Let f , g : R → R be functions defined by f x = x , x < 0 1 - x , x ≥ 0 and g x = e x - x , x < 0 x - 1 2 - 1 , x ≥ 0 where x denote the greatest integer less than or equal to x . Then, the function fog is discontinuous at exactly

Step-by-step solution

Let f , g : R → R be functions defined by f x = x ,    x < 0 1 - x ,    x ≥ 0 and g x = e x - x , x < 0 x - 1 2 - 1 , x ≥ 0 ⇒ f o g x = g x ,   g x < 0 1 - g x ,   g x ≥ 0 After taking intersection we get, ⇒ f o g x = 1 + x - e x ,   x < 0 x - 1 2 - 1 ,     0 ≤ x < 2 2 - x - 1 2 ,   x ≥ 2 Now, checking continuity at x = 0 and at x = 2 We get, f o g 0 + = - 1 , f o g 0 = 0 Hence, discontinuous at x =