Question

Let$\mathrm{g}\left(\mathrm{x}\right)={\int }_0^{\mathrm{t}}\mathrm{f}\left(\mathrm{t}\right)\mathrm{dt}$ where $\mathrm{f}$ is a continuous function in $[0,3]$ such that $(1/3)\le \mathrm{f}\left(\mathrm{t}\right)\le 1$ for all $\mathrm{t}\in [0,1]$ and $0\le \mathrm{f}\left(\mathrm{t}\right)\le \frac{1}{2}$ for all $\mathrm{t}\in (1,3]$. The largest possible interval in which $\mathrm{g}\left(3\right)$ lies is:

• A. $[1,3]$
• B. $\left[-1,\frac{–1}{2}\right]$
• C. $\left[\frac{-3}{2},–1\right]$
• D. $\left[\frac{1}{3},2\right]$

Answer 1

The correct option is D

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

Explanation for correct answer:

Finding the largest possible interval:

We are given with following data:

$(1/3)\le \mathrm{f}\left(\mathrm{t}\right)\le 1\forall \mathrm{t}\in [0,1]$ and

$0\le \mathrm{f}\left(\mathrm{t}\right)\forall \mathrm{t}\in (1,3]$

Also $\mathrm{g}\left(\mathrm{x}\right)={\int }_0^{\mathrm{t}}\mathrm{f}\left(\mathrm{t}\right)\mathrm{dt}$.

Now calculating $\mathrm{g}\left(3\right)$, we get:

$$\begin{gathered} \mathrm{g}\left(3\right)={\int }_0^3\mathrm{f}\left(\mathrm{t}\right)d\mathrm{t} \\ \Rightarrow \mathrm{g}\left(3\right)={\int }_0^1\mathrm{f}\left(\mathrm{t}\right)d\mathrm{t}+{\int }_1^3\mathrm{f}\left(\mathrm{t}\right)d\mathrm{t} \end{gathered}$$

Also, we have,

$$\begin{gathered} {\int }_0^1\frac{1}{3}d\mathrm{t}\le {\int }_0^1\mathrm{f}\left(\mathrm{t}\right)d\mathrm{t}\le {\int }_0^{1}1d\mathrm{t} \\ \Rightarrow \frac{1}{3}\left(1-0\right)\le {\int }_0^{\mathrm{t}}\mathrm{f}\left(\mathrm{t}\right)d\mathrm{t}\le 1\left(1-0\right) \\ \Rightarrow \frac{1}{3}\le {\int }_0^1\mathrm{f}\left(\mathrm{t}\right)d\mathrm{t}\le 1\left(1\right) \end{gathered}$$

Again we have,

$$\begin{gathered} {\int }_1^{3}0d\mathrm{t}\le {\int }_1^3\mathrm{f}\left(\mathrm{t}\right)d\mathrm{t}\le {\int }_1^3\frac{1}{2}d\mathrm{t} \\ \Rightarrow 0\le {\int }_1^3\mathrm{f}\left(\mathrm{t}\right)d\mathrm{t}\le \frac{1}{2}(3-1) \\ \Rightarrow 0\le {\int }_1^3\mathrm{f}\left(\mathrm{t}\right)d\mathrm{t}\le 1\left(2\right) \end{gathered}$$

Adding $\left(1\right)$ and $\left(2\right)$ we get:

$$\begin{gathered} \frac{1}{3}\le {\int }_0^3\mathrm{f}\left(\mathrm{t}\right)d\mathrm{t}\le 2 \\ \Rightarrow \frac{1}{3}\le \mathrm{g}\left(3\right)\le 2 \end{gathered}$$

Therefore, the largest possible interval in which the function $\mathrm{g}\left(3\right)$ lies is $\left[\frac{1}{3},2\right]$.

Hence, option (D) is the correct answer.