Let $\mathrm{g}(\mathrm{x})=\int_{0}^{x} f(\mathrm{t}) \mathrm{dt}$, 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 $\mathrm{g}(3)$ lies is:
Correct Option: , 3
$\frac{1}{3} \leq f(\mathrm{t}) \leq 1 \forall \mathrm{t} \in[0,1]$
$0 \leq f(\mathrm{t}) \leq \frac{1}{2} \forall \mathrm{t} \in(1,3]$
Now, $\mathrm{g}(3)=\int_{0}^{3} f(\mathrm{t}) \mathrm{d} \mathrm{t}=\int_{0}^{1} f(\mathrm{t}) \mathrm{d} \mathrm{t}+\int_{1}^{3} f(\mathrm{t}) \mathrm{d} \mathrm{t}$
$\because \int_{0}^{1} \frac{1}{3} \mathrm{~d} \mathrm{t} \leq \int_{0}^{1} f(\mathrm{t}) \mathrm{dt} \leq \int_{0}^{1} 1 . \mathrm{dt} \quad \ldots(1)$
and $\int_{1}^{3} 0 \mathrm{dt} \leq \int_{1}^{3} f(1) \mathrm{dt} \leq \int_{1}^{3} \frac{1}{2} \mathrm{dt} \quad \ldots(2)$
Adding, we get
$\frac{1}{3}+0 \leq g(3) \leq 1+\frac{1}{2}(3-1)$
$\frac{1}{3} \leq g(3) \leq 2$
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