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Chapter 14: Problem 1329

If \(\left(\mathrm{a}, \mathrm{a}^{2}\right)\) falls inside the angle made by the lines \(\mathrm{y}=(\mathrm{x} / 2), \mathrm{x}>0\) and \(\mathrm{y}=3 \mathrm{x}, \mathrm{x}>0\) the 'a' belongs to (a) \((3, \infty)\) (b) \([(1 / 2), 3]\) (c) \([(-3),-(1 / 2)]\) (d) \([0,(1 / 2)]\)

Short Answer

Expert verified
The 'a' belongs to the interval \((\frac{1}{2}, 3)\), which corresponds to answer choice (b).

Step by step solution

01

Graph and Intersection of Lines

We are given two lines: \(y = \frac{x}{2}\) where \(x > 0\) \(y = 3x\) where \(x > 0\) These lines both pass through the origin (0, 0). To find the intersection of these lines, we'll set the equations equal to each other: \(\frac{x}{2} = 3x\) Solve for x: \(x = \frac{1}{5}\) Now, we find the y-coordinate of the intersection. \(y = 3(\frac{1}{5}) = \frac{3}{5}\) So the point of intersection is (\(\frac{1}{5}, \frac{3}{5}\)).
02

Determine Which Line Is Above the Other

This will help us understand the angle between the lines. To do this, we can evaluate how y-values change as x-values increase for each line. We can see that as \(x\) increases, the slope of the line \(y = 3x\) also increases faster than the line \(y = \frac{x}{2}\). Therefore, the line \(y = 3x\) is above the line \(y = \frac{x}{2}\) when \(x > 0\).
03

Determine the Conditions for the Point to Fall Inside the Angle

For the point \((a, a^2)\) to fall inside the angle, it needs to be above the lower line \(\frac{x}{2}\) and below the upper line \(3x\). This means that the following conditions must be met: \(a^2 > \frac{a}{2}\) \(a^2 < 3a\)
04

Solving the Inequalities

First inequality: \(a^2 > \frac{a}{2}\) Multiplying both sides of the inequality by 2, we get: \(2a^2 > a\) Now, subtract a from both sides: \(2a^2 - a > 0\) Second inequality: \(a^2 < 3a\) Subtracting 3a from both sides, we get: \(a^2 - 3a < 0\) Now, we will analyze both inequalities considering the \(x\) values.
05

Interval Based on the Inequalities

For the inequality \(2a^2 - a > 0\), we can factor out a: \(a(2a-1) > 0\) This inequality holds true for \(a < 0\) and \(a > \frac{1}{2}\). For the inequality \(a^2 - 3a < 0\), we can factor out a: \(a(a-3) < 0\) This inequality holds true for \(0 < a < 3\).
06

Find the Intersection of the Intervals

To determine the interval 'a' belongs to where the point falls inside the angle, we need to find the intersection of the intervals from both inequalities. The intersection of \(a < 0\) and \(0 < a < 3\) is empty, but the intersection of \(a > \frac{1}{2}\) and \(0 < a < 3\) is \(\frac{1}{2} < a < 3\). Therefore, the 'a' belongs to the interval \((\frac{1}{2}, 3)\), which corresponds to answer choice (b).

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Most popular questions from this chapter

The equation of line passing through the point of intersection of the lines \(3 \mathrm{x}-2 \mathrm{y}=0\) and \(5 \mathrm{x}+\mathrm{y}-2=0\) and making the angle of measure \(\tan ^{-1}(-5)\) with the positive direction of \(\mathrm{x}\) -axis is \(\ldots \ldots\) (a) \(3 x-2 y=0\) (b) \(5 x+y-2=0\) (c) \(5 \mathrm{x}+\mathrm{y}=0\) (d) \(3 \mathrm{x}+2 \mathrm{y}+1=0\)

A square of side 'a' lies above the \(\mathrm{x}\) -axis and has one vertex at the origin. The side passing through the origin makes an angle o. \([0<\alpha<(\pi / 4)]\) with the positive direction of \(\mathrm{x}\) -axis. The equation of the diagonal not passing through the origin is (a) \(y(\cos \alpha-\sin \alpha)-x(\sin \alpha-\cos \alpha)=a\) (b) \(y(\cos \alpha+\sin \alpha)+x(\sin \alpha-\cos \alpha)=a\) (c) \(y(\cos \alpha-\sin \alpha)-x(\sin \alpha+\cos \alpha)=a\) (d) \(y(\cos \alpha+\sin \alpha)+x(\cos \alpha-\sin \alpha)=a\)

If \(\mathrm{a}_{1}, \mathrm{a}_{2}, \mathrm{a}_{3}\) and \(\mathrm{b}_{1}, \mathrm{~b}_{2}, \mathrm{~b}_{3}\) are in geometric progression and their common ratios are equal, then the points \(\mathrm{A}\left(\mathrm{a}_{1}, \mathrm{~b}_{1}\right)\) \(\mathrm{B}\left(\mathrm{a}_{2}, \mathrm{~b}_{2}\right)\) and \(\mathrm{C}\left(\mathrm{a}_{3}, \mathrm{~b}_{3}\right) \ldots \ldots\) (a) lie on the same line (b) lie on a circle (c) lie on an ellipse (d) None of these

If \(5 \mathrm{x}+12 \mathrm{y}+13=0\) is transformed into \(x \cos \alpha+\mathrm{ysin} \alpha=\mathrm{p}\) then \(\alpha=? \alpha \in[-\pi, \pi]\) (a) \(\cos ^{-1}[-(5 / 13)]\) (b) \(\sin ^{-1}[-(12 / 13)]\) (c) \(\tan ^{-1}(12 / 5)-\pi\) (d) \(\tan ^{-1}(12 / 5)\)

If two vertices of a triangle are \((5,-1)\) and \((-2,3)\) and if its orthocenter lies at the origin then the coordinates of the third vertex are (a) \((4,7)\) (b) \((-4,-7)\) (c) \((2,-3)\) (d) \((5,-1)\)
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