Explanation

(b)
Here, $g(2) = \int_{0}^{2}{f(t)dt}$
$= \int_{0}^{1}{f(t)dt +}\int_{1}^{2}{f(t)dt}$
As $\frac{1}{2} \leq f(t) \leq 1\ for\ 0 \leq t \leq 1$
$\Rightarrow \ \int_{0}^{1}{\frac{1}{2}dt \leq}\int_{0}^{1}{f(t)dt \leq}\int_{0}^{1}{dt}$
$\Rightarrow \ \frac{1}{2} \leq \int_{0}^{1}{f(t)dt \leq 1}$ …(i)
As $0 \leq f(t) \leq \frac{1}{2}for\ 1 < t \leq 2$
$\Rightarrow \ \int_{1}^{2}{0\ dt \leq \int_{1}^{2}{f(t)dt \leq}\int_{1}^{2}{\frac{1}{2}dt}}$
$\therefore\ \ 0 \leq \int_{1}^{2}{f(t)dt \leq \frac{1}{2}}$ …(ii)
On adding Eqs.(i) and (ii), we get
$\frac{1}{2} \leq g(2) \leq \frac{3}{2}$
$\therefore g(2)$ satisfies the inequality $0 \leq g(2) < 2$