Let g (x) = ∫0x f (t) dt, where f is such that 1/2 ≤ f(t) ≤ 1 for t ∈ [0,1] and

0 ≤ f(t) ≤1/2 for t ∈[1,2]. Then g (2) satisfies the inequality:

g(2) = ∫₀² f (t) dt = ∫₀¹ f (t) dt   +  ∫₁² f (t) dt

1/2 ≤ f(t) ≤ 1 for t ∈ [0,1] and 0 ≤ f(t) ≤1/2 for t ∈[1,2].

 ∫₀¹1/2.dt ≤ ∫₀¹ f (t).dt ≤ ∫₀¹ 1.dt   and ∫₁² 0.dt ≤ ∫₁² f (t) dt ≤∫₁² 1/2.dt

1/2 ≤ ∫₀¹ f (t).dt ≤  1   and     0 ≤ ∫₁² f (t) dt ≤ 1/2

on adding above gives,

1/2 ≤ ∫₀¹ f (t).dt + ∫₁² f (t) dt ≤ 3/2

1/2 ≤ g(2) ≤ 3/2

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IIT Kgp - 05 batch