SATHEE: Indefinite Integration 2 Question 3

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) $0 \leq g (2) < 2$

(d) $2 < g (2) < 4$

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Answer:

Correct Answer: 3. (b)

Solution:

Given, $g (x) = \int_{0}^{x} f (t) d t$

$\Rightarrow g (2) = \int_{0}^{2} f (t) d t = \int_{0}^{1} f (t) d t + \int_{1}^{2} f (t) d t$

Now, $\frac{1}{2} \leq f (t) \leq 1$ for $t \in [0, 1]$ We get $\int_{0}^{1} \frac{1}{2} d t \leq \int_{0}^{1} f (t) d t \leq \int_{0}^{1} 1 d t$

$\Rightarrow \frac{1}{2} \leq \int_{0}^{1} f (t) d t \leq 1$

Again, $0 \leq f (t) \leq \frac{1}{2}$ 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

$\begin{aligned} \frac{1}{2} \leq \int_{0}^{1} f (t) d t + \int_{1}^{2} f (t) d t \leq \frac{3}{2} \\ \Rightarrow & \frac{1}{2} & \leq g (2) \leq \frac{3}{2} \\ \Rightarrow & 0 & \leq g (2) < 2 \end{aligned}$

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

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Indefinite Integration 2 Question 3

3. Let g(x)=∫0xf(t)dt, where f is such that 12≤f(t)≤1 for t∈[0,1] and 0≤f(t)≤12 for t∈[1,2]. Then, g(2) satisfies the inequality

Analytical & Descriptive Questions

Answer:

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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