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Question: If \(\left( {a,{a^2}} \right)\) falls inside the angle made by the lines \(y = \dfrac{x}{2}\), \(x >...

If (a,a2)\left( {a,{a^2}} \right) falls inside the angle made by the lines y=x2y = \dfrac{x}{2}, x>0x > 0 and y=3xy = 3x, x>0x > 0, then aa belong to
(a) (0,12)\left( {0,\dfrac{1}{2}} \right)
(b) (3,)\left( {3,\infty } \right)
(c) (12,3)\left( {\dfrac{1}{2},3} \right)
(d) (3,12)\left( { - 3, - \dfrac{1}{2}} \right)

Explanation

Solution

We will first draw the lines y=x2y = \dfrac{x}{2} and y=3xy = 3x. Now the point (a,a2)\left( {a,{a^2}} \right) must lie between the lines. Hence, the value of aa must satisfy the inequalities a2a2>0{a^2} - \dfrac{a}{2} > 0 and a23a<0{a^2} - 3a < 0. Plot the roots on the number line and find the intersection of the values that satisfy the value of aa.

Complete step by step solution:
We are given that there are two lines, y=x2y = \dfrac{x}{2} and y=3xy = 3x.
Let us first draw these lines on the graph.

We have to find the values for which the point (a,a2)\left( {a,{a^2}} \right) should lie between the lines y=x2y = \dfrac{x}{2} and y=3xy = 3x.
If we substitute aa for xx and a2{a^2} for yy
Then, a2a2>0{a^2} - \dfrac{a}{2} > 0 and a23a<0{a^2} - 3a < 0 for the point to lie between the lines.
We have to solve both the inequalities.
If a2a2>0{a^2} - \dfrac{a}{2} > 0, this implies, a(a12)>0a\left( {a - \dfrac{1}{2}} \right) > 0
Plot the roots on a number line.

Take the testing point between 12\dfrac{1}{2} and \infty , say 1
Then, 1(112)>01\left( {1 - \dfrac{1}{2}} \right) > 0 is true, then it is positive.
In the similar manner, we have,

Hence, the required region here is (,0)(12,)\left( { - \infty ,0} \right) \cup \left( {\dfrac{1}{2},\infty } \right)
Next, find the region of a23a<0{a^2} - 3a < 0 which is a(a3)<0a\left( {a - 3} \right) < 0
Plot the roots and determine the sign.
Let 4 be the point between 3 and \infty , say 4.
4(43)>04\left( {4 - 3} \right) > 0

And the required area is (0,3)\left( {0,3} \right)
We have to take the intersection of both the intervals.
((,0)(12,))(0,3)=(12,3)\left( {\left( { - \infty ,0} \right) \cup \left( {\dfrac{1}{2},\infty } \right)} \right) \cap \left( {0,3} \right) = \left( {\dfrac{1}{2},3} \right)
Thus, aa belongs to (12,3)\left( {\dfrac{1}{2}, 3} \right)

Hence, option (c) is correct.

Note:
Formation of the inequalities should be correct. Also, one must know how to find values of a given inequality. At last, do not forget to take intersection of the values of both the inequalities.