Students can go through AP Inter 1st Year Maths Notes 5th Lesson Linear Inequalities will help students in revising the entire concepts quickly.
Linear Inequalities Class 11 Notes AP Inter 1st Year Maths 5th Lesson
→ Two real numbers or two algebraic expressions related by the symbols <, >, ≤ or ≥ form an inequality.
→ Equal numbers may be added to (or subtracted from) both sides of an inequality.
→ Both sides of an inequality can be multiplied (or divided) by the same positive number.
→ But when both sides are multiplied (or divided) by a negative number, then the inequality is reversed.
→ The values of x, which make an inequality a true statement, are called solutions of the inequality.
→ To represent x < a (or x > a) on a number line, put a circle on the number a and a dark line to the left (or right) of the number a.
→ To represent x ≤ a (or x ≥ a) on a number line, put a dark circle on the number a and darken the line to the left (or right) of the number as.
Example Problems
Question 1.
Solve 30x < 200 when
(i) x is a natural number.
(ii) x is an integer.
Solution:
We are given 30x < 200
or \(\frac{30 x}{30}<\frac{200}{30}\) (Rule 2)
i.e., x < \(\frac {20}{3}\)
(i) When x is a natural number, in this case, the following values of x make the statement true.
1, 2, 3, 4, 5, 6
The solution set of the inequality is {1, 2, 3, 4, 5, 6}.
(ii) When x is an integer, the solutions of the given inequality are ….., -3, -2, -1, 0, 1, 2, 3, 4, 5, 6
The solution set of the inequality is {…, -3, -2, -1, 0, 1, 2, 3, 4, 5, 6}
Question 2.
Solve 5x – 3 < 3x + 1 when
(i) x is an integer.
(ii) x is a real number.
Solution:
We have, 5x – 3 < 3x + 1
or 5x – 3 + 3 < 3x + 1 + 3 (Rule 1)
or 5x < 3x + 4
or 5x – 3x < 3x + 4 – 3x (Rule 1)
or 2x < 4
or x < 2 (Rule 2)
(i) When x is an integer, the solutions of the given inequality are ….., -4, -3, -2, -1, 0, 1
(ii) When x is a real number, the solutions of the inequality are given by x < 2.
i.e., all real numbers x which are less than 2.
Therefore, the solution set of the inequality is x ∈ (-∞, 2).
Question 3.
Solve 4x + 3 < 6x + 7.
Solution:
We have, 4x + 3 < 6x + 7
or 4x – 6x < 6x + 4 – 6x
or -2x < 4 or x > -2
i.e., all the real numbers which are greater than -2 are the solutions of the given inequality.
Hence, the solution set is (-2, ∞).
Question 4.
Solve \(\frac{5-2 x}{3} \leq \frac{x}{6}-5\)
Solution:
We have \(\frac{5-2 x}{3} \leq \frac{x}{6}-5\)
or 2(5 – 2x) ≤ x – 30
or 10 – 4x ≤ x – 30
or -5x ≤ -40
i.e., x ≥ 8
Thus, all real numbers x which are greater than or equal to 8 are the solutions of the given inequality, i.e., x ∈ [8, ∞).
Question 5.
Solve 7x + 3 < 5x + 9. Show the graph of the solution on the number line.
Solution:
We have 7x + 3 < 5x + 9
or 2x < 6
or x < 3
The graphical representation of the solutions is given in Figure.
Question 6.
Solve \(\frac{3 x-4}{2} \geq \frac{x+1}{4}-1\). Show the graph of the solutions on the number line.
Solution:
We have \(\frac{3 x-4}{2} \geq \frac{x+1}{4}-1\)
or \(\frac{3 x-4}{2} \geq \frac{x-3}{4}\)
or 2 (3x – 4) ≥ (x – 3)
or 6x – 8 ≥ x – 3
or 5x ≥ 5
or x ≥ 1
The graphical representation of solutions is given in Figure.
Question 7.
The marks obtained by a student of Class XI in the first and second terminal examinations are 62 and 48, respectively. Find the minimum marks he should get in the annual examination to have an average of at least 60 marks.
Solution:
Let x be the marks obtained by the student in the annual examination.
Then \(\frac{62+48+x}{3}\) ≥ 60
or 110 + x ≥ 180
or x ≥ 70
Thus, the student must obtain a minimum of 70 marks to get an average of at least 60 marks.
Question 8.
Find all pairs of consecutive odd natural numbers, both of which are larger than 10, such that their sum is less than 40.
Solution:
Let x be the smaller of the two consecutive odd natural numbers, so that the other one is x + 2.
Then, we should have x > 10 ………(1)
and x + (x + 2) < 40 ………..(2)
Solving (2), we get
2x + 2 < 40
i.e., x < 19 ……….(3)
From (1) and (3), we get
10 < x < 19
Since x is an odd number, x can take the values 11, 13, 15, and 17.
Therefore, the possible pairs that are required are (11, 13), (13, 15), (15, 17), and (17, 19).
Question 9.
Solve -8 ≤ 5x – 3 < 7.
Solution:
In this case, we have two inequalities, -8 ≤ 5x – 3 and 5x – 3 < 7, which we will solve simultaneously.
We have -8 ≤ 5x – 3 < 7
or -5 ≤ 5x < 10
or -1 ≤ x < 2
Question 10.
Solve -5 ≤ \(\frac{5-3 x}{2}\) ≤ 8.
Solution:
We have -5 ≤ \(\frac{5-3 x}{2}\) ≤ 8
or -10 ≤ 5 – 3x ≤ 16
or -15 ≤ -3x ≤ 11
or 5 ≥ x ≥ \(-\frac {11}{3}\)
which can be written as \(-\frac {11}{3}\) ≤ x ≤ 5
Question 11.
Solve the system of inequalities:
3x – 7 < 5 + x ………….(1)
11 – 5x ≤ 1 …………..(2)
and represent the solutions on the number line.
Solution:
From inequality (1), we have
3x – 7 < 5 + x
or x < 6 ………(3)
Also, from inequality (2), we have
11 – 5x ≤ 1
or -5x ≤ -10
i.e., x ≥ 2 ……..(4)
If we draw the graph of inequalities (3) and (4) on the number line,
We see that the values of x, which are common to both, are shown by a bold line in the Figure.
Thus, solutions of the system are real numbers x lying between 2 and 6, including 2.
i.e., 2 ≤ x < 6
Question 12.
In an experiment, a solution of hydrochloric acid is to be kept between 30° and 35° Celsius. What is the range of temperature in degrees Fahrenheit if the conversion formula is given by C = \(\frac {5}{9}\)(F – 32), where C and F represent temperature in degrees Celsius and degrees Fahrenheit, respectively.
Solution:
It is given that 30 < C < 35.
Putting C = \(\frac {5}{9}\)(F – 32), we get
30 < \(\frac {5}{9}\)(F – 32) < 35
or \(\frac {9}{5}\)(30) < (F – 32) < \(\frac {9}{5}\)(35)
or 54 < (F – 32) < 63
or 86 < F < 95
Thus, the required range of temperature is between 86° F and 95° F.
Question 13.
A manufacturer has 600 litres of a 12% solution of acid. How many litres of a 30% acid solution must be added to it so that the acid content in the resulting mixture will be more than 15% but less than 18%?
Solution:
Let x litres of 30% acid solution is required to be added.
Then Total mixture = (x + 600) litres
Therefore 30% x + 12% of 600 > 15% of (x + 600) and 30% x + 12% of 600 < 18% of (x + 600)
or \(\frac{30 x}{100}+\frac{12}{100}(600)>\frac{15}{100}(x+600)\) and \(\frac{30 x}{100}+\frac{12}{100}(600)<\frac{18}{100}(x+600)\)
or 30x + 7200 > 15x + 9000 and 30x + 7200 < 18x + 10800
or 15x > 1800 and 12x < 3600
or x > 120 and x < 300,
i.e., 120 < x < 300.
Thus, the number of litres of the 30% solution of acid will have to be more than 120 litres but less than 300 litres.
Multiple Choice Questions
Question 1.
The length of a rectangle is four times its breadth. If the minimum perimeter of the rectangle is 160 cm, then
(1) breadth > 16 cm
(2) length < 20 cm
(3) breadth ≥ 16 cm
(4) length ≤ 20 cm
Answer:
(3) breadth ≥ 16 cm
Question 2.
Consider a linear inequality -3x + 15 < -12, then
(1) x ∈ (9, ∞)
(2) x ∈ [9, ∞)
(3) x ∈ (-∞, 9]
(4) x ∈ [-9, 10)
Answer:
(1) x ∈ (9, ∞)
Question 3.
Suppose we have three real numbers, which include x, y, and b, given by x > y, b < 0. Determine the relation between the three variables.
(1) x/b < y
(2) x ≤ y
(3) x/b < y/b
(4) x ≥ y/b
Answer:
(3) x/b < y/b
Question 4.
If |x – 2| > 6, then
(1) x ∈ (-4, 8)
(2) x ∈ [-4, 8]
(3) x ∈ (-∞, -4) ∪ (8, ∞)
(4) x ∈ (-∞, -4) ∪ [8, ∞)
Answer:
(3) x ∈ (-∞, -4) ∪ (8, ∞)
Question 5.
If \(\frac{|x-8|}{(x-8)}\) ≥ 0, then
(1) x ∈ [8, ∞)
(2) x ∈ (7, 8)
(3) x ∈ (-∞, 8)
(4) x ∈ (8, ∞)
Answer:
(4) x ∈ (8, ∞)
Question 6.
If |x + 5| ≥ 10, then
(1) x ∈ (-15, 15]
(2) x ∈ (-15, 5]
(3) x ∈ (-∞, -15] ∪ [5, ∞)
(4) x ∈ (-∞, -15] ∪ [15, ∞)
Answer:
(3) x ∈ (-∞, -15] ∪ [5, ∞)
Question 7.
If 4x + 4 < 6x + 5, then x belongs to the interval
(1) (2, ∞)
(2) (\(-\frac {1}{2}\), ∞)
(3) (-∞, \(\frac {1}{2}\))
(4) (-4, ∞)
Answer:
(2) (\(-\frac {1}{2}\), ∞)
Question 8.
The solution of the linear inequality 2x – 17 > 3 – 8x is
(1) x ≥ 2
(2) x < 2
(3) x ≤ 2
(4) x > 2
Answer:
(4) x > 2
Question 9.
Isha scored 90 and 90 marks in the first two class tests. The minimum marks x she should get in the third test to have an average of atleast 90 marks, is
(1) x ≥ 90
(2) x > 90
(3) x ≤ 90
(4) x < 90
Answer:
(1) x ≥ 90
Question 10.
The set of values of x in 5x + 2 > 12 is (x is a real number)
(1) (-2, 2)
(2) (-1, ∞)
(3) (2, ∞)
(4) (-2, 1)
Answer:
(3) (2, ∞)
Question 11.
When x is an integer, the set of values of x satisfying the linear inequality 5x – 2 ≤ 3x + 2 are
(1) {…, -2, -1, 0, 1}
(2) {…, -3, -2, -1, 0, 1}
(3) {…, -2, -1, 0, 1, 2}
(4) {…, -2, -1, 0, 1, 2, 3}
Answer:
(3) {…, -2, -1, 0, 1, 2}
Question 12.
The set of values of x satisfying 50x ≤ 150, where x is a natural number, is
(1) {1, 2, 3, 4}
(2) {1, 2}
(3) {0, 1, 2}
(4) {1, 2, 3}
Answer:
(2) {1, 2}
Question 13.
If 9x > -36, then the least integral value of x is
(1) -3
(2) -4
(3) -5
(4) -6
Answer:
(1) -3