Higher Mathematics 2nd Paper · Classes 11-12 · Chapter 1 · 1.7

Absolute Value Inequalities.

|x − 3| < 5 looks hard, but it says something plain: the distance between x and 3 is less than 5. Think in distances and every absolute value inequality becomes a picture on the number line, either one stretch in the middle or two rays going outward. Almost every board paper has a question from this part.

1. Inside, or outside

|x| is the distance of x from the origin. So for a > 0, |x| < a means −a < x < a: x sits between −a and a. A distance is never negative, so when a = 0 or a < 0, |x| < a has no solution at all. With an equals sign the two ends come in as well: for a ≥ 0, |x| ≤ a holds if and only if −a ≤ x ≤ a.

The book draws this as the V of y = |x| against a level line y = a. Move the line up and down and watch where the V dips below it.

Turn the sign round and the picture turns round too. |x| > a means x is more than a away from the origin, so x < −a or x > a. And for a ≥ 0, |x| ≥ a gives x ≤ −a or x ≥ a. The word here is "or": x cannot be on both sides at once.

The origin can be replaced by any number. |x − 3| is the distance between x and 3. So |x − 3| = 5 has the solution {−2, 8}, the two points exactly 5 from 3. |x − 3| < 5 gives −2 < x < 8. |x − 3| > 5 gives (−∞, −2) ∪ (8, ∞). And in 0 < |x − 3| < 5 the distance may not be zero either, so the centre 3 itself drops out: (−2, 3) ∪ (3, 8). Below, set the centre a and the distance b yourself.

Now just recognise the shape. |x| ≥ 3 has the solution (−∞, −3] ∪ [3, ∞), two rays with filled ends. |2x − 9| > 7 gives (−∞, 1) ∪ (8, ∞). |x − 1| ≥ 1 gives (−∞, 0] ∪ [2, ∞). Sort each inequality by the shape of its answer.

2. Taking the modulus off

There is one trick: write |…| < k as −k < … < k, then do the same thing to all three parts. Example 4(i): |x − 2| < 5 gives −5 < x − 2 < 5, and adding 2 throughout gives −3 < x < 7. Example 6: |2x + 1| < 3 gives −3 < 2x + 1 < 3, then −4 < 2x < 2, and finally −2 < x < 1.

The same mould: |2x − 7| ≤ 3 gives 2 ≤ x ≤ 5; |3x − 4| < 2 gives 2/3 < x < 2; |2x − 5| < 3 gives 1 < x < 4; |x − 1| ≤ 1 gives 0 ≤ x ≤ 2; |x − 3| ≤ 1 gives 2 ≤ x ≤ 4; |3 − 2x| ≤ 1 gives 1 ≤ x ≤ 2. With > you split in two: |2x − 7| > 5 means 2x − 7 < −5 or 2x − 7 > 5, so x < 1 or x > 6. And |x − 1| < 1/5 means 4/5 < x < 6/5, where the only natural number is 1.

3. Modulus under a fraction: flip it, flip the sign

When the modulus is in a denominator, flip the fraction first. Both sides are positive, so flipping turns the sign round. Example 4(ii): 1/|2x − 1| ≥ 7 (x ≠ 1/2) gives |2x − 1| ≤ 1/7, that is −1/7 ≤ 2x − 1 ≤ 1/7, so 3/7 ≤ x ≤ 4/7. A denominator cannot be zero, so x ≠ 1/2 must go in the answer.

Example 7: 1/|3x − 5| > 2 (x ≠ 5/3) gives |3x − 5| < 1/2, then 9/2 < 3x < 11/2, so 3/2 < x < 11/6, x ≠ 5/3. The book prints 3/2 < x ≤ 11/6; the correct answer is 3/2 < x < 11/6, x ≠ 5/3, because at x = 11/6 the value of 1/|3x − 5| is exactly 2, which is not more than 2. Only a ≥ in the question would let the two ends in.

A few the boards like: 1/|5x − 1| > 1/9 (x ≠ 1/5) gives |5x − 1| < 9, so −8/5 < x < 2, x ≠ 1/5. 1/|x − 1| ≥ 2 gives 1/2 ≤ x ≤ 3/2, x ≠ 1. 1/|5x − 3| ≥ 2 (x ≠ 3/5) gives 1/2 ≤ x ≤ 7/10, x ≠ 3/5. 1/|x − 5| ≥ 3 gives 14/3 ≤ x ≤ 16/3, x ≠ 5. 1/|x − 2| ≥ 2 gives 3/2 ≤ x ≤ 5/2, x ≠ 2.

Example 24(a): 2 ≤ 1/|x − 1| (x ≠ 1) gives |x − 1| ≤ 1/2, so 1/2 ≤ x ≤ 3/2, x ≠ 1. The book prints 1/2 < x < 3/2; the correct answer is 1/2 ≤ x ≤ 3/2, x ≠ 1, because at x = 1/2 the value of 1/|x − 1| is 2, and 2 ≤ 2 is true. When two conditions are given together the answer is the overlap: |x − 2| ≤ 1 gives 1 ≤ x ≤ 3, |x − 3| ≤ 1 gives 2 ≤ x ≤ 4, and both together give 2 ≤ x ≤ 3. Below, see the steps and the number line for each.

4. The other way: from an interval to a modulus

Boards often ask the reverse: write a ≤ bx + c ≤ d with a modulus. The trick is to subtract the mean of the two ends, (a + d)/2, from all three parts. The ends become equal and opposite, and you can write |bx + k| ≤ (d − a)/2, where k = c − (a + d)/2. Example 5: the midpoint of −2 < x < 6 is 2, so taking 2 from every part gives −4 < x − 2 < 4, that is |x − 2| < 4.

The midpoint becomes the centre and the half-length becomes the distance. Drag the two ends and see.

The plain ones: −7 < x < −1 gives |x + 4| < 3; −1 < x < 3 gives |x − 1| < 2; −2 < x < 8 gives |x − 3| < 5. When the middle holds more than x, open it up first, or subtract the mean straight away. −2 < 5 − x < 8 gives −3 < x < 7, so |x − 2| < 5. −1 < 2x − 3 < 5 gives 2 < 2x < 8, midpoint 5, so |2x − 5| < 3. −4 < 2x − 1 < 12 has mean 4, so |2x − 5| < 8. And with f(x) = x − 1, −2 < 2 − f(x) < 8 gives −5 < x < 5, that is |x| < 5.

5. From |x − a| < δ to |f(x) − L| < ε

This pattern keeps coming up in part b of the creative questions: if |x − 1| < 1/10, show that |x² − 1| < 21/100. The route has four steps. Open the modulus: 9/10 < x < 11/10. Square: 81/100 < x² < 121/100. Subtract 1: −19/100 < x² − 1 < 21/100. Now compare the sizes of the two ends: −19/100 > −21/100, so everything lies between −21/100 and 21/100, that is |x² − 1| < 21/100. In the same way |x − 1| < 1/11 gives |x² − 1| < 23/121.

The others are the same four steps. |x − 3| < 1/5 gives 196/25 < x² < 256/25, so −4/25 < x² − 8 < 56/25, that is |x² − 8| < 56/25; and −29/25 < x² − 9 < 31/25, so |x² − 9| < 31/25. |x − 3| < 1/7 gives −41/49 < x² − 9 < 43/49, so |x² − 9| < 43/49. |x − 2| < 1/4 gives −15/16 < x² − 4 < 17/16, so |x² − 4| < 17/16. |x − 1| < 1/2 gives 1/8 < x³ < 27/8, so |x³ − 1| < 19/8. |x − 1| < 1/5 gives |x² − 1| < 11/25. And |x − 5| < 1/2 gives 729 < 8x³ < 1331, adding 29 gives 758 < 8x³ + 29 < 1360, so |8x³ + 29| < 1360.

The fourth step is where most mistakes happen, and the book slips here too. Example 14: |x − 4| < 1/6 gives 23/6 < x < 25/6, cubing gives 12167/216 < x³ < 15625/216, so −3169/216 < x³ − 71 < 289/216. The larger of the two end sizes is 3169/216, so |x³ − 71| < 3169/216. The book raises the lower end to −289/216 and writes 289/216, which is wrong: −3169/216 < −289/216, so the lower end cannot be raised. Move x below and see for yourself which window covers both ends.

6. Two moduli, two conditions

With a modulus on both sides, square. Both sides are non-negative and |x|² = x², so |A| < |B| holds exactly when A² < B². |x + 2| < |4x + 1| gives (x + 2)² < (4x + 1)², which rearranges to 15x² + 4x − 3 > 0, that is (3x − 1)(5x + 3) > 0. So x < −3/5 or x > 1/3.

With a bound on each side, solve the two parts separately and take the overlap. In 1 < |2x − 3| < 7, |2x − 3| < 7 gives −2 < x < 5, and |2x − 3| > 1 gives x < 1 or x > 2. The overlap is (−2, 1) ∪ (2, 5). Over ℝ its sup is 5 and its inf −2. If x must be an integer the set is {−1, 0, 3, 4}, with sup 4 and inf −1.

7. Check yourself

Three questions in the exam's style.

Source: Higher Mathematics 2nd Paper, Classes 11-12 (NCTB), Chapter 1 “Real Numbers and Inequalities”, section 1.7.