NCERT Solutions for Class 11 Maths Linear Inequalities Chapter 6
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LINEAR INEQUALITIES
A statement involving the symbols ‘>’, ‘<’, ‘ ≥’, ‘≤’ is called an inequality. For example 5 > 4, x ≤ 4, x + y ≥ 7.
A solution to inequality:
The value(s) of the variable(s) which makes the inequality a true statement is called its solutions. The set of all solutions of an inequality is called the solution set of the inequality.
Rules to solve inequality:
- Add(or subtract) the same quantity to (from) both sides without changing the sign of inequality.
- Multiply (or divide) both sides by the same positive quantity without changing the sign of inequality. However, if both sides of the inequality are multiplied (or divided)by the same negative quantity the sign of inequality is reversed, i.e., ‘>’ change into ‘<’ and vice versa.
Result
If a is any positive real number, i.e., a > 0, then
(i) | x| < a ⇔– a <x <a
| x| ≤ a ⇔– a ≤ x ≤ a
(ii) | x| > a ⇔x < – a or x >a
| x| ≥ a ⇔x ≤ – a or x ≥ a
Example 1: Solve the inequality 3x– 5 <x + 7, when
(i) x is a natural number (ii) x is a whole number
(iii) x is an integer (iv) x is a real number
Solution: We have 3x – 5 <x + 7
⇒3x <x + 12 (Add 5 to both sides)
⇒2x < 12 (Subtractx from both sides)
⇒x< 6 (Dividing by 2 on both sides)
(i) Solution set is {1, 2, 3, 4, 5}
(ii) Solution set is {0, 1, 2, 3, 4, 5}
(iii) Solution set is {….– 3, – 2, –1, 0, 1, 2, 3, 4, 5}
(iv) Solution set is {x :x ∈ R and x < 6}, i.e., any real number less than 6.
Example 2: The cost and revenue functions of a product are given by C(x) = 20 x + 4000 and R(x) = 60x + 2000, respectively, where x is the number of items produced and sold. How many items must be sold to get some profit?
Solution: We have profit = Revenue – Cost
= (60x + 2000) – (20x + 4000)
= 40x – 2000
To earn some profit, 40x – 2000 > 0
⇒x> 50
Hence, the manufacturer must sell more than 50 items to get some profit.
Example 3: Find all pairs of consecutive odd natural numbers, both of which are larger than 10, such that their sum is less than 38.
Solution: Let x be the smaller of the two consecutive odd natural number so that the other one is x +2. Then, according to given condition we have
X > 10 and
x + ( x+ 2) < 38
Solving we get
2x + 2 < 38
2x < 36
i.e., x < 18
So , we get
10 < x < 18
Since x is an odd number, x can take the values 11, 13, 15, and 17. So, the required possible pairs will be
(11, 13), (13, 15), (15, 17), (17, 19)
Example 4: The length of a rectangle is three times the breadth. If the minimum perimeter of the rectangle is 160 cm, then
(A) breadth > 20 cm (B) length < 20 cm
(C) breadth ≥ 20 cm (D) length ≤ 20 cm
Solution: (C) is the correct choice.
Let x is breadth then the length is 3x and the perimeter is 2(3x+x).
Acc to given condition
2 (3x + x) ≥ 160 ⇒x ≥ 20.
Example 5: Solve │x+2│≤ 9
Solution:
-9≤ x+2 ≤ 9 (Since | x| ≤ a ⇔– a ≤ x ≤ a)
-11≤ x ≤ 7
Example 6: Solve │3-4x│≥9
Solution:
3 – 4x ≤ – 9 or3 – 4x ≥ 9 (Since | x| ≥ a ⇔ x≤ – a or x ≥ a)
– 4x ≤ – 12or – 4x ≥ 6
X ≥ 3 or x ≤ -3/2
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