Let g x = ∫ 0 x f t d t , where f is continuous function in [ 0 , 3 ] such that 1 3 ≤ f t ≤ 1 for all t ∈ [ 0 , 1 ] and 0 ≤ f t ≤ 1 2 for all t ∈ ( 1 , 3 ] . The largest possible interval in which g ( 3 ) lies is :

Step-by-step solution

1 3 ≤ f t ≤ 1 ∀ t ∈ 0 , 1 0 ≤ f t ≤ 1 2 ∀ t ∈ ( 1 , 3 ] Now, g 3 = ∫ 0 3 f t d t = ∫ 0 1 f t d t + ∫ 1 3 f t d t ∵ ∫ 0 1 1 3 d t ≤ ∫ 0 1 f t d t ≤ ∫ 0 1 1 . d t    . . . 1 and ∫ 1 3 0 d t ≤ ∫ 1 3 f 1 d t ≤ ∫ 1 3 1 2 d t    . . . 2 Adding, we get 1 3 + 0 ≤ g 3 ≤ 1 + 1 2 3 - 1 1 3 ≤ g 3 ≤ 2