31≤f(t)≤1∀t∈[0,1]
0≤f(t)≤21∀t∈(1,3]
Now, g(3)=∫03f(t)dt=∫01f(t)dt+∫13f(t)dt
∵∫0131dt≤∫01f(t)dt≤∫011.dt...(1)
and ∫130dt≤∫13f(1)dt≤∫1321dt...(2)
Adding, we get
31+0≤g(3)≤1+21(3−1)
31≤g(3)≤2
Let g(x)=∫0xf(t)dt, where f is continuous function in [0,3] such that 31≤f(t)≤1 for all t∈[0,1] and 0≤f(t)≤21 for all t∈(1,3].
The largest possible interval in which g(3) lies is :
Held on 18 Mar 2021 · Verified 6 Jul 2026.
[−1,−21]
[−23,−1]
[31,2]
[1,3]
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