If \[\left( {a,{a^2}} \right)\] falls inside the angle made by the lines \[y = \dfrac{x}{2}, x > 0\] and \[y = 3x, x > 0\]. Then find the interval of \[a\].
A. \[\left( {3,\infty } \right)\]
B. \[\left( {\dfrac{1}{2},3} \right)\]
C. \[\left( { - 3, - \left( {\dfrac{1}{2}} \right)} \right)\]
D. \[\left( {0,\dfrac{1}{2}} \right)\]

Hint In the given question, two equations of lines and a point are given. We substitute the coordinates of the point in both equations and apply the conditions of greater than 0 or less than 0. In the end, simplify both inequality equations and get the interval of \[a\].

Formula used
If a point \[\left( {{x_1},{y_1}} \right)\] lies above the line \[ax + by = c\], then the inequality equation is \[a{x_1} + b{y_1} > c\].
If a point \[\left( {{x_1},{y_1}} \right)\] lies below the line \[ax + by = c\], then the inequality equation is \[a{x_1} + b{y_1} < c\].

Complete step by step solution:
The given equations of lines are \[y = \dfrac{x}{2}, x > 0\] and \[y = 3x, x > 0\].
The point \[\left( {a,{a^2}} \right)\] lies inside the angle made by the given two lines.

The point \[\left( {a,{a^2}} \right)\] lies above the line \[y - \dfrac{x}{2} = 0, x > 0\].
So, the inequality equation is
\[{a^2} - \dfrac{a}{2} > 0\]
\[ \Rightarrow \]\[a\left( {a - \dfrac{1}{2}} \right) > 0\]
\[ \Rightarrow \]\[a > 0\] and \[a > \dfrac{1}{2}\] \[.....\left( 1 \right)\]

The point \[\left( {a,{a^2}} \right)\] lies below the line \[y - 3x = 0, x > 0\].
So, the inequality equation is
\[{a^2} - 3a < 0\]
\[ \Rightarrow \]\[a\left( {a - 3} \right) < 0\]
\[ \Rightarrow \]\[a < 0\] and \[a < 3\] \[.....\left( 2 \right)\]

From equation \[\left( 1 \right)\] and \[\left( 2 \right)\], we get
\[a > \dfrac{1}{2}\] or \[a < 0\] and \[0 < a < 3\]
\[ \Rightarrow \]\[\dfrac{1}{2} < a < 3\]
\[ \Rightarrow \]\[a \in \left( {\dfrac{1}{2},3} \right)\]

Hence the correct option is B.

Note:The “and” word represents the intersection of two sets and the “or” word denotes the union of two sets. Students are often confused by these two words.
Students often get confused in the conversion of the inequality into an interval.  Use the following rules to understand the conversion.
\[a < x < b \Rightarrow x \in \left( {a,b} \right)\]
\[a < x \le b \Rightarrow x \in \left( {a,b} \right]\]
\[a \le x < b \Rightarrow x \in \left[ {a,b} \right)\]
\[a \le x \le b \Rightarrow x \in \left[ {a,b} \right]\]