Solve this following

Question:

Let $\mathrm{g}(\mathrm{x})=\int_{0}^{\mathrm{x}} f(\mathrm{t}) \mathrm{dt}$, where $f$ is continuous

function in $[0,3]$ such that $\frac{1}{3} \leq f(\mathrm{t}) \leq 1$ for all

$\mathrm{t} \in[0,1]$ and $0 \leq f(\mathrm{t}) \leq \frac{1}{2}$ for all $\mathrm{t} \in(1,3]$

The largest possible interval in which $\mathrm{g}(3)$ lies is :

 

  1. $\left[-1,-\frac{1}{2}\right]$

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

  3. $\left[\frac{1}{3}, 2\right]$

  4. $[1,3]$


Correct Option: 3,

Solution:

$\frac{1}{3} \leq f(\mathrm{t}) \leq 1 \forall \mathrm{t} \in[0,1]$

$0 \leq f(\mathrm{t}) \leq \frac{1}{2} \forall \mathrm{t} \in(1,3]$

Now, $\mathrm{g}(3)=\int_{0}^{3} f(\mathrm{t}) \mathrm{dt}=\int_{0}^{1} f(\mathrm{t}) \mathrm{dt}+\int_{1}^{3} f(\mathrm{t}) \mathrm{dt}$

$\because \int_{0}^{1} \frac{1}{3} \mathrm{dt} \leq \int_{0}^{1} f(\mathrm{t}) \mathrm{dt} \leq \int_{0}^{1} 1 . \mathrm{dt}$ ..............(1)

and $\int_{1}^{3} 0 \mathrm{dt} \leq \int_{1}^{3} f(1) \mathrm{dt} \leq \int_{1}^{3} \frac{1}{2} \mathrm{dt}$ ..............(2)

Adding, we get

$\frac{1}{3}+0 \leq g(3) \leq 1+\frac{1}{2}(3-1)$

$\frac{1}{3} \leq \mathrm{g}(3) \leq 2$

 

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