OP Malhotra Class 11 Maths Solutions Chapter 11 Inequalities Exercise 11 (A)

S Chand Class 11 ICSE Maths Solutions Chapter 11 Inequalities Ex 11(A)

 

Question 1. You are given the following numbers : -2.6 5.1 -3 0.4 1.2 -3.1 4.7. Fill in the blanks.
(i) A = \( \{x : x \geq -3\} = \{...\} \)
(ii) B = \( \{x : x \leq 1\} = \{...\} \)
Answer:
(i) For set A, we need numbers from the given list that are greater than or equal to -3. The numbers that fit this rule are -3, -2.6, 0.4, 1.2, 4.7, and 5.1.
So, \( A = \{-3, -2.6, 0.4, 1.2, 4.7, 5.1\} \)
(ii) For set B, we need numbers from the list that are less than or equal to 1. The numbers that fit this rule are -2.6, -3, 0.4, and -3.1.
So, \( B = \{-2.6, -3, 0.4, -3.1\} \). A number line can help visualize which numbers fall into these ranges easily.
In simple words: For set A, pick numbers that are -3 or bigger. For set B, pick numbers that are 1 or smaller.

🎯 Exam Tip: When filling sets based on inequalities, carefully check each number against the condition. Pay attention to "greater than or equal to" (\(\geq\)) versus "greater than" (\(>\)), as this determines if the boundary number is included.

 

Question 2. If the replacement set is \( \{-2, -1, +1, +2, +4, +5, +9\} \), what is the solution set of each of the following mathematical sentences?
(i) \( x + \frac { 3 }{ 2 } > \frac { 5 }{ 2 } \)
(ii) \( 2x - 5 \geq 10 \)
(iii) \( 3y + 2 \leq \frac { 5 }{ 2 } \)
Answer: The given replacement set is \( \{-2, -1, 1, 2, 4, 5, 9\} \).
(i) We need to solve \( x + \frac { 3 }{ 2 } > \frac { 5 }{ 2 } \).
Subtract \( \frac{3}{2} \) from both sides:
\( x > \frac { 5 }{ 2 } - \frac { 3 }{ 2 } \)
\( x > \frac { 2 }{ 2 } \)
\( x > 1 \)
From the replacement set, the numbers greater than 1 are 2, 4, 5, and 9. So the solution set is \( \{2, 4, 5, 9\} \).
(ii) We need to solve \( 2x - 5 \geq 10 \).
Add 5 to both sides:
\( 2x \geq 10 + 5 \)
\( 2x \geq 15 \)
Divide by 2:
\( x \geq \frac { 15 }{ 2 } \)
\( x \geq 7.5 \)
From the replacement set, the only number greater than or equal to 7.5 is 9. So the solution set is \( \{9\} \).
(iii) We need to solve \( 3y + 2 \leq \frac { 5 }{ 2 } \).
Subtract 2 from both sides:
\( 3y \leq \frac { 5 }{ 2 } - 2 \)
\( 3y \leq \frac { 5 }{ 2 } - \frac { 4 }{ 2 } \)
\( 3y \leq \frac { 1 }{ 2 } \)
Divide by 3:
\( y \leq \frac { 1 }{ 6 } \)
\( y \leq 0.166... \)
From the replacement set, the numbers less than or equal to \( \frac{1}{6} \) are -2 and -1. So the solution set is \( \{-2, -1\} \). It's always important to use the given replacement set to find the final answers.
In simple words: For each part, solve the inequality to find the range for x or y. Then, look at the given list of numbers (the replacement set) and pick out only those numbers that fit the range you found.

🎯 Exam Tip: Always state the replacement set clearly at the beginning of your solution. Remember to consider all values from the replacement set that satisfy the inequality, not just positive ones or integers.

 

Question 3. List the solution set of \( 30 - 4(2x - 1) < 30 \), given that x is a positive integer.
Answer: We are given the inequality \( 30 - 4(2x - 1) < 30 \).
Subtract 30 from both sides:
\( -4(2x - 1) < 0 \)
Divide by -4. Remember to reverse the inequality sign when dividing by a negative number:
\( 2x - 1 > 0 \)
Add 1 to both sides:
\( 2x > 1 \)
Divide by 2:
\( x > \frac{1}{2} \)
We are told that x is a positive integer (\( x \in N \)). This means x can be 1, 2, 3, and so on. Since x must be greater than \( \frac{1}{2} \), the positive integers that satisfy this condition are 1, 2, 3, and all subsequent positive integers.
Thus, the solution set is \( \{1, 2, 3, ......\} \).
In simple words: First, solve the math problem to find what x must be bigger than. Since x has to be a whole number that is positive, start counting from the smallest whole number that fits your answer.

🎯 Exam Tip: Always remember to reverse the inequality sign when multiplying or dividing by a negative number. Also, pay close attention to the specified domain for x (e.g., positive integers, real numbers, etc.).

 

Question 4.
(i) \( x \in \{2, 4, 6, 9\} \) and \( y \in \{4, 6, 18, 27, 54\} \). Form all ordered pairs \( (x,y) \) such that x is a factor of y and \( x < y \).
(ii) Find the truth set of the inequality \( x > y + 2 \) where \( (x, y) \in \{(1, 2), (2,3), (5,1), (7,3), (5, 6), (6, 5)\} \).
Answer:
(i) We are given \( x \in \{2, 4, 6, 9\} \) and \( y \in \{4, 6, 18, 27, 54\} \). We need to find ordered pairs \( (x,y) \) where x divides y evenly, and x is smaller than y.
Let's check each value of x:
- If \( x = 2 \), possible \( y \) values from its factors are 4, 6, 18, 54. All these are greater than 2. So pairs are \( (2,4), (2,6), (2,18), (2,54) \).
- If \( x = 4 \), possible \( y \) values from its factors are 4. But \( x < y \) is required, so \( (4,4) \) is not included.
- If \( x = 6 \), possible \( y \) values from its factors are 6, 18, 54. Only 18 and 54 are greater than 6. So pairs are \( (6,18), (6,54) \).
- If \( x = 9 \), possible \( y \) values from its factors are 9, 18, 27, 54. Only 18, 27, and 54 are greater than 9. So pairs are \( (9,18), (9,27), (9,54) \).
Combining these, the solution set A is:
\( A = \{(2,4), (2,6), (2,18), (2,54), (6,18), (6,54), (9,18), (9,27), (9,54)\} \)
(ii) We need to find which pairs \( (x,y) \) from the given list satisfy \( x > y + 2 \).
Let's test each pair:
- For \( (1,2) \): \( 1 > 2 + 2 \implies 1 > 4 \) (False)
- For \( (2,3) \): \( 2 > 3 + 2 \implies 2 > 5 \) (False)
- For \( (5,1) \): \( 5 > 1 + 2 \implies 5 > 3 \) (True)
- For \( (7,3) \): \( 7 > 3 + 2 \implies 7 > 5 \) (True)
- For \( (5,6) \): \( 5 > 6 + 2 \implies 5 > 8 \) (False)
- For \( (6,5) \): \( 6 > 5 + 2 \implies 6 > 7 \) (False)
So, the truth set, containing all pairs that make the inequality true, is \( \{(5,1), (7,3)\} \). This problem helps understand how conditions apply to sets of numbers.
In simple words: For the first part, find pairs where the first number divides the second number evenly, and the first number is smaller. For the second part, check each given pair to see if the first number is more than the second number plus two.

🎯 Exam Tip: When forming ordered pairs, systematically go through each element of the first set and match it with elements of the second set, checking all conditions carefully. For truth sets, test each given pair against the inequality one by one.

 

Question 5. P is the solution set of \( 8x - 1 > 5x + 2 \) and Q is the solution set of \( 7x - 2 \geq 3 (x + 6) \), where \( x \in N \). Find the set \( P \cap Q \).
Answer: We need to find the intersection of two solution sets, P and Q, where x is a natural number (positive integer).

First, let's find solution set P:
Given inequality: \( 8x - 1 > 5x + 2 \)
Subtract \( 5x \) from both sides and add 1 to both sides:
\( 8x - 5x > 2 + 1 \)
\( 3x > 3 \)
Divide by 3:
\( x > 1 \)
Since \( x \in N \) (natural numbers), the values for x are 2, 3, 4, 5, and so on.
So, \( P = \{2, 3, 4, 5, ......\} \).

Next, let's find solution set Q:
Given inequality: \( 7x - 2 \geq 3 (x + 6) \)
First, distribute 3 on the right side:
\( 7x - 2 \geq 3x + 18 \)
Subtract \( 3x \) from both sides and add 2 to both sides:
\( 7x - 3x \geq 18 + 2 \)
\( 4x \geq 20 \)
Divide by 4:
\( x \geq 5 \)
Since \( x \in N \), the values for x are 5, 6, 7, 8, and so on.
So, \( Q = \{5, 6, 7, 8, ......\} \).

Finally, find the intersection \( P \cap Q \). This means finding the numbers that are in both set P and set Q.
\( P = \{2, 3, 4, 5, 6, 7, 8, ......\} \)
\( Q = \{5, 6, 7, 8, ......\} \)
The common numbers are 5, 6, 7, 8, and all subsequent natural numbers.
Therefore, \( P \cap Q = \{5, 6, 7, 8, ......\} \). Finding intersections is a common task in set theory.
In simple words: Solve each inequality separately to find the range of x. Since x must be a natural number, list the numbers that fit each range. Then, find the numbers that are present in both lists.

🎯 Exam Tip: When finding the intersection of two sets, look for the elements that satisfy both conditions simultaneously. For natural numbers, remember to start from 1, and for inequalities, be careful with the greater than/less than or equal to signs.

 

Question 6. Find the solution of the inequality \( 2 \leq 2p - 3 \leq 5 \), \( p \in R \). Hence, graph the solution set on the number line.
Answer: We need to solve the compound inequality \( 2 \leq 2p - 3 \leq 5 \).
Add 3 to all parts of the inequality:
\( 2 + 3 \leq 2p - 3 + 3 \leq 5 + 3 \)
\( 5 \leq 2p \leq 8 \)
Divide all parts by 2:
\( \frac{5}{2} \leq p \leq \frac{8}{2} \)
\( 2.5 \leq p \leq 4 \)
Since \( p \in R \) (real numbers), the solution set is all real numbers between 2.5 and 4, including 2.5 and 4.
The solution set in interval notation is \( [2.5, 4] \).
To graph this on a number line, we draw a thick line from 2.5 to 4. We use dark (filled) dots at 2.5 and 4 to show that these exact points are part of the solution. This is a common way to show intervals for real numbers.

🎯 Exam Tip: When graphing inequalities with "less than or equal to" or "greater than or equal to" signs, always use a closed (dark) circle at the boundary points to show that those values are included in the solution set. For "less than" or "greater than", use an open (hollow) circle.

 

Question 7. Given that x is a negative integer, find the solution set of \( \frac { 2 }{ 3 } + \frac { 1 }{ 3 } (x + 1) > 0 \).
Answer: We need to solve the inequality \( \frac { 2 }{ 3 } + \frac { 1 }{ 3 } (x + 1) > 0 \).
First, expand the term \( \frac{1}{3}(x+1) \):
\( \frac { 2 }{ 3 } + \frac { x }{ 3 } + \frac { 1 }{ 3 } > 0 \)
Combine the constant terms:
\( \frac { x }{ 3 } + \frac { 3 }{ 3 } > 0 \)
\( \frac { x }{ 3 } + 1 > 0 \)
Subtract 1 from both sides:
\( \frac { x }{ 3 } > -1 \)
Multiply both sides by 3:
\( x > -3 \)
We are also given that x is a negative integer. This means x can be -1, -2, -3, and so on. Since x must be greater than -3, the negative integers that satisfy this condition are -2 and -1. Understanding integer properties is key here.
Thus, the solution set is \( \{-2, -1\} \).
In simple words: Solve the math problem to find x must be bigger than -3. Then, since x has to be a negative whole number, the only options are -2 and -1.

🎯 Exam Tip: When dealing with integers, especially negative ones, carefully check the boundary conditions. "Greater than -3" means -3 is not included, so you start from -2, -1 for negative integers.

 

Question 8. Solve the inequality: \( 3 - 2x \geq x - 12 \), given that \( x \in N \).
Answer: We need to solve the inequality \( 3 - 2x \geq x - 12 \), where \( x \in N \) (x is a natural number).
Add \( 2x \) to both sides and add 12 to both sides:
\( 3 + 12 \geq x + 2x \)
\( 15 \geq 3x \)
This can also be written as:
\( 3x \leq 15 \)
Divide by 3:
\( x \leq 5 \)
Since \( x \in N \), the natural numbers that are less than or equal to 5 are 1, 2, 3, 4, and 5. Natural numbers start from 1. Therefore, the solution set is \( \{1, 2, 3, 4, 5\} \).
In simple words: Solve the problem to find x must be 5 or less. Since x must be a natural number (a positive whole number), the answers are 1, 2, 3, 4, and 5.

🎯 Exam Tip: Always remember what kind of numbers the variable can be (e.g., natural numbers, integers, real numbers) as this affects how you determine the final solution set.

 

Question 9. Find the range of values x which satisfies \( -2\frac { 2 }{ 3 } \leq x + \frac { 1 }{ 3 } < 9\frac { 1 }{ 3 } \), \( x \in R \). Graph these values of x on the number line.
Answer: We need to solve the compound inequality \( -2\frac { 2 }{ 3 } \leq x + \frac { 1 }{ 3 } < 9\frac { 1 }{ 3 } \).
First, convert the mixed numbers to improper fractions:
\( -2\frac { 2 }{ 3 } = -\frac{8}{3} \)
\( 9\frac { 1 }{ 3 } = \frac{28}{3} \)
So the inequality becomes:
\( -\frac{8}{3} \leq x + \frac { 1 }{ 3 } < \frac{28}{3} \)
Subtract \( \frac{1}{3} \) from all parts of the inequality:
\( -\frac{8}{3} - \frac{1}{3} \leq x + \frac { 1 }{ 3 } - \frac{1}{3} < \frac{28}{3} - \frac{1}{3} \)
\( -\frac{

 

Question 21. Solve and graphs the solution set
(i) \( | x - 3 | < 4 \)
(ii) \( | x - 3 | > 4 \)
(iii) \( | x + 3 | > 4 \)
(iv) \( | 11 - 3x | < 4 \)
(v) \( 2 + 3 | 2y - 1 | > 8 \)
Answer:
(iii) Given \( | x + 3 | \ge 4 \)
This means \( x + 3 \ge 4 \) or \( x + 3 \le -4 \).
Solving these, we get \( x \ge 4 - 3 \implies x \ge 1 \) or \( x \le -4 - 3 \implies x \le -7 \).
So, the solution set is \( \{x : x \ge 1 \text{ or } x \le -7, x \in R\} = (-\infty, -7] \cup [1, \infty) \).
The solution set is shown on the number line below. The solid dots at -7 and 1 show that these numbers are part of the solution, and the lines extend infinitely in both directions from these points.

(iv) Given \( | 1 - 3x | < 4 \)
This inequality can be rewritten as \( -4 < 1 - 3x < 4 \).
First, subtract 1 from all parts: \( -4 - 1 < -3x < 4 - 1 \implies -5 < -3x < 3 \).
Next, divide all parts by -3. Remember to flip the inequality signs when dividing by a negative number: \( \frac{-5}{-3} > x > \frac{3}{-3} \implies \frac{5}{3} > x > -1 \).
This can be written as \( -1 < x < \frac{5}{3} \).
So, the solution set is \( \{x : -1 < x < \frac{5}{3}, x \in R\} = (-1, \frac{5}{3}) \).
The solution is shown on the number line below. The empty circles (hollow dots) at -1 and \( \frac{5}{3} \) mean these numbers are not part of the solution set.

(v) Given \( 2 + 3 | 2y - 1 | > 8 \)
First, isolate the absolute value term: \( 3 | 2y - 1 | > 8 - 2 \implies 3 | 2y - 1 | > 6 \).
Divide by 3: \( | 2y - 1 | > 2 \).
Remember, for an inequality like \( |X| > a \), the solution is \( X > a \) or \( X < -a \).
So, we have \( 2y - 1 > 2 \) or \( 2y - 1 < -2 \).
Solve the first part: \( 2y > 2 + 1 \implies 2y > 3 \implies y > \frac{3}{2} \).
Solve the second part: \( 2y < -2 + 1 \implies 2y < -1 \implies y < -\frac{1}{2} \).
So, the solution set includes all 'y' values that are greater than \( \frac{3}{2} \) or less than \( -\frac{1}{2} \).
This is written as \( (-\infty, -\frac{1}{2}) \cup (\frac{3}{2}, \infty) \).
The solution is shown on the number line below. The hollow dots at \( -\frac{1}{2} \) and \( \frac{3}{2} \) show that these points themselves are not part of the solution, and the lines extend infinitely outwards.
In simple words: When solving inequalities with absolute values, remember to separate them into two conditions. If it's "less than", combine them into a single range. If it's "greater than", keep them as two separate ranges connected by "or". Always flip the inequality sign if you multiply or divide by a negative number.

🎯 Exam Tip: For absolute value inequalities with 'greater than' (\( |X| > a \)), split it into two separate inequalities connected by 'or': \( X > a \) or \( X < -a \). Always show open circles (hollow dots) on the number line for strict inequalities (not including the endpoints).

 

Question 22. Solve the inequality: \( 3|x-6|-4 \le 11 \)
Answer:
Given the inequality: \( 3|x-6|-4 \le 11 \)
First, add 4 to both sides to isolate the absolute value term: \( 3|x-6| \le 11 + 4 \implies 3|x-6| \le 15 \).
Next, divide both sides by 3: \( |x-6| \le 5 \).
Remember, if \( |X| \le a \), then the solution is \( -a \le X \le a \).
So, we write: \( -5 \le x-6 \le 5 \).
Now, add 6 to all parts of the inequality: \( -5 + 6 \le x - 6 + 6 \le 5 + 6 \).
This simplifies to \( 1 \le x \le 11 \).
The solution set includes all real numbers 'x' such that 'x' is greater than or equal to 1 and less than or equal to 11. This can be written as the interval \( [1, 11] \).
In simple words: To solve this, first get the absolute value part by itself. Then, because it's "less than or equal to", put the expression inside the absolute value between the negative and positive of the other number. Solve for 'x' across all parts.

🎯 Exam Tip: For absolute value inequalities with 'less than or equal to' (\( |X| \le a \)), always express it as a single compound inequality: \( -a \le X \le a \). Remember to perform operations on all three parts of the compound inequality simultaneously to maintain balance.

 

Question 23. Solve the inequality: \( 2|3p-5|+1 > 7 \)
Answer:
Given the inequality: \( 2|3p-5|+1 > 7 \)
First, subtract 1 from both sides: \( 2|3p-5| > 7 - 1 \implies 2|3p-5| > 6 \).
Next, divide both sides by 2: \( |3p-5| > 3 \).
Using the rule for absolute value inequalities where \( |X| > a \), we get \( X > a \) or \( X < -a \).
So, we have two separate inequalities: \( 3p - 5 > 3 \) or \( 3p - 5 < -3 \).
Solving the first inequality: \( 3p > 3 + 5 \implies 3p > 8 \implies p > \frac{8}{3} \).
Solving the second inequality: \( 3p < -3 + 5 \implies 3p < 2 \implies p < \frac{2}{3} \).
So, the solution set contains all values of 'p' that are greater than \( \frac{8}{3} \) or less than \( \frac{2}{3} \).
In simple words: Get the absolute value part alone first. Since it's "greater than", you split it into two separate problems: one where the inside is greater than the number, and another where the inside is less than the negative of that number. Then solve both parts.

🎯 Exam Tip: When dealing with 'greater than' absolute value inequalities, correctly splitting into two 'or' conditions is crucial. Ensure you isolate the absolute value term before applying the split rule.

 

Question 24. Solve the inequality: \( | 5 - (m - 3) | + 8 < 15 \)
Answer:
Given the inequality: \( | 5 - (m - 3) | + 8 < 15 \).
First, simplify the expression inside the absolute value: \( 5 - (m - 3) = 5 - m + 3 = 8 - m \).
So, the inequality becomes: \( | 8 - m | + 8 < 15 \).
Now, subtract 8 from both sides to isolate the absolute value term: \( | 8 - m | < 15 - 8 \implies | 8 - m | < 7 \).
This means that \( 8 - m \) must be between -7 and 7: \( -7 < 8 - m < 7 \).
Subtract 8 from all parts of the inequality: \( -7 - 8 < -m < 7 - 8 \implies -15 < -m < -1 \).
Next, multiply all parts by -1. Remember to flip the inequality signs when multiplying by a negative number: \( (-15) \times (-1) > m > (-1) \times (-1) \implies 15 > m > 1 \).
This can be more conventionally written as \( 1 < m < 15 \).
The solution set includes all 'm' values strictly between 1 and 15.
In simple words: Start by tidying up the inside of the absolute value. Then get the absolute value by itself. Since it's "less than", place the expression inside the absolute value between the negative and positive of the other number. Solve this combined inequality, being careful to flip signs if you multiply or divide by a negative number.

🎯 Exam Tip: Always simplify the expression inside the absolute value first. Be extra careful when multiplying or dividing by a negative number across an inequality, as this requires flipping all the inequality signs.