Let g(x)= _0^xf(t)\,dt, where f is continuous function in [0,3] such that 13 f(t) 1 for all t (0,1) and 0 f(t) 12 for all t (1,3). The largest possible interval in which g(3) lies:

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Accepted Answer

Correct answer: C. (c) JEE Main 2021 13 f (t ) 1 ∀t ∈ (0 ,1 ) 0 f (t ) 12 ∀t ∈ (1 ,3 ) Now, g (3 )= f (t ) dt= f (t ) dt+ f (t ) dt ∵ 13 dt f (t ) dt 1 ∙ dt … (i) And 0 dt f (1 ) dt 12 dt … (ii) Adding, we get 13+0 g (3 ) 1+ 12(3-1) 13 g (3 ) 2

$Let g (x) = \int_{0}^{x} f (t) d t, where f is continuous function in [0,3] such that \frac{1}{3} \leq f (t) \leq 1 for all t \in (0, 1) and 0 \leq f (t) \leq \frac{1}{2} for all t \in (1, 3) . The largest possible interval in which g (3) lies:$

Solution

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