Indefinite Integration 2 Question 3
3. Let $g(x)=\int _0^{x} f(t) d t$, where $f$ is such that $\frac{1}{2} \leq f(t) \leq 1$ for $t \in[0,1]$ and $0 \leq f(t) \leq \frac{1}{2}$ for $t \in[1,2]$. Then, $g(2)$ satisfies the inequality
$(2000,2 M)$
(a) $-\frac{3}{2} \leq g(2)<\frac{1}{2}$
(b) $0 \leq g(2)<2$
(c) $\frac{3}{2}<g(2) \leq 5 / 2$
(d) $2<g(2)<4$
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Answer:
Correct Answer: 3. (b)
Solution:
- Given, $g(x)=\int _0^{x} f(t) d t$
$\Rightarrow \quad g(2)=\int _0^{2} f(t) d t=\int _0^{1} f(t) d t+\int _1^{2} f(t) d t$
Now, $\quad \frac{1}{2} \leq f(t) \leq 1$ for $t \in[0,1]$ We get $\int _0^{1} \frac{1}{2} d t \leq \int _0^{1} f(t) d t \leq \int _0^{1} 1 d t$
$\Rightarrow \quad \frac{1}{2} \leq \int _0^{1} f(t) d t \leq 1$
Again, $\quad 0 \leq f(t) \leq \frac{1}{2}$ for $t \in[1,2]$
$$ \begin{aligned} \Rightarrow & \int _1^{2} 0 d t & \leq \int _1^{2} f(t) d t \leq \int _1^{2} d t \\ \Rightarrow & 0 & \leq \int _1^{2} f(t) d t \leq \frac{1}{2} \end{aligned} $$
From Eqs. (i) and (ii), we get
$$ \begin{aligned} & & \frac{1}{2} \leq \int _0^{1} f(t) d t+\int _1^{2} f(t) d t \leq \frac{3}{2} \\ \Rightarrow & \frac{1}{2} & \leq g(2) \leq \frac{3}{2} \\ \Rightarrow & 0 & \leq g(2)<2 \end{aligned} $$
