If x - P x 2 - 3 x + 2 takes all real values for x ∈ ℝ , then the range of P is

Step-by-step solution

Consider, y = x - P x 2 - 3 x + 2 and it is provided that x 2 - 3 x + 2 ≠ 0 x ≠ 1 ,   2 Now solve the above equation. y x 2 - ( 3 y + 1 ) x + ( 2 y + P ) = 0 ∵ x ∈ ℝ , So D ≥ 0 ⇒ ( 3 y + 1 ) 2 - 4 y ( 2 y + P ) ≥ 0 ⇒ y 2 + ( 6 - 4 P ) y + 1 ≥ 0 ∵ y ∈ ℝ , so D ≤ 0 ⇒ ( 6 - 4 P ) 2 - 4 ≤ 0 ⇒ ( 4 P - 6 + 2 ) ( 4 P - 6 - 2 ) ≤ 0 ⇒ ( P - 1 ) ( P - 2 ) ≤ 0 P ∈ [ 1 , 2 ] but P ≠ 1 , &#160