Let g(x) = int_0^x f(t) , dt where frac{1}{2} | vedclass

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(B) Given $g(2) = \int_0^2 f(t) \, dt = \int_0^1 f(t) \, dt + \int_1^2 f(t) \, dt$.For $t \in [0, 1]$,we have $\frac{1}{2} \le f(t) \le 1$. Integratin

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$0 \le g(2) < 2$

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(B) Given $g(2) = \int_0^2 f(t) \, dt = \int_0^1 f(t) \, dt + \int_1^2 f(t) \, dt$.For $t \in [0, 1]$,we have $\frac{1}{2} \le f(t) \le 1$. Integrating this over $[0, 1]$:$\int_0^1 \frac{1}{2} \, dt \le \int_0^1 f(t) \, dt \le \int_0^1 1 \, dt$$\frac{1}{2} \le \int_0^1 f(t) \, dt \le 1 \quad \dots (i)$For $t \in (1, 2]$,we have $0 \le f(t) \le \frac{1}{2}$. Integrating this over $(1, 2]$:$\int_1^2 0 \, dt \le \int_1^2 f(t) \, dt \le \int_1^2 \frac{1}{2} \, dt$$0 \le \int_1^2 f(t) \, dt \le \frac{1}{2} \quad \dots (ii)$Adding $(i)$ and $(ii)$:$\frac{1}{2} + 0 \le \int_0^1 f(t) \, dt + \int_1^2 f(t) \, dt \le 1 + \frac{1}{2}$$\frac{1}{2} \le g(2) \le \frac{3}{2}$.Since $\frac{1}{2} \le g(2) \le \frac{3}{2}$,the value of $g(2)$ lies within the interval $[0, 2)$. Thus,$0 \le g(2) < 2$ is the correct inequality.

Let $g(x) = \int_0^x f(t) \, dt$ where $\frac{1}{2} \le f(t) \le 1$ for $t \in [0, 1]$ and $0 \le f(t) \le \frac{1}{2}$ for $t \in (1, 2]$. Then which of the following is true for $g(2)$?

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