Let $g(x)=\int\limits_0^x f(t) d t$, 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,

$0 < g(2)<2$

We have,

$g(x)=\int\limits_0^x f(t) d t$

$\Rightarrow g(2)=\int\limits_0^2 f(t) d t=\int\limits_0^1 f(t) d t+\int\limits_1^2 f(t) d t$

We know that

$m \leq f'(x) \leq M$ for $x \in[a, b]$

$\Rightarrow m(b-a) \leq \int\limits_a^b f(x) d x \leq M(b-a)$

∴  $\frac{1}{2} \leq f(t) \leq 1$ for $t \in[0,1]$ and, $0 \leq f(t) \leq \frac{1}{2}$ for $t \in[1,2]$

$\Rightarrow \frac{1}{2}(1-0) \leq \int\limits_0^1 f(t) d t \leq 1(1-0)$

and, $0(2-1) \leq \int\limits_1^2 f(t) d t \leq \frac{1}{2}(2-1)$

$\Rightarrow \frac{1}{2} \leq \int\limits_0^1 f(t) d t \leq 1$ and, $0 \leq \int\limits_1^2 f(t) d t \leq \frac{1}{2}$

$\Rightarrow \frac{1}{2} \leq \int\limits_0^1 f(t) d t+\int\limits_1^2 f(t) d t \leq \frac{3}{2}$

$\Rightarrow \frac{1}{2} \leq \int\limits_0^1 f(t) d t+\int\limits_1^2 f(t) d t \leq \frac{3}{2}$

$\Rightarrow \frac{1}{2} \leq g(2) \leq \frac{3}{2} \Rightarrow 0 < g(2)<2$