Complete Exercise 1.3 Class 11 Solutions: Logic, Sets and Real Number System

BASIC MATHEMATICS GRADE XI

Exercise 1.3 - Absolute Values, Inequalities, and Sets

Approved by Curriculum Development Centre (CDC), Nepal

Problem 1: Evaluate

a) |-2| + 4

|-2| = 2

2 + 4 = 6

Answer: 6

b) |-5| + |-2| - 3

|-5| = 5

|-2| = 2

5 + 2 - 3 = 4

Answer: 4

c) 2 + |-3| - |-5|

|-3| = 3

|-5| = 5

2 + 3 - 5 = 0

Answer: 0

d) |3 - |-5||

First, evaluate |-5| = 5

Then, 3 - 5 = -2

Finally, |-2| = 2

Answer: 2

Problem 2: Verify Properties

Let (i) x=2, y=3 (ii) x=2, y=-3 verify each of the following:

a) |x+y| ≤ |x| + |y|

Case (i): x=2, y=3

|2+3| = |5| = 5

|2| + |3| = 2 + 3 = 5

5 ≤ 5 ✓

Case (ii): x=2, y=-3

|2+(-3)| = |-1| = 1

|2| + |-3| = 2 + 3 = 5

1 ≤ 5 ✓

Verified in both cases

b) |x-y| ≥ |x| - |y|

Case (i): x=2, y=3

|2-3| = |-1| = 1

|2| - |3| = 2 - 3 = -1

1 ≥ -1 ✓

Case (ii): x=2, y=-3

|2-(-3)| = |5| = 5

|2| - |-3| = 2 - 3 = -1

5 ≥ -1 ✓

Verified in both cases

c) |xy| = |x|·|y|

Case (i): x=2, y=3

|2×3| = |6| = 6

|2| × |3| = 2 × 3 = 6

6 = 6 ✓

Case (ii): x=2, y=-3

|2×(-3)| = |-6| = 6

|2| × |-3| = 2 × 3 = 6

6 = 6 ✓

Verified in both cases

d) |x/y| = |x|/|y|

Case (i): x=2, y=3

|2/3| = 2/3

|2|/|3| = 2/3

2/3 = 2/3 ✓

Case (ii): x=2, y=-3

|2/(-3)| = |-2/3| = 2/3

|2|/|-3| = 2/3

2/3 = 2/3 ✓

Verified in both cases

Problem 3: Solve Inequalities

a) x - 1 > 2

x > 2 + 1

x > 3

Solution: x ∈ (3, ∞)

b) x - 3 ≤ 5

x ≤ 5 + 3

x ≤ 8

Solution: x ∈ (-∞, 8]

c) -1 < x - 2 < 3

Add 2 to all parts: -1+2 < x < 3+2

1 < x < 5

Solution: x ∈ (1, 5)

d) -3 ≤ 2x - 1 ≤ 5

Add 1 to all parts: -2 ≤ 2x ≤ 6

Divide by 2: -1 ≤ x ≤ 3

Solution: x ∈ [-1, 3]

e) x² - 2x > 0

Factor: x(x-2) > 0

Critical points: x = 0, x = 2

Test intervals: (-∞, 0), (0, 2), (2, ∞)

For x < 0: positive × negative = negative ✗

For 0 < x < 2: positive × negative = negative ✗

For x > 2: positive × positive = positive ✓

Solution: x ∈ (-∞, 0) ∪ (2, ∞)

f) 6 + 5x - x² ≥ 0

Multiply by -1 (reverse inequality): x² - 5x - 6 ≤ 0

Factor: (x-6)(x+1) ≤ 0

Critical points: x = -1, x = 6

Test intervals: (-∞, -1), (-1, 6), (6, ∞)

For x < -1: negative × negative = positive ✗

For -1 < x < 6: negative × positive = negative ✓

For x > 6: positive × positive = positive ✗

Solution: x ∈ [-1, 6]

g) x(x+2)/(x-1) ≤ 0

Critical points: x = -2, x = 0, x = 1

Test intervals: (-∞, -2), (-2, 0), (0, 1), (1, ∞)

For x < -2: negative × negative ÷ negative = negative ✓

For -2 < x < 0: negative × positive ÷ negative = positive ✗

For 0 < x < 1: positive × positive ÷ negative = negative ✓

For x > 1: positive × positive ÷ positive = positive ✗

Include points where expression equals 0: x = -2, x = 0

Exclude x = 1 (undefined)

Solution: x ∈ (-∞, -2] ∪ [0, 1)

Problem 4: Set Operations

a) Let A = [-3, 1) and B = [-2, 4]

i) A ∪ B = [-3, 4]

ii) A ∩ B = [-2, 1)

iii) A - B = [-3, -2)

iv) B - A = [1, 4]

b) If A = (-1, 4) and B = [3, 5)

A ∪ B = (-1, 5)

A ∩ B = [3, 4)

A - B = (-1, 3)

c) If U = R = (-∞, ∞), A = (-1, 3] and B = [0, 5)

A ∪ B = (-1, 5)

A ∩ B = [0, 3]

A - B = (-1, 0)

B - A = (3, 5)

A' = (-∞, -1] ∪ (3, ∞)

B' = (-∞, 0) ∪ [5, ∞)

Problem 5: Write Without Absolute Value

a) |x| < 4

-4 < x < 4

x ∈ (-4, 4)

b) |x-3| < 2

-2 < x-3 < 2

1 < x < 5

x ∈ (1, 5)

c) |2x+1| ≤ 3

-3 ≤ 2x+1 ≤ 3

-4 ≤ 2x ≤ 2

-2 ≤ x ≤ 1

x ∈ [-2, 1]

d) |2x-1| ≤ 5

-5 ≤ 2x-1 ≤ 5

-4 ≤ 2x ≤ 6

-2 ≤ x ≤ 3

x ∈ [-2, 3]

Problem 6: Rewrite Using Absolute Value

a) -5 < x < 7

Midpoint: (-5+7)/2 = 1

Distance: 7-1 = 6

|x-1| < 6

b) -3 ≤ x ≤ -1

Midpoint: (-3+(-1))/2 = -2

Distance: -1-(-2) = 1

|x+2| ≤ 1

c) -3 < x < 4

Midpoint: (-3+4)/2 = 0.5

Distance: 4-0.5 = 3.5

|x-0.5| < 3.5

d) -4 ≤ x ≤ -1

Midpoint: (-4+(-1))/2 = -2.5

Distance: -1-(-2.5) = 1.5

|x+2.5| ≤ 1.5

Problem 7: Solve and Graph Inequalities

a) |x + 2| < 4

-4 < x+2 < 4

-6 < x < 2

Solution: x ∈ (-6, 2)

b) |x - 1| ≤ 2

-2 ≤ x-1 ≤ 2

-1 ≤ x ≤ 3

Solution: x ∈ [-1, 3]

c) |2x + 3| ≤ 1

-1 ≤ 2x+3 ≤ 1

-4 ≤ 2x ≤ -2

-2 ≤ x ≤ -1

Solution: x ∈ [-2, -1]

d) |x - 1| > 1

x-1 > 1 or x-1 < -1

x > 2 or x < 0

Solution: x ∈ (-∞, 0) ∪ (2, ∞)

e) |2x + 1| ≥ 3

2x+1 ≥ 3 or 2x+1 ≤ -3

2x ≥ 2 or 2x ≤ -4

x ≥ 1 or x ≤ -2

Solution: x ∈ (-∞, -2] ∪ [1, ∞)

BASIC MATHEMATICS GRADE XI

ALGEBRA - Absolute Values, Inequalities, and Sets

Approved by Curriculum Development Centre (CDC), Nepal