Let E 1 = x ∈ ℝ : x ≠ 1 and x x - 1 > 0 and E 2 = x ∈ E 1 : sin - 1 log e x x - 1 is a real number (Here the inverse trigonometric function sin - 1 x assumes values in - π 2 . π 2 .) Let f : E 1 → ℝ be the function defined by f x = log e ⁡ x x - 1 And g : E 2 → R be the function defined by g x = sin - 1 ⁡ log e ⁡ x x - 1 . List-I List-II A. The

Step-by-step solution

E 1 : x x - 1 > 0 ⇒ E 1 : x ∈ - ∞ , 0 ∪ 1 , ∞ E 2 : - 1 ≤ ln ⁡ x x + 1 ≤ 1 1 e ≤ x x - 1 ≤ e Now x x - 1 - 1 e ≥ 0 ⇒ e - 1 x + 1 e x - 1 ≥ 0 ⇒ x ϵ - ∞ , 1 1 - e ∪ 1 , ∞ Also x x - 1 - e ≤ 0 e - 1 x - e x - 1 ≥ 0 ⇒ x ∈ - ∞ , 1 1 - e ∪ e e - 1 , ∞ As Range of x x - 1 is R + - 1 ⇒ Range of f is R - 0 or - ∞ , 0 ∪ 0 , ∞ Range of g is – π 2 , π 2 - 0 or – π 2 , 0 ∪ 0 , π 2 Now P → 4 , Q → 2 , R → 1 , S → 1 Hence A is correct.