CRE 3 - Modulus | Algebra - Inequalities & Modulus

Answer: Option C

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Explanation :

If |ax + b| ≤ c, then
- c ≤ ax + b ≤ c

⇒ -6 ≤ 2x - 4 ≤ 6

⇒ -2 ≤ 2x ≤ 10

⇒ -1 ≤ x ≤ 5

Hence, option (c).

Answer: Option B

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Explanation :

If |ax + b| ≥ c, then
- c ≥ ax + b ≥ c

⇒ -3 ≥ 3x - 5 ≥ 3

⇒ 2 ≥ 3x ≥ 8

⇒ 2/3 ≥ x ≥ 8/3

Hence, option (b).

Answer: Option A

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Explanation :

Given, ||x - 1| - 2| > 3

⇒ -3 > |x - 1| - 2 > 3

⇒ -1 > |x - 1| > 5

Case 1: |x - 1| > 5
⇒ -5 > x – 1 > 5
⇒ -4 > x > 6
⇒ x ∈ (-∞, -4) ∪ (6, ∞)

Case 2: |x - 1| < -1
Now |x - 1| is always positive, hence no solution.

⇒ x ∈ (-∞, -4) ∪ (6, ∞)

Hence, option (a).

Answer: Option C

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Explanation :

In such questions we first calculate the critical points and then takes cases.

Critical points are when the expressions within the mod are zero.

∴ 2x + 4 = 0 when x = -2, &
3x – 9 = 0 when x = 3.

∴ The critical points are -2 and 3.

Now let us take cases for x.

Case 1: x > 3
Both 2x + 4 and 3x – 9 are positive
∴ 2x + 4 ≥ 3x – 9
⇒ x ≤ 13

∴ x ∈ (3, 13]

Case 2: -2 < x < 3
2x + 4 is positive while 3x – 9 is negative
∴ 2x + 4 ≥ -(3x – 9)
⇒ 5x ≥ 5
⇒ x ≥ 1

∴ x ∈ [1, 3)

Case 3: x < -2
Both 2x + 4 and 3x – 9 are negative
∴ -(2x + 4) ≥ -(3x – 9)
⇒ 2x + 4 ≤ 3x – 9
⇒ x ≥ 13

∴ No solution

⇒ x ∈ [1, 3) ∪ (3, 13]

Now we need to check for critical points as well. The original inequality is true for x = -2 and 3. Hence, we need to include critical points in our answer.

∴ x ∈ [1, 13]

Hence, option (c).

Answer: Option D

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Explanation :

f(x) = |3x + 7| + 16

To minimize f(x) we need to minimize |3x + 7|

We know |3x + 7| ≥ 0
∴ least possible value of |3x + 7| = 0
⇒ 3x + 7 = 0
⇒ x = -7/3

Hence, option (d).

Answer: Option D

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Explanation :

To minimize any expression of the form |x + a| + |x + b| + …
we need to find critical points and assume a value of x equal to the critical point in the middle.

f(x) = |x - 1| + |x + 1| + |x + 3|

The critical points are: -3, -1 and 1.
The critical point in the middle is -1, hence we substitute x = -1 to find the least possible value of f(x).

∴ f(-1) = |-1 - 1| + |- 1 + 1| + |-1 + 3| = 2 + 0 + 2 = 4

Hence, option (d).

Answer: Option C

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Explanation :

To minimize any expression of the form |x + a| + |x + b| + …
we need to find critical points and assume a value of x between the 2 middle critical points.

f(x) = |x - 1| + |x + 1| + |x + 3| +  |x + 5|

The critical points are: -5, -3, -1 and 1.
The 2 critical points in the middle are -3 and -1, hence we assume a value of x in between -3 and -1. Let x = -2.

∴ f(-2) = |-2 - 1| + |-2 + 1| + |-2 + 3| + |-2 + 5| = 3 + 1 + 1 + 3 = 8

Hence, option (c).