CBSE Class 11 Mathematics Linear Inequalities Assignment Set 01

Chapter: Linear Inequalities, Mathematical Reasoning, Sets and Statistics

Question 1. The cost and revenue functions of a product are given by C(x) = 20x + 4000 and R(x) = 60x + 2000, respectively, where x is the number of items produced and sold. How many items must be sold to realise some profit?
Answer: For a product to realize a profit, the total revenue must be strictly greater than the total cost.
\(R(x) > C(x)\)
\(60x + 2000 > 20x + 4000\)
\(60x - 20x > 4000 - 2000\)
\(40x > 2000\)
\(x > \frac{2000}{40}\)
\(x > 50\)
In simple words: The manufacturer needs to sell more than 50 items to start making a profit.

Exam Tip: Profit is defined as Revenue minus Cost, so set up the inequality \(R(x) - C(x) > 0\) for quick verification.

Question 2. Solve the following system of inequalities: \(\frac{x}{2x+1} \geq \frac{1}{4}\), \(\frac{6x}{4x-1} < \frac{1}{2}\)
Answer: Solve the first inequality:
\(\frac{x}{2x+1} - \frac{1}{4} \geq 0\)
\(\frac{4x - (2x+1)}{4(2x+1)} \geq 0\)
\(\frac{2x - 1}{4(2x+1)} \geq 0\)
Critical points are \(x = \frac{1}{2}\) and \(x = -\frac{1}{2}\).
Solution for first inequality: \(x \in \left(-\infty, -\frac{1}{2}\right) \cup \left[\frac{1}{2}, \infty\right)\).

Solve the second inequality:
\(\frac{6x}{4x-1} - \frac{1}{2} < 0\)
\(\frac{12x - (4x-1)}{2(4x-1)} < 0\)
\(\frac{8x + 1}{2(4x-1)} < 0\)
Critical points are \(x = -\frac{1}{8}\) and \(x = \frac{1}{4}\).
Solution for second inequality: \(x \in \left(-\frac{1}{8}, \frac{1}{4}\right)\).

Since there is no common region overlapping between the two solution sets, there is no common solution.
In simple words: The conditions required by the two inequalities contradict each other, hence no solution exists.

Exam Tip: Always find the intersection of individual solution sets when solving a system of inequalities.

Question 3. Solve for x, If \(\frac{x-2}{x-4} \geq 0\)
Answer: The rational expression is non-negative when the numerator and denominator have the same sign and the denominator is non-zero.
Critical points are \(x = 2\) and \(x = 4\) (where denominator is zero).
Using the sign method on intervals \((-\infty, 2]\), \([2, 4)\), and \((4, \infty)\):
The expression is positive or zero in \((-\infty, 2]\) and strictly positive in \((4, \infty)\).
Therefore, \(x \in (-\infty, 2] \cup (4, \infty)\).
In simple words: The fraction is positive when both top and bottom are positive or both are negative.

Exam Tip: Always exclude values that make the denominator zero from the solution domain.

Question 4. Solve for x, If \(\frac{x+6}{x+2} \leq 0\)
Answer: Critical points are \(x = -6\) and \(x = -2\) (where denominator \(x \neq -2\)).
For the fraction to be less than or equal to zero, the numerator and denominator must have opposite signs.
Testing intervals: when \(x \in [-6, -2)\), the expression is less than or equal to zero.
Therefore, \(x \in [-6, -2)\).
In simple words: Find the range where one part is positive and the other is negative.

Question 5. Solve for x, If \(\frac{x+3}{x-7} < 0\)
Answer: Critical points are \(x = -3\) and \(x = 7\) (with \(x \neq 7\)).
For a negative quotient, numerator and denominator must have opposite signs.
Testing intervals yields negative values between the critical points.
Therefore, \(x \in (-3, 7)\).
In simple words: The quotient is negative between its roots.

Exam Tip: For expressions of the form \(\frac{x-a}{x-b} < 0\), the solution always lies between \(a\) and \(b\).

Question 6. Solve for x, If \(|x| < 4\)
Answer: By the property of absolute values, \(|x| < a\) is equivalent to \(-a < x < a\).
Here, \(-4 < x < 4\).
Therefore, \(x \in (-4, 4)\).
In simple words: The distance of x from zero is less than 4 units.

Exam Tip: Remember the golden rule: \(|x| < k \iff -k < x < k\) for positive \(k\).

Question 7. Solve for x, If \(|3x - 7| \leq 2\)
Answer: Using absolute value properties:
\(-2 \leq 3x - 7 \leq 2\)
Add 7 to all parts:
\(-2 + 7 \leq 3x \leq 2 + 7\)
\(5 \leq 3x \leq 9\)
Divide by 3:
\(\frac{5}{3} \leq x \leq 3\)
Therefore, \(x \in \left[\frac{5}{3}, 3\right]\).
In simple words: Isolate x in the middle by applying operations to all three parts.

Exam Tip: Compound inequalities can be solved simultaneously by performing identical operations on all parts.

Question 8. Solve for x, If \(|x+1| + |x| > 3\)
Answer: We analyze the critical points where expressions inside absolute values become zero: \(x = -1\) and \(x = 0\).
Case 1: \(x < -1\)
\(-(x+1) - x > 3 \implies -2x - 1 > 3 \implies -2x > 4 \implies x < -2\).
Intersection with case condition gives \(x < -2\), so \(x \in (-\infty, -2)\).

Case 2: \(-1 \leq x < 0\)
\((x+1) - x > 3 \implies 1 > 3\) (which is false, no solution).

Case 3: \(x \geq 0\)
\((x+1) + x > 3 \implies 2x + 1 > 3 \implies 2x > 2 \implies x > 1\).
Intersection with case condition gives \(x > 1\), so \(x \in (1, \infty)\).

Combining all cases: \(x \in (-\infty, -2) \cup (1, \infty)\).
In simple words: Break the number line into intervals based on critical points and test each region.

Exam Tip: Critical points split the real line into distinct zones; solve each zone independently.

Question 9. Solve for x, If \(\frac{|x+3| + x}{x+2} > 1\)
Answer: Critical point for absolute value is \(x = -3\).
Case 1: \(x < -3\)
\(\frac{-(x+3) + x}{x+2} > 1 \implies \frac{-3}{x+2} > 1 \implies \frac{-3}{x+2} - 1 > 0 \implies \frac{-3 - x - 2}{x+2} > 0 \implies \frac{-x - 5}{x+2} > 0\)
\(\frac{x+5}{x+2} < 0 \implies x \in (-5, -2)\).
Intersecting with \(x < -3\) gives \(x \in (-5, -3)\).

Case 2: \(x \geq -3\)
\(\frac{(x+3) + x}{x+2} > 1 \implies \frac{2x+3}{x+2} > 1 \implies \frac{2x+3}{x+2} - 1 > 0 \implies \frac{x+1}{x+2} > 0\)
Solution gives \(x \in (-\infty, -2) \cup (-1, \infty)\).
Intersecting with \(x \geq -3\) gives \(x \in [-3, -2) \cup (-1, \infty)\).

Combining both cases: \(x \in (-5, -2) \cup (-1, \infty)\).
In simple words: Analyze the inequality separately for values below and above -3.

Question 10. Solve the following system of linear inequalities: \(3x+2y \geq 24\), \(3x+y \leq 15\), \(x \geq 4\)
Answer: From \(x \geq 4\) and the inequality \(3x + y \leq 15\):
Substitute the minimum value of \(x\): \(3(4) + y \leq 15 \implies 12 + y \leq 15 \implies y \leq 3\).
Now test these values in the first inequality \(3x + 2y \geq 24\):
Minimum value of the left side is when \(x = 4\) and \(y = 3\):
\(3(4) + 2(3) = 12 + 6 = 18\), which is not greater than or equal to 24.
Since the maximum possible value under the other constraints is 18, the inequality \(3x + 2y \geq 24\) can never be satisfied.
Therefore, No solution.
In simple words: The given conditions pull the variables in opposite directions, leaving no overlapping feasible region.

Exam Tip: Test extreme corner points of the restricted domain to quickly check for empty feasible regions.

Question 11. Show that the solution set of the following system of linear inequalities is an unbounded region: \(2x+y \geq 8\), \(x+2y \geq 10\), \(x \geq 0, y \geq 0\)
Answer: 1. Graph the line \(2x + y = 8\) (intercepts: (0, 8) and (4, 0)) with region pointing away from origin.
2. Graph the line \(x + 2y = 10\) (intercepts: (0, 5) and (10, 0)) with region pointing away from origin.
3. The non-negativity constraints \(x \geq 0\) and \(y \geq 0\) restrict the region to the first quadrant.
4. The intersection of these half-planes forms a region bounded below and to the left by the lines and axes, but extending indefinitely toward the top-right (infinitely large values of \(x\) and \(y\)).
Therefore, the solution set is an unbounded region.
In simple words: The shaded feasible area stretches infinitely toward the top-right direction.

Exam Tip: Feasible regions with all 'greater than or equal to' constraints alongside positive quadrant restrictions are typically unbounded.

Question 12. In drilling world’s deepest hole it was found that the temperature T in degree Celsius, x km below the earth’s surface was given by \(T = 30 + 25(x-3)\), \(3 \leq x \leq 15\), at what depth will the temperature be between \(155^\circ\text{C}\) and \(255^\circ\text{C}\)?
Answer: We are given the condition \(155 < T < 255\).
Substitute the expression for \(T\):
\(155 < 30 + 25(x-3) < 255\)
Subtract 30 from all parts:
\(125 < 25(x-3) < 225\)
Divide all parts by 25:
\(5 < x - 3 < 9\)
Add 3 to all parts:
\(8 < x < 12\)
Therefore, the depth will be between 8 km and 12 km.
In simple words: Set up a compound inequality with the temperature limits and solve for depth x.

Exam Tip: Ensure the calculated depth range falls within the valid domain \(3 \leq x \leq 15\).

Question 13. Give an example of (I) statement (II) Simple Statement (III) Compound Statement
Answer: (I) Statement: "The sum of angles in a triangle is \(180^\circ\)." (A declarative sentence that is definitively true or false).
(II) Simple Statement: "Paris is the capital of France." (A single statement without any connective words).
(III) Compound Statement: "Paris is the capital of France and London is the capital of England." (Combined using logical connectives like 'and').
In simple words: Statements have a truth value; simple ones have single ideas, while compound ones connect multiple ideas.

Exam Tip: Remember that imperative, interrogative, or exclamatory sentences are never mathematical statements.

Question 14. Form the conjunction of the following simple statements: p: Ram is a boy, q: Sita is a girl
Answer: Conjunction is represented by the logical connective AND (\(\land\)).
\(p \land q\): Ram is a boy and Sita is a girl.
In simple words: Join both statements using the word 'and'.

Exam Tip: Conjunction is true only when both constituent statements are true simultaneously.

Question 15. Form the disjunction of the following simple statements: p: the sun shines, q: It rains
Answer: Disjunction is represented by the logical connective OR (\(\lor\)).
\(p \lor q\): The sun shines or it rains.
In simple words: Join both statements using the word 'or'.

Exam Tip: Disjunction is true if at least one of the statements is true.

Question 16. Write the negation of the statement: p: New Delhi is a city
Answer: Negation is denoted by \(\sim p\).
\(\sim p\): New Delhi is not a city.
In simple words: Insert 'not' to reverse the truth value of the statement.

Exam Tip: The negation of a true statement is always false, and vice versa.

Question 17. Calculate the Mean deviation about Median for the following data:
Class interval: 0-6, 6-12, 12-18, 18-24, 24-30
Frequency: 4, 5, 3, 6, 2
Answer:

Here \(N = 20\), so \(\frac{N}{2} = 10\). Median class is 12-18.
Median \(\text{Med} = L + \frac{\frac{N}{2} - cf}{f} \times h = 12 + \frac{10 - 9}{3} \times 6 = 12 + 2 = 14\).
Mean Deviation about Median = \(\frac{\sum f_i |x_i - \text{Med}|}{N} = \frac{140}{20} = 7\).
In simple words: Find the median first, calculate absolute deviations from it, multiply by frequencies, and find the weighted average.
Exam Tip: Always construct a clean cumulative frequency table to locate the exact median class interval.

Question 18. Find the Mean, Variance and S.D. for the following frequency distribution
Classes: 0-10, 10-20, 20-30, 30-40, 40-50, 50-60
Frequency: 11, 29, 18, 4, 5, 3
Answer:

Mean \(\bar{x} = \frac{\sum f_i x_i}{N} = \frac{1470}{70} = 21\).
Variance \(\sigma^2 = \frac{1}{N}\sum f_i x_i^2 - (\bar{x})^2 = \frac{42150}{70} - (21)^2 = 602.14 - 441 = 161.14\).
Standard Deviation \(\text{S.D.} = \sqrt{161.14} = 12.7\).
In simple words: Use summary columns for frequency times midpoints and frequency times squared midpoints to compute mean and variance easily.
Exam Tip: Using the formula \(\sigma^2 = \frac{\sum f_i x_i^2}{N} - \left(\frac{\sum f_i x_i}{N}\right)^2\) avoids calculating individual deviations.

Question 19. Calculate the Mean deviation about Mean for the following data
Class interval: 0-4, 4-8, 8-12, 12-16, 16-20
Frequency: 4, 6, 8, 5, 2
Answer:

Mean \(\bar{x} = \frac{230}{25} = 9.2\).
Mean Deviation about Mean = \(\frac{\sum f_i |x_i - \bar{x}|}{N} = \frac{96}{25} = 3.84\).
In simple words: Find the arithmetic mean, take absolute differences with midpoints, and compute the weighted average.
Exam Tip: Double-check midpoint calculations before computing deviations to maintain accuracy.

Question 20. Mean and S.D of 100 items are 50 and 4 respectively, find the sum of all the items and the sum of the square of the items.
Answer: Given \(N = 100\), Mean \(\bar{x} = 50\), and S.D \(\sigma = 4\).
Sum of all items \(\sum x_i = N \times \bar{x} = 100 \times 50 = 5000\).
Variance \(\sigma^2 = 16\).
Using variance formula: \(\sigma^2 = \frac{\sum x_i^2}{N} - (\bar{x})^2\)
\(16 = \frac{\sum x_i^2}{100} - (50)^2\)
\(16 = \frac{\sum x_i^2}{100} - 2500\)
\(\frac{\sum x_i^2}{100} = 2516\)
Sum of squares of items \(\sum x_i^2 = 251600\).
In simple words: Use standard statistical formulas backward to find sums and sum of squares.

Exam Tip: Remember that variance is the square of standard deviation (\(\sigma^2\)).

Question 21. Define following with example: Set, Set Theory, Roaster/Tabular form, Set builder form, Empty set, Singleton set, Finite set, Infinite set, Equal set, Subset, Super set, Open interval, Close interval, Half close interval, Half open interval, Power set, Universal set, Venn diagram, Union of set, Intersection of set, Complement of a set, Difference of sets and Disjoint set.
Answer:
- Set: A well-defined collection of objects. Example: \(A = \{1, 2, 3\}\).
- Set Theory: The mathematical branch studying sets and their properties.
- Roaster/Tabular form: Listing elements inside curly braces. Example: \(\{a, e, i, o, u\}\).
- Set builder form: Defining elements by a common property. Example: \(\{x : x \text{ is a vowel}\}\).
- Empty set: A set containing no elements, denoted by \(\phi\). Example: \(\{x : x^2 = 4 \text{ and } x \text{ is odd}\}\).
- Singleton set: A set with exactly one element. Example: \(\{5\}\).
- Finite set: A set with a countable number of elements. Example: \(\{1, 2, 3\}\).
- Infinite set: A set with endless elements. Example: \(\mathbb{N} = \{1, 2, 3, \dots\}\).
- Equal set: Sets with identical elements. Example: \(\{1, 2\}\) and \(\{2, 1\}\).
- Subset: A set whose elements all belong to another set (\(A \subseteq B\)). Example: \(\{1\} \subseteq \{1, 2\}\).
- Super set: If \(A \subseteq B\), then \(B\) is a superset of \(A\) (\(B \supseteq A\)).
- Open interval: Interval excluding endpoints, denoted \((a, b)\). Example: \((1, 3)\).
- Close interval: Interval including endpoints, denoted \([a, b]\). Example: \([1, 3]\).
- Half close interval: Closed at one end and open at the other, e.g., \((a, b]\).
- Half open interval: Open at one end and closed at the other, e.g., \([a, b)\).
- Power set: The set of all subsets of a set. Example: For \(\{1\}\), power set is \(\{\phi, \{1\}\}\).
- Universal set: A master set containing all objects under consideration, denoted \(U\).
- Venn diagram: A diagram representing sets using overlapping geometric circles.
- Union of set: Combining all elements of sets, denoted \(A \cup B\).
- Intersection of set: Common elements between sets, denoted \(A \cap B\).
- Complement of a set: Elements in universal set not in \(A\), denoted \(A'\) or \(A^c\).
- Difference of sets: Elements in \(A\) but not in \(B\), denoted \(A - B\).
- Disjoint set: Sets with no common elements, i.e., \(A \cap B = \phi\).
In simple words: These fundamental definitions form the building blocks of set theory and logic.

Exam Tip: Memorize standard notations and symbols for set operations to score full marks in definitions.

Question 22. If \(\Omega = \{a, b, c, d, e, f, g, h\}\), \(A = \{b, c\}\), \(B = \{c, d, e\}\), \(C = \{c, g, h\}\), then State and verify De Morgan’s rule for two and three sets
Answer:
De Morgan's Laws for two sets:
1. \((A \cup B)' = A' \cap B'\)
2. \((A \cap B)' = A' \cup B'\)

De Morgan's Laws for three sets:
1. \((A \cup B \cup C)' = A' \cap B' \cap C'\)
2. \((A \cap B \cap C)' = A' \cup B' \cup C'\)

Verification for two sets (\(A \cup B\)):
\(A \cup B = \{b, c, d, e\}\)
\((A \cup B)' = \Omega - \{b, c, d, e\} = \{a, f, g, h\}\)
\(A' = \{a, d, e, f, g, h\}\), \(B' = \{a, b, f, g, h\}\)
\(A' \cap B' = \{a, f, g, h\}\). Hence verified.
In simple words: The complement of a union is the intersection of complements, and vice versa.

Exam Tip: Always compute individual complements and unions carefully step-by-step.

Question 23. In a survey of 25 students, it was found that 15 had taken Mathematics, 12 had taken Physics and 11 had taken Chemistry, 5 had taken Mathematics and Chemistry, 9 had taken Mathematics and Physics, 4 had taken Physics and Chemistry and 3 had taken all the three subjects. Find the number of students that had taken: (i) Only Chemistry, (ii) only Mathematics, (iii) Only Physics, (iv) Physics and Chemistry but not Mathematics, (v) Mathematics and Physics but not Chemistry, (vi) Mathematics and Chemistry but not Physics, (vii) Only one of the subjects, (viii) At least one of the three subjects, (ix) None of the subjects.
Answer: Let \(M, P, C\) denote Mathematics, Physics, and Chemistry respectively.
Given: \(n(U) = 25\), \(n(M) = 15\), \(n(P) = 12\), \(n(C) = 11\),
\(n(M \cap C) = 5\), \(n(M \cap P) = 9\), \(n(P \cap C) = 4\), \(n(M \cap P \cap C) = 3\).

(i) Only Chemistry: \(n(C) - n(M \cap C) - n(P \cap C) + n(M \cap P \cap C) = 11 - 5 - 4 + 3 = 5\).
(ii) Only Mathematics: \(15 - (9-3) - (5-3) - 3 = 15 - 6 - 2 - 3 = 4\).
(iii) Only Physics: \(12 - (9-3) - (4-3) - 3 = 12 - 6 - 1 - 3 = 2\).
(iv) Physics & Chemistry not Mathematics: \(4 - 3 = 1\).
(v) Mathematics & Physics not Chemistry: \(9 - 3 = 6\).
(vi) Mathematics & Chemistry not Physics: \(5 - 3 = 2\).
(vii) Only one subject: \(4 (\text{only M}) + 2 (\text{only P}) + 5 (\text{only C}) = 11\).
(viii) At least one subject: \(n(M \cup P \cup C) = n(M) + n(P) + n(C) - n(M \cap P) - n(P \cap C) - n(M \cap C) + n(M \cap P \cap C) = 15 + 12 + 11 - 9 - 4 - 5 + 3 = 23\).
(ix) None of the subjects: \(25 - 23 = 2\).
In simple words: Draw a 3-circle Venn diagram and fill numbers from innermost intersection outward.

Exam Tip: Always start filling a Venn diagram from the center (intersection of all three sets).

Question 24. If \(A=\{1,2,3,4,5\}\), \(B=\{1,3,5,8\}\), \(C=\{2,5,7,8\}\), then verify the following:
(i) \(A - (B \cup C) = (A - B) \cap (A - C)\)
(ii) \(A - (B \cap C) = (A - B) \cup (A - C)\)
(iii) \(A \cap (B - C) = (A \cap B) - (A \cap C)\)
(iv) \(A \cup B = (A - B) \cup B\)
(v) \(A \cup (B \cap C) = (A \cup B) \cap (A \cup C)\)
(vi) \(A \cap (B \cup C) = (A \cap B) \cup (A \cap C)\)
Answer:
Given sets: \(A = \{1, 2, 3, 4, 5\}\), \(B = \{1, 3, 5, 8\}\), \(C = \{2, 5, 7, 8\}\).

(i) \(B \cup C = \{1, 2, 3, 5, 7, 8\} \implies A - (B \cup C) = \{4\}\).
\(A - B = \{2, 4\}\), \(A - C = \{1, 3, 4\} \implies (A - B) \cap (A - C) = \{4\} \text{ (Verified).}\)

(ii) \(B \cap C = \{5, 8\} \implies A - (B \cap C) = \{1, 2, 3, 4\}\).
\((A - B) \cup (A - C) = \{2, 4\} \cup \{1, 3, 4\} = \{1, 2, 3, 4\} \text{ (Verified).}\)

(iii) \(B - C = \{1, 3\} \implies A \cap (B - C) = \{1, 3\}\).
\(A \cap B = \{1, 3, 5\}\), \(A \cap C = \{2, 5\} \implies (A \cap B) - (A \cap C) = \{1, 3\} \text{ (Verified).}\)

(iv) \(A \cup B = \{1, 2, 3, 4, 5, 8\}\).
\((A - B) \cup B = \{2, 4\} \cup \{1, 3, 5, 8\} = \{1, 2, 3, 4, 5, 8\} \text{ (Verified).}\)

(v) \(B \cap C = \{5, 8\} \implies A \cup (B \cap C) = \{1, 2, 3, 4, 5, 8\}\).
\((A \cup B) \cap (A \cup C) = \{1, 2, 3, 4, 5, 8\} \cap \{1, 2, 3, 4, 5, 7, 8\} = \{1, 2, 3, 4, 5, 8\} \text{ (Verified).}\)

(vi) \(B \cup C = \{1, 2, 3, 5, 7, 8\} \implies A \cap (B \cup C) = \{1, 2, 3, 5\}\).
\((A \cap B) \cup (A \cap C) = \{1, 3, 5\} \cup \{2, 5\} = \{1, 2, 3, 5\} \text{ (Verified).}\)
In simple words: Compute left-hand side and right-hand side sets separately and confirm they contain identical elements.

Exam Tip: Check each set operation meticulously to avoid missing individual elements.

Question 25. Life of bulbs produced by two factories A and B are given below:
Length of life (in hrs): 550-650, 650-750, 750-850, 850-950, 950-1000
Factory A (No of bulbs): 10, 22, 52, 20, 16
Factory B (No of bulbs): 8, 60, 24, 16, 12
The bulbs of which factory are more consistent from the point of view of length of life?
Answer: Consistency is measured by the coefficient of variation (\(\text{C.V.} = \frac{\sigma}{\bar{x}} \times 100\)). The factory with the smaller coefficient of variation is more consistent.
Calculating mean and standard deviation for Factory A and Factory B reveals that Factory A has a smaller coefficient of variation (or standard deviation relative to mean).
Therefore, Factory A bulbs are more consistent.
In simple words: Lower variability means higher consistency among products.

Exam Tip: Coefficient of variation is the standard metric used for comparing consistency between datasets.

Question 26. Determine mean and S.D of first n terms of an A.P. Whose first term is 'a' and common difference is 'd'
Answer: The terms of the A.P. are \(a, a+d, a+2d, \dots, a+(n-1)d\).
Sum of first \(n\) terms = \(\frac{n}{2}[2a + (n-1)d]\).
Mean \(\bar{x} = \frac{\sum x_i}{n} = \frac{\frac{n}{2}[2a + (n-1)d]}{n} = a + \frac{(n-1)d}{2}\).
Standard Deviation \(\text{S.D.} = |d| \sqrt{\frac{n^2 - 1}{12}}\).
In simple words: Use standard formulas for summation of arithmetic progressions to derive mean and variance.

Exam Tip: Remember the standard variance formula for the first \(n\) natural numbers when simplifying A.P. variance.

Question 27. A and B are two sets such that \(n(A) = 3\) and \(n(B) = 6\), find (i) minimum value of \(n(A \cup B)\), (ii) maximum value of \(n(A \cup B)\)
Answer:
(i) Minimum value of \(n(A \cup B)\) occurs when \(A\) is a subset of \(B\). Hence, minimum value = \(n(B) = 6\).
(ii) Maximum value of \(n(A \cup B)\) occurs when \(A\) and \(B\) are disjoint sets. Hence, maximum value = \(n(A) + n(B) = 3 + 6 = 9\).
In simple words: The union size is smallest when one set is inside the other, and largest when they share no common elements.

Exam Tip: Use the formula \(n(A \cup B) = n(A) + n(B) - n(A \cap B)\) to determine bounds.

Question 28. Two finite sets have m and n elements. The total number of subsets of the first set is 56 more than the total number of subsets of the second set. Find the value of m and n.
Answer: Number of subsets of a set with \(k\) elements is \(2^k\).
Given: \(2^m - 2^n = 56\)
\(2^n(2^{m-n} - 1) = 8 \times 7 = 2^3(7)\)
Comparing factors: \(2^n = 2^3 \implies n = 3\).
\(2^{m-3} - 1 = 7 \implies 2^{m-3} = 8 = 2^3 \implies m - 3 = 3 \implies m = 6\).
Therefore, \(m = 6\) and \(n = 3\).
In simple words: Equate powers of two by factoring 56 into a power of 2 and an odd number.

Exam Tip: A set with \(k\) elements always has exactly \(2^k\) subsets.

Question 29. Let A and B be two sets such that \(n(A) = 20\), \(n(A \cup B) = 42\), and \(n(A \cap B) = 4\), find (i) \(n(B)\), (ii) \(n(A-B)\), (iii) \(n(B-A)\)
Answer:
(i) Using union formula: \(n(A \cup B) = n(A) + n(B) - n(A \cap B)\)
\(42 = 20 + n(B) - 4\)
\(42 = 16 + n(B) \implies n(B) = 42 - 16 = 26\).

(ii) \(n(A - B) = n(A) - n(A \cap B) = 20 - 4 = 16\).

(iii) \(n(B - A) = n(B) - n(A \cap B) = 26 - 4 = 22\).
In simple words: Use standard set cardinality formulas to find unknown counts.

Exam Tip: Memorize fundamental relations like \(n(A - B) = n(A) - n(A \cap B)\) for quick calculations.