Past Exam Questions

To solve the inequality \(2|2x - a| < |x + 3a|\), we first consider the non-modular inequality:

\(2^2(2x - a)^2 < (x + 3a)^2\).

This simplifies to solving the quadratic inequality:

\(4(2x - a)^2 < (x + 3a)^2\).

Expanding both sides, we have:

\(4(4x^2 - 4ax + a^2) < x^2 + 6ax + 9a^2\).

Simplifying further, we get:

\(16x^2 - 16ax + 4a^2 < x^2 + 6ax + 9a^2\).

Rearranging terms gives:

\(15x^2 - 22ax - 5a^2 < 0\).

Solving the quadratic equation \(15x^2 - 22ax - 5a^2 = 0\) using the quadratic formula \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\), we find the critical values:

\(x = \frac{5}{3}a\) and \(x = -\frac{1}{5}a\).

The solution to the inequality is the interval between these critical values:

\(-\frac{1}{5}a < x < \frac{5}{3}a\).

To solve the inequality \(|5x - 3| < 2|3x - 7|\), we first consider the non-modular inequality:

\((5x - 3)^2 < 2^2(3x - 7)^2\).

This simplifies to:

\(11x^2 - 138x + 187 > 0\).

We solve the quadratic equation \(11x^2 - 138x + 187 = 0\) to find the critical values:

\(x = \frac{17}{11}\) and \(x = 11\).

The solution to the inequality is:

\(x < \frac{17}{11}\) or \(x > 11\).

To solve the inequality \(|x - 3| < 3x - 4\), we consider two cases based on the definition of absolute value.

Case 1: Assume \(x - 3 \, \geq \, 0\), which implies \(x \, \geq \, 3\).

In this case, the inequality becomes:

\(x - 3 < 3x - 4\)

Solving for \(x\):

\(x - 3 < 3x - 4\)

\(x - 3x < -4 + 3\)

\(-2x < -1\)

\(x > \frac{1}{2}\)

Since \(x \, \geq \, 3\), the solution for this case is \(x > 3\).

Case 2: Assume \(x - 3 \, < \, 0\), which implies \(x \, < \, 3\).

In this case, the inequality becomes:

\(-(x - 3) < 3x - 4\)

\(-x + 3 < 3x - 4\)

Solving for \(x\):

\(-x + 3 < 3x - 4\)

\(-x - 3x < -4 - 3\)

\(-4x < -7\)

\(x > \frac{7}{4}\)

Since \(x \, < \, 3\), the solution for this case is \(\frac{7}{4} < x < 3\).

Combining both cases, the solution to the inequality is \(x > \frac{7}{4}\).

To solve the inequality \(|2x + 1| < 3|x - 2|\), we consider the critical points where each expression inside the absolute value is zero.

1. Solve \(2x + 1 = 0\):

\(2x + 1 = 0 \Rightarrow x = -\frac{1}{2}\)

2. Solve \(x - 2 = 0\):

\(x - 2 = 0 \Rightarrow x = 2\)

These critical points divide the number line into intervals. We test each interval:

Interval 1: \(x < -\frac{1}{2}\)

Interval 2: \(-\frac{1}{2} < x < 2\)

Interval 3: \(x > 2\)

For each interval, remove the absolute values and solve the inequality:

1. For \(x < -\frac{1}{2}\), both expressions are negative:

\(-(2x + 1) < -3(x - 2)\)

\(2x + 1 > 3x - 6\)

\(1 + 6 > 3x - 2x\)

\(7 > x\)

2. For \(-\frac{1}{2} < x < 2\), \(2x + 1\) is positive and \(x - 2\) is negative:

\(2x + 1 < -3(x - 2)\)

\(2x + 1 < -3x + 6\)

\(2x + 3x < 6 - 1\)

\(5x < 5\)

\(x < 1\)

3. For \(x > 2\), both expressions are positive:

\(2x + 1 < 3(x - 2)\)

\(2x + 1 < 3x - 6\)

\(1 + 6 < 3x - 2x\)

\(7 < x\)

Combining the results, the solution is \(x < 1\) and \(x > 7\).

Start by considering the inequality \(|x - 4| < 2|3x + 1|\).

Rewrite it as \((x - 4)^2 < (2(3x + 1))^2\).

This simplifies to \((x - 4)^2 < 4(3x + 1)^2\).

Expand both sides: \(x^2 - 8x + 16 < 36x^2 + 24x + 4\).

Rearrange to form a quadratic inequality: \(0 < 35x^2 + 32x - 12\).

Find the critical points by solving \(35x^2 + 32x - 12 = 0\).

Using the quadratic formula, \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\), where \(a = 35\), \(b = 32\), \(c = -12\).

Calculate the discriminant: \(b^2 - 4ac = 32^2 - 4 \times 35 \times (-12) = 1024 + 1680 = 2704\).

\(x = \frac{-32 \pm \sqrt{2704}}{70}\).

\(\sqrt{2704} = 52\), so \(x = \frac{-32 \pm 52}{70}\).

This gives critical points \(x = \frac{20}{70} = \frac{2}{7}\) and \(x = \frac{-84}{70} = -\frac{6}{5}\).

The solution to the inequality is \(x < -\frac{6}{5}\) or \(x > \frac{2}{7}\).

To solve the inequality \(2|x - 2| > |3x + 1|\), we consider the critical points where each absolute value expression changes sign.

1. Solve \(2(x - 2) = 3x + 1\) and \(2(x - 2) = -(3x + 1)\).

2. For \(2(x - 2) = 3x + 1\):

\(2x - 4 = 3x + 1\)

\(-4 - 1 = 3x - 2x\)

\(-5 = x\)

3. For \(2(x - 2) = -(3x + 1)\):

\(2x - 4 = -3x - 1\)

\(2x + 3x = -1 + 4\)

\(5x = 3\)

\(x = \frac{3}{5}\)

4. The critical values are \(x = -5\) and \(x = \frac{3}{5}\).

5. Test intervals around the critical points to determine where the inequality holds:

- For \(x < -5\), choose \(x = -6\):

\(2|-6 - 2| = 16\) and \(|3(-6) + 1| = 17\), so \(16 \not> 17\).

- For \(-5 < x < \frac{3}{5}\), choose \(x = 0\):

\(2|0 - 2| = 4\) and \(|3(0) + 1| = 1\), so \(4 > 1\).

- For \(x > \frac{3}{5}\), choose \(x = 1\):

\(2|1 - 2| = 2\) and \(|3(1) + 1| = 4\), so \(2 \not> 4\).

6. The solution is \(-5 < x < \frac{3}{5}\).

To solve the equation \(2|x - 1| = 3|x|\), consider the cases for the absolute values.

Case 1: \(x \geq 1\)

In this case, \(|x - 1| = x - 1\) and \(|x| = x\).

The equation becomes \(2(x - 1) = 3x\).

Simplifying gives \(2x - 2 = 3x\).

Rearranging, we get \(-2 = x\), which is not valid for \(x \geq 1\).

Case 2: \(0 \leq x < 1\)

Here, \(|x - 1| = 1 - x\) and \(|x| = x\).

The equation becomes \(2(1 - x) = 3x\).

Simplifying gives \(2 - 2x = 3x\).

Rearranging, we get \(2 = 5x\).

Thus, \(x = \frac{2}{5}\), which is valid for \(0 \leq x < 1\).

Case 3: \(x < 0\)

In this case, \(|x - 1| = 1 - x\) and \(|x| = -x\).

The equation becomes \(2(1 - x) = 3(-x)\).

Simplifying gives \(2 - 2x = -3x\).

Rearranging, we get \(2 = -x\).

Thus, \(x = -2\), which is valid for \(x < 0\).

Therefore, the solutions are \(x = -2\) and \(x = \frac{2}{5}\).

To solve the inequality \(|2x - 5| > 3|2x + 1|\), we consider two cases based on the definition of absolute value.

Case 1: \(2x - 5 > 3(2x + 1)\)

Simplifying, we get:

\(2x - 5 > 6x + 3\)

\(-5 - 3 > 6x - 2x\)

\(-8 > 4x\)

\(x < -2\)

Case 2: \(2x - 5 < -3(2x + 1)\)

Simplifying, we get:

\(2x - 5 < -6x - 3\)

\(2x + 6x < -3 + 5\)

\(8x < 2\)

\(x < \frac{1}{4}\)

Combining both cases, the solution is \(-2 < x < \frac{1}{4}\).

To solve the inequality \(|x - 2| > 2x - 3\), we consider two cases based on the definition of absolute value.

Case 1: When \(x - 2 \geq 0\), i.e., \(x \geq 2\).

The inequality becomes \(x - 2 > 2x - 3\).

Simplifying, we get:

\(x - 2 > 2x - 3\)

\(-2 + 3 > 2x - x\)

\(1 > x\)

Since \(x \geq 2\) and \(1 > x\) cannot both be true, there is no solution in this case.

Case 2: When \(x - 2 < 0\), i.e., \(x < 2\).

The inequality becomes \(-(x - 2) > 2x - 3\).

Simplifying, we get:

\(-x + 2 > 2x - 3\)

\(2 + 3 > 2x + x\)

\(5 > 3x\)

\(x < \frac{5}{3}\)

Since \(x < 2\) and \(x < \frac{5}{3}\) are consistent, the solution is \(x < \frac{5}{3}\).

To solve the inequality \(|3x - 1| < |2x + 5|\), we consider the cases for the absolute values.

Case 1: Both expressions are positive or both are negative.

1. Solve \(3x - 1 = 2x + 5\):

\(3x - 2x = 5 + 1\)

\(x = 6\)

2. Solve \(3x - 1 = -(2x + 5)\):

\(3x - 1 = -2x - 5\)

\(3x + 2x = -5 + 1\)

\(5x = -4\)

\(x = -\frac{4}{5}\)

Thus, the solution is \(-\frac{4}{5} < x < 6\).

To solve the inequality \(|x + 2a| > 3|x - a|\), we consider the non-modular inequality \((x + 2a)^2 > (3(x - a))^2\).

Expanding both sides, we have:

\((x + 2a)^2 = x^2 + 4ax + 4a^2\)

\((3(x - a))^2 = 9(x^2 - 2ax + a^2) = 9x^2 - 18ax + 9a^2\)

Setting up the inequality:

\(x^2 + 4ax + 4a^2 > 9x^2 - 18ax + 9a^2\)

Simplifying, we get:

\(-8x^2 + 22ax - 5a^2 > 0\)

Solving the quadratic equation \(-8x^2 + 22ax - 5a^2 = 0\) using the quadratic formula:

\(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\)

Here, \(a = -8\), \(b = 22a\), \(c = -5a^2\).

\(x = \frac{-22a \pm \sqrt{(22a)^2 - 4(-8)(-5a^2)}}{2(-8)}\)

\(x = \frac{-22a \pm \sqrt{484a^2 - 160a^2}}{-16}\)

\(x = \frac{-22a \pm \sqrt{324a^2}}{-16}\)

\(x = \frac{-22a \pm 18a}{-16}\)

\(x = \frac{-22a + 18a}{-16}\) or \(x = \frac{-22a - 18a}{-16}\)

\(x = \frac{-4a}{-16} = \frac{1}{4}a\) or \(x = \frac{-40a}{-16} = \frac{5}{2}a\)

Thus, the solution to the inequality is \(\frac{1}{4}a < x < \frac{5}{2}a\).

To solve the inequality \(|4x + 3| > |x|\), we consider two cases based on the definition of absolute value.

Case 1: \(4x + 3 > x\)

Simplify to get \(3x + 3 > 0\), which gives \(x > -1\).

Case 2: \(4x + 3 < -x\)

Simplify to get \(5x < -3\), which gives \(x < -\frac{3}{5}\).

Combining both cases, the solution is \(x < -1\) or \(x > -\frac{3}{5}\).

(a) The graph of \(y = |2x + 3|\) is a V-shaped graph with the vertex at the point where \(2x + 3 = 0\), which is \(x = -\frac{3}{2}\). The graph is symmetric about the vertical line through the vertex.

(b) To solve \(3x + 8 > |2x + 3|\), consider the two cases for the absolute value:

Case 1: \(3x + 8 > 2x + 3\)

Simplifying gives \(x > -5\).

Case 2: \(3x + 8 > -(2x + 3)\)

Simplifying gives \(3x + 8 > -2x - 3\), which leads to \(5x > -11\) and \(x > -\frac{11}{5}\).

The solution is the intersection of these two cases, which is \(x > -\frac{11}{5}\).

To solve \(|x - 2| = \left|\frac{1}{3}x\right|\), consider the cases for the absolute values.

Case 1: \(x - 2 = \frac{1}{3}x\)

Rearrange to get: \(x - \frac{1}{3}x = 2\)

\(\frac{2}{3}x = 2\)

\(x = 3\)

Case 2: \(x - 2 = -\frac{1}{3}x\)

Rearrange to get: \(x + \frac{1}{3}x = 2\)

\(\frac{4}{3}x = 2\)

\(x = \frac{3}{2}\)

Thus, the solutions are \(x = 3\) or \(x = \frac{3}{2}\).

To solve \(|4x - 1| = |x - 3|\), consider the cases for the absolute values.

Case 1: \(4x - 1 = x - 3\)

Solve for \(x\):

\(4x - x = -3 + 1\)

\(3x = -2\)

\(x = -\frac{2}{3}\)

Case 2: \(4x - 1 = -(x - 3)\)

Solve for \(x\):

\(4x - 1 = -x + 3\)

\(4x + x = 3 + 1\)

\(5x = 4\)

\(x = \frac{4}{5}\)

Thus, the solutions are \(x = -\frac{2}{3}\) and \(x = \frac{4}{5}\).

To solve the inequality \(3|x - 1| < |2x + 1|\), we consider the non-modular inequality \((3(x - 1))^2 < (2x + 1)^2\).

Expanding both sides, we have:

\((3(x - 1))^2 = 9(x^2 - 2x + 1) = 9x^2 - 18x + 9\)

\((2x + 1)^2 = 4x^2 + 4x + 1\)

Setting the inequality:

\(9x^2 - 18x + 9 < 4x^2 + 4x + 1\)

Simplifying, we get:

\(5x^2 - 22x + 8 < 0\)

Solving the quadratic inequality \(5x^2 - 22x + 8 = 0\) gives the critical points:

\(x = \frac{2}{5}\) and \(x = 4\)

Testing intervals, we find the solution to the inequality is:

\(\frac{2}{5} < x < 4\)

To solve \(|x| < |5 + 2x|\), consider the critical points where \(x = 0\) and \(5 + 2x = 0\).

First, solve \(5 + 2x = 0\):

\(2x = -5\)

\(x = -\frac{5}{2}\)

Now, consider the intervals determined by the critical points \(x = 0\) and \(x = -\frac{5}{2}\).

1. For \(x < -5\), both \(x\) and \(5 + 2x\) are negative, so \(|x| = -x\) and \(|5 + 2x| = -(5 + 2x)\).

2. For \(-5 < x < -\frac{5}{3}\), \(x\) is negative and \(5 + 2x\) is positive, so \(|x| = -x\) and \(|5 + 2x| = 5 + 2x\).

3. For \(x > -\frac{5}{3}\), both \(x\) and \(5 + 2x\) are positive, so \(|x| = x\) and \(|5 + 2x| = 5 + 2x\).

Check each interval:

- For \(x < -5\), \(-x < -(5 + 2x)\) simplifies to \(x < -5\).

- For \(-5 < x < -\frac{5}{3}\), \(-x < 5 + 2x\) simplifies to \(x > -\frac{5}{3}\).

- For \(x > -\frac{5}{3}\), \(x < 5 + 2x\) simplifies to \(x < -5\), which is not possible.

Thus, the solution is \(x < -5\) or \(x > -\frac{5}{3}\).

To solve the inequality \(2|x - 3| > |3x + 1|\), we consider the critical points where the expressions inside the absolute values are zero.

1. Solve \(x - 3 = 0\) to get \(x = 3\).

2. Solve \(3x + 1 = 0\) to get \(x = -\frac{1}{3}\).

These critical points divide the number line into intervals: \((-\infty, -\frac{1}{3})\), \((-\frac{1}{3}, 3)\), and \((3, \infty)\).

Test each interval:

- For \(x < -\frac{1}{3}\), choose \(x = -1\):

\(2|-1 - 3| = 2|4| = 8\) and \(|3(-1) + 1| = |-3 + 1| = 2\).

Since \(8 > 2\), this interval satisfies the inequality.

- For \(-\frac{1}{3} < x < 3\), choose \(x = 0\):

\(2|0 - 3| = 2|3| = 6\) and \(|3(0) + 1| = |1| = 1\).

Since \(6 > 1\), this interval satisfies the inequality.

- For \(x > 3\), choose \(x = 4\):

\(2|4 - 3| = 2|1| = 2\) and \(|3(4) + 1| = |12 + 1| = 13\).

Since \(2 < 13\), this interval does not satisfy the inequality.

Thus, the solution is \(-7 < x < 1\).

To solve \(|x - 3| > 2|x + 1|\), consider the critical points where each expression inside the absolute value is zero: \(x = 3\) and \(x = -1\).

Also consider where the expressions \(x - 3\) and \(x + 1\) change sign: \(x = 3\) and \(x = -1\).

Test intervals determined by these points: \((-\infty, -1)\), \((-1, 3)\), and \((3, \infty)\).

1. For \(x < -1\), both \(x - 3\) and \(x + 1\) are negative. The inequality becomes \(-(x - 3) > 2(-(x + 1))\), simplifying to \(x - 3 > 2x + 2\), which gives \(-3 > x + 2\) or \(-5 > x\). This is not possible in this interval.

2. For \(-1 < x < 3\), \(x - 3\) is negative and \(x + 1\) is positive. The inequality becomes \(-(x - 3) > 2(x + 1)\), simplifying to \(x - 3 < -2x - 2\), which gives \(3x < 1\) or \(x < \frac{1}{3}\).

3. For \(x > 3\), both \(x - 3\) and \(x + 1\) are positive. The inequality becomes \(x - 3 > 2(x + 1)\), simplifying to \(x - 3 > 2x + 2\), which gives \(-3 > x + 2\) or \(-5 > x\). This is not possible in this interval.

Combining the valid intervals, the solution is \(-5 < x < \frac{1}{3}\).

To solve the inequality \(|x + 3a| > 2|x - 2a|\), we consider the non-modular inequality \((x + 3a)^2 > (2(x - 2a))^2\).

Expanding both sides, we have:

\((x + 3a)^2 = x^2 + 6ax + 9a^2\)

\((2(x - 2a))^2 = 4(x^2 - 4ax + 4a^2) = 4x^2 - 16ax + 16a^2\)

Thus, the inequality becomes:

\(x^2 + 6ax + 9a^2 > 4x^2 - 16ax + 16a^2\)

Rearranging terms gives:

\(-3x^2 + 22ax - 7a^2 > 0\)

Solving the quadratic equation \(-3x^2 + 22ax - 7a^2 = 0\), we find the critical values:

\(x = \frac{1}{3}a\) and \(x = 7a\).

Therefore, the solution to the inequality is:

\(\frac{1}{3}a < x < 7a\).