Let the function f x = x + 1 . The number of values of x ∈ - 2,2 for which f x - 3 , f x - 1 and f x + 1 are in the arithmetic progression is

Step-by-step solution

f x = x + 1 ⇒ f x - 3 = x - 2 f x - 1 = x , f x + 1 = x + 2 Given f x - 3 , f x - 1 , f ( x + 1 ) are in A.P. ⇒ 2 x = x - 2 + x + 2 Case 1 : x < - 2 - x + 2 - x - 2 = - 2 x ⇒ 0 = 0 ⇒ x < - 2 Case 2 : - 2 ≤ x < 0 - x + 2 + x + 2 = - 2 x ⇒ x = - 2 ⇒ x = - 2 Case 3 : 0 ≤ x < 2 - x + 2 + x + 2 = 2 x ⇒ x = 2 ⇒ no solution Case 4 : x ≥ 2 x - 2 + x + 2 = 2 x ⇒ 0 = 0 ⇒ x ≥ 2 Taking union, we get x ≤ - 2 or x ≥ 2 But. x ∈ - 2,2 Taking intersection, we get, x = - 2 and 2