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Chapter 18: Problem 81

If \(\left(a, a^{2}\right)\) falls inside the angle made by the lines \(y=\frac{x}{2}\), \(x>0\) and \(y=3 x, x>0\), then \(a\) belongs to (A) \(\left(0, \frac{1}{2}\right)\) (B) \((3, \infty)\) (C) \(\left(\frac{1}{2}, 3\right)\) (D) \(\left(-3,-\frac{1}{2}\right)\)

Short Answer

Expert verified
(C) \(\left(\frac{1}{2}, 3\right)\)

Step by step solution

01

Understanding the Boundary Lines

The lines represented by the equations \(y = \frac{x}{2}\) and \(y = 3x\) define the boundaries of the angle within which the point \((a, a^2)\) must fall. Both lines have a positive slope, which means the angle is in the first quadrant where both \(x > 0\) and \(y > 0\).
02

Set the Boundary Conditions for Point \((a, a^2)\)

For the point \((a, a^2)\) to be inside the angle formed by the lines, the y-coordinate must satisfy: \(y = a^2 > \frac{x}{2}\) and \(y = a^2 < 3x\). Substituting \(x = a\), we get the inequalities: \(a^2 > \frac{a}{2}\) and \(a^2 < 3a\).
03

Solve the Inequality \(a^2 > \frac{a}{2}\)

Rearrange the inequality to isolate terms involving \(a\): \(a^2 - \frac{a}{2} > 0\). This simplifies to \((2a^2 - a) > 0\), further factorized to \(a(2a - 1) > 0\). Solve \(a = 0\) and \(2a - 1 = 0\) to find the critical points, yielding roots \(a = 0\) and \(a = \frac{1}{2}\). Testing intervals reveals that \(a > \frac{1}{2}\) satisfies the inequality.
04

Solve the Inequality \(a^2 < 3a\)

Rearrange and factor to find critical points: \(a^2 - 3a < 0\), which gives \(a(a - 3) < 0\). Solving \(a = 0\) and \(a - 3 = 0\), critical points are \(a = 0\) and \(a = 3\). Testing intervals, the inequality holds for \(0 < a < 3\).
05

Determine the Intersection of Solutions

Combine the solutions from Step 3, \(a > \frac{1}{2}\), and Step 4, \(0 < a < 3\). The intersection of these conditions is \(\frac{1}{2} < a < 3\).
06

Select the Correct Option

Match the range found in Step 5, \(\frac{1}{2} < a < 3\), to the given options: (A) \(\left(0, \frac{1}{2}\right)\), (B) \((3, \infty)\), (C) \(\left(\frac{1}{2}, 3\right)\), (D) \(\left(-3,-\frac{1}{2}\right)\). The correct choice is (C).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Inequalities
In the realm of coordinate geometry, inequalities serve as a powerful tool to determine relationships between expressions. When it comes to understanding how points fit within regions bounded by lines, such as our exercise, inequalities play a pivotal role. We delve into this through:
  • Identifying the inequality: Consider the function and the boundary it must adhere to. In our exercise, the point \(a, a^2\) must satisfy inequalities that position it between two lines: \(y = \frac{x}{2}\) and \(y = 3x\).
  • Simplifying inequalities: Often, you'll need to manipulate an inequality to find solutions more easily. This involves algebraic operations such as adding, subtracting, multiplying, or dividing. For instance, in our problem, \(a^2 > \frac{a}{2}\) transforms into \((2a^2 - a) > 0\).
  • Solving and testing intervals: Once simplified, we seek the critical values (where inequality equals zero) and test intervals between these points to determine where the inequality holds true. For example, solving \(a(2a - 1) > 0\), critical values were found, leading to the determination that \(a > \frac{1}{2}\).
  • Intersecting solutions: It's common to solve multiple inequalities and find their intersection, as we did with \(a > \frac{1}{2}\) and \(0 < a < 3\), revealing the final range for \(a\).
Understanding these steps is crucial for effectively solving inequalities in problems related to coordinates and geometry, ensuring that all conditions are met for the given region.
Quadratic Functions
Quadratic functions are essential components in coordinate geometry given their common occurrence. They have a standard form: \(ax^2 + bx + c = 0\). Understanding how to manipulate and solve these functions is valuable, especially in relation to inequality problems.Quadratic equations, such as \(a^2 - \frac{a}{2} > 0\), showcase a parabola on the graph. Solving such inequalities usually involves:
  • Finding roots: Identify where the quadratic crosses the x-axis by setting the equation to zero. This is done via factorization, completing the square, or using the quadratic formula. In our example, solving \(a(2a - 1) > 0\) provided roots at \(a = 0\) and \(a = \frac{1}{2}\).
  • Identifying intervals: Once roots are found, they divide the x-axis into intervals you can test to see where the quadratic is positive or negative. Testing values within these intervals helps determine where the inequality is satisfied.
  • Understanding concavity: Quadric functions typically form a U-shaped curve, either \(鈭猏) opening upwards or \(鈭‐) opening downwards depending on the coefficient of \(a^2\). This affects the solution set for the inequality testing.
Mastering the behavior and characteristics of quadratic functions is critical, as they often provide the frameworks and constraints needed for solving more complex coordinate geometry problems.
Graphical Analysis
Graphical analysis is a crucial aspect of solving coordinate geometry problems. It aids in visualizing equations, inequalities, and the areas they define within a given plane.Visualizing the elements from the exercise:
  • Understanding line graphs: Graphs of \(y = \frac{x}{2}\) and \(y = 3x\) are essential. Both lines commence from the origin but with different slopes, defining a wedge in the first quadrant.
  • Locating points: Points like \(a, a^2\) need to be analyzed in relation to these lines. By substituting \(x = a\), we were able to evaluate whether they fall within the region.
  • Interpreting graph positions: Knowing whether a point lies above or below a line aids in solving inequalities. For example, confirming \(a^2 > \frac{a}{2}\) places the parabola above the line \(y = \frac{x}{2}\).
  • Demonstrating solutions on the graph: Once inequalities are solved algebraically, representing them visually confirms the understanding of the area of solutions in relation to the plotted lines and curves.
Using graphical methods not only supports the algebraic approach but also enhances understanding of inequalities and quadratic functions by visual confirmation. It is a vital skill for anyone learning coordinate geometry, providing clarity to complex problems.

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Most popular questions from this chapter

The line \(x+y=1\) meets \(x\)-axis at \(A\) and \(y\)-axis at \(B \cdot P\) is the mid- point of \(A B \cdot P_{1}\) is the foot of the perpendicular from \(P\) to \(O A ; M_{1}\) is that from \(P_{1}\) to \(O P ; P_{2}\) is that from \(M_{1}\) to \(O A\) and so on. If \(P_{n}\) denotes the foot of the \(n\)th perpendicular on \(O A\) from \(M_{n-1}\), then \(O P_{n}\) is equal to (A) \(\frac{1}{2^{n}}\) (B) \(\frac{1}{2^{n-1}}\) (C) \(\frac{1}{2^{n-2}}\) (D) none of these

The equation of a family of lines is given by \((2+3 t)\) \(x+(1-2 t) y+4=0\), where \(t\) is the parameter. The equation of a straight line, belonging to this family, at the maximum distance from the point \((2,3)\) is (A) \(21 x+14 y=0\) (B) \(21 x-14 y=0\) (C) \(14 x-21 y=0\) (D) none of these

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The coordinates of a point \(P\) on the line \(3 x+2 y+10\) \(=0\) such that \(|P A-P B|\) is maximum where \(A\) is \((4,2)\) and \(B\) is \((2,4)\), are (A) \((22,28)\) (B) \((22,-28)\) (C) \((-22,28)\) (D) \((-22,-28)\)

If the sum of the slopes of the lines given by \(x^{2}-\) \(2 c x y-7 y^{2}=0\) is four times their product, then \(c\) has the value (A) 1 (B) \(-1\) (C) 2 (D) \(-2\)
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