Question
Let $$g\left( x \right) = \int\limits_0^x {f\left( t \right)dt,} $$ where $$f$$ is such that $$\frac{1}{2} \leqslant f\left( t \right) \leqslant 1,$$ for $$t \in \left[ {0,\,1} \right]$$ and $$0 \leqslant f\left( t \right) \leqslant \frac{1}{2},$$ for $$t \in \left[ {1,\,2} \right]$$
Then $$g\left( 2 \right)$$ satisfies the inequality -
A.
$$ - \frac{3}{2} \leqslant g\left( 2 \right) < \frac{1}{2}$$
B.
$$0 \leqslant g\left( 2 \right) < 2$$
C.
$$\frac{3}{2} < g\left( 2 \right) \leqslant \frac{5}{2}$$
D.
$$2 < g\left( 2 \right) < 4$$
Answer :
$$0 \leqslant g\left( 2 \right) < 2$$
Solution :
$$\eqalign{
& g\left( x \right) = \int\limits_0^x {f\left( t \right)dt} \cr
& \Rightarrow g\left( 2 \right) = \int\limits_0^2 {f\left( t \right)dt} = \int\limits_0^1 {f\left( t \right)dt} + \int\limits_1^2 {f\left( t \right)dt} \cr
& {\text{Now,}}\,\frac{1}{2} \leqslant f\left( t \right) \leqslant 1\,{\text{for }}t \in \left[ {0,\,1} \right] \cr
& {\text{We get }}\int\limits_0^1 {\frac{1}{2}dt} \leqslant \int\limits_0^1 {f\left( t \right)dt} \leqslant \int\limits_0^1 {1dt} \cr} $$
(applying line integral on inequality)
$$\eqalign{
& \Rightarrow \frac{1}{2} \leqslant \int\limits_0^1 {f\left( t \right)dt} \leqslant 1.....(1) \cr
& {\text{Again, 0}} \leqslant f\left( t \right) \leqslant \frac{1}{2}\,{\text{for }}t \in \left[ {1,\,2} \right] \cr
& {\text{We get }}\int\limits_1^2 {0dt} \leqslant \int\limits_1^2 {f\left( t \right)dt} \leqslant \int\limits_1^2 {\frac{1}{2}dt} \cr} $$
(applying line integral on inequality)
$$ \Rightarrow 0 \leqslant \int\limits_1^2 {f\left( t \right)dt} \leqslant \frac{1}{2}.....(2)$$
From (1) and (2), we get
$$\frac{1}{2} \leqslant \int\limits_0^1 {f\left( t \right)dt} + \int\limits_1^2 {f\left( t \right)dt} \leqslant \frac{3}{2}\,\,{\text{or}}\,\,\frac{1}{2} \leqslant g\left( 2 \right) \leqslant \frac{3}{2}$$
$$ \Rightarrow 0 \leqslant g\left( 2 \right) \leqslant 2$$ is the most appropriate solution.