0 x 2 - x - 3 = 0 x = 1 ± 1 + 12 2 = 1 + 13 2 , 1 - 13 2 We know,  3 < 13 < 4 So, both roots lie in the interval  ( - ∞ , - 1 ] ∪ [ 1 , ∞ ) Hence, 4  real roots.