Letg(x)=∫0xf(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 g3 lies is :

∫0113dt+∫130.dt≤g(3)≤∫011dt+∫1312·dt ⇒13≤g(3)≤2

Letg(x)=∫0xf(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 g3 lies is :

∫0113dt+∫130.dt≤g(3)≤∫011dt+∫1312·dt ⇒13≤g(3)≤2

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

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$[- 1, - \frac{1}{2}]$

$[1, 3]$

$[\frac{1}{3}, 2]$

$[- \frac{3}{2}, - 1]$

answer is C.

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$\int_{0}^{1} \frac{1}{3} d t + \int_{1}^{3} 0 . d t \leq g (3) \leq \int_{0}^{1} 1 d t + \int_{1}^{3} \frac{1}{2} \cdot d t \Rightarrow \frac{1}{3} \leq g (3) \leq 2$

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Letg(x)=∫0xf(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 g3 lies is :

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