CBSE Class 11 Mathematics Revision Worksheet Set 02

SECTION A

Question 1. Find the number of non empty subsets of the set \(\{ 1, 2, 3, 4 \}\)
Answer: Let the given set be \(A = \{ 1, 2, 3, 4 \}\).
The total number of elements in set \(A\) is \(n = 4\).
The total number of subsets of a set containing \(n\) elements is given by \(2^n\).
Total number of subsets = \(2^4 = 16\).
A non-empty subset excludes the empty set (\(\emptyset\)).
Number of non-empty subsets = \(16 - 1 = 15\).
In simple words: The total number of subsets is 16, and subtracting the empty set gives 15 non-empty subsets.

Exam Tip: Remember the formula for the number of subsets of a finite set is \(2^n\), and always remember to subtract 1 for the empty set when asked for non-empty subsets.

Question 2. A wheel makes 180 revolutions per minute. Through how many radians does it turn in 1 second?
Answer: Number of revolutions made in 1 minute (60 seconds) = 180.
Number of revolutions made in 1 second = \(\frac{180}{60} = 3\) revolutions.
In 1 complete revolution, a wheel turns through an angle of \(2\pi\) radians.
Angle turned in 3 revolutions = \(3 \times 2\pi = 6\pi\) radians.
In simple words: Find revolutions per second first, then multiply by 2\pi to get the total radians.

Exam Tip: Pay careful attention to units like minutes and seconds. Convert time units correctly before calculating angles in radians.

Question 3. Solve \(|4x - 3| < 27\)
Answer: The given inequality is \(|4x - 3| < 27\).
Using the property of absolute value, if \(|x| < a\), then \(-a < x < a\).
\(-27 < 4x - 3 < 27\)
Add 3 to all parts of the inequality:
\(-27 + 3 < 4x < 27 + 3\)
\(-24 < 4x < 30\)
Divide by 4:
\(\frac{-24}{4} < x < \frac{30}{4}\)
\(-6 < x < \frac{15}{2}\)
Solution in interval notation: \(\left(-6, \frac{15}{2}\right)\).
In simple words: Remove the modulus by placing the expression between negative and positive values of the number on the right.

Exam Tip: Always rewrite modulus inequalities into standard compound linear inequalities before solving for \(x\).

Question 4. Write the general solution of \(\sin^2 2\theta = 0\)
Answer: The given equation is \(\sin^2 2\theta = 0\).
Taking the square root on both sides:
\(\sin 2\theta = 0\)
The general solution for \(\sin x = 0\) is \(x = n\pi\), where \(n \in \mathbb{Z}\).
Therefore, substitute \(x = 2\theta\):
\(2\theta = n\pi\)
\(\theta = \frac{n\pi}{2}\), where \(n \in \mathbb{Z}\).
In simple words: Simplify the squared sine equation to standard sine form and use standard trigonometric general solutions.

Exam Tip: Do not forget to state that \(n\) is an integer (\(n \in \mathbb{Z}\) or \(n \in \mathbb{N}_0\)) when writing general solutions.

SECTION B

Question 5. Prove that \(5^{2n} - 1\) is divisible by 24, using principle of mathematical induction for all \(n \in \mathbb{N}\)
Answer: Let \(P(n) : 5^{2n} - 1\) is divisible by 24.
Step 1: Check for \(n = 1\).
\(P(1) = 5^{2(1)} - 1 = 25 - 1 = 24\).
Since 24 is divisible by 24, \(P(1)\) is true.
Step 2: Assume \(P(k)\) is true for some positive integer \(k\).
That is, \(5^{2k} - 1 = 24m\), where \(m \in \mathbb{Z}\).
\(5^{2k} = 24m + 1\)  --- (Equation 1)
Step 3: We need to prove that \(P(k+1)\) is true, i.e., \(5^{2(k+1)} - 1\) is divisible by 24.
Consider \(5^{2(k+1)} - 1 = 5^{2k + 2} - 1 = 5^{2k} \cdot 5^2 - 1 = 25 \cdot 5^{2k} - 1\).
Substitute value of \(5^{2k}\) from Equation 1:
\(= 25(24m + 1) - 1\)
\(= 25 \cdot 24m + 25 - 1\)
\(= 25 \cdot 24m + 24\)
\(= 24(25m + 1)\).
This expression is a multiple of 24 and hence divisible by 24. Thus, \(P(k+1)\) is true whenever \(P(k)\) is true.
By the principle of mathematical induction, \(P(n)\) is true for all \(n \in \mathbb{N}\).
In simple words: Show it works for 1, assume it works for k, and prove it works for k + 1 using algebra.

Exam Tip: Always substitute the induction hypothesis cleanly in the step for \(k+1\) by isolating terms like \(5^{2k}\).

Question 6. Prove by using principle of mathematical induction \(\left(1 + \frac{1}{1}\right)\left(1 + \frac{1}{2}\right)\left(1 + \frac{1}{3}\right)\cdots\left(1 + \frac{1}{n}\right) = n + 1\) for all \(n \in \mathbb{N}\)
Answer: Let \(P(n) : \left(1 + \frac{1}{1}\right)\left(1 + \frac{1}{2}\right)\left(1 + \frac{1}{3}\right)\cdots\left(1 + \frac{1}{n}\right) = n + 1\).
Step 1: Check for \(n = 1\).
LHS = \(1 + \frac{1}{1} = 2\).
RHS = \(1 + 1 = 2\).
LHS = RHS, so \(P(1)\) is true.
Step 2: Assume \(P(k)\) is true for some positive integer \(k\).
\(\left(1 + \frac{1}{1}\right)\left(1 + \frac{1}{2}\right)\cdots\left(1 + \frac{1}{k}\right) = k + 1\)  --- (Equation 1)
Step 3: Prove for \(n = k + 1\).
LHS for \(k + 1\) is:
\(\left(1 + \frac{1}{1}\right)\left(1 + \frac{1}{2}\right)\cdots\left(1 + \frac{1}{k}\right)\left(1 + \frac{1}{k+1}\right)\)
Using Equation 1, substitute the product up to \(k\):
\(= (k + 1) \left(1 + \frac{1}{k+1}\right)\)
\(= (k + 1) \left(\frac{k + 1 + 1}{k+1}\right)\)
\(= (k + 1) \left(\frac{k + 2}{k+1}\right)\)
\(= k + 2 = (k + 1) + 1\), which is equal to RHS.
Thus, \(P(k+1)\) is true whenever \(P(k)\) is true.
By the principle of mathematical induction, the statement is true for all \(n \in \mathbb{N}\).
In simple words: Substitute the assumed result for k into the k + 1 term product and simplify algebraically.

Exam Tip: In product series proofs using induction, group the terms up to \(k\) to directly substitute your induction hypothesis.

Question 7. Prove that \(\cos 20^\circ \cos 30^\circ \cos 40^\circ \cos 80^\circ = \frac{\sqrt{3}}{16}\)
Answer: Consider the LHS:
\(\text{LHS} = \cos 20^\circ \cos 30^\circ \cos 40^\circ \cos 80^\circ\)
Substitute the exact value of \(\cos 30^\circ = \frac{\sqrt{3}}{2}\):
\(\text{LHS} = \frac{\sqrt{3}}{2} (\cos 20^\circ \cos 40^\circ \cos 80^\circ)\)
Multiply and divide the remaining terms by 2 to use the formula \(2\cos A \cos B = \cos(A+B) + \cos(A-B)\):
\(= \frac{\sqrt{3}}{2 \times 2} [2 \cos 40^\circ \cos 20^\circ] \cos 80^\circ\)
\(= \frac{\sqrt{3}}{4} [\cos(40^\circ + 20^\circ) + \cos(40^\circ - 20^\circ)] \cos 80^\circ\)
\(= \frac{\sqrt{3}}{4} [\cos 60^\circ + \cos 20^\circ] \cos 80^\circ\)
Since \(\cos 60^\circ = \frac{1}{2}\):
\(= \frac{\sqrt{3}}{4} \left[ \frac{1}{2} + \cos 20^\circ \right] \cos 80^\circ\)
\(= \frac{\sqrt{3}}{8} [ \cos 80^\circ + 2 \cos 20^\circ \cos 80^\circ ]\)
Use the product-to-sum formula again for \(2 \cos 20^\circ \cos 80^\circ = \cos(20^\circ + 80^\circ) + \cos(20^\circ - 80^\circ) = \cos 100^\circ + \cos(-60^\circ)\):
\(= \frac{\sqrt{3}}{8} [ \cos 80^\circ + \cos 100^\circ + \cos 60^\circ ]\)
Since \(\cos 100^\circ = \cos(180^\circ - 80^\circ) = -\cos 80^\circ\) and \(\cos 60^\circ = \frac{1}{2}\):
\(= \frac{\sqrt{3}}{8} [ \cos 80^\circ - \cos 80^\circ + \frac{1}{2} ]\)
\(= \frac{\sqrt{3}}{8} \left[ \frac{1}{2} \right] = \frac{\sqrt{3}}{16} = \text{RHS}\).
In simple words: Use product-to-sum trigonometric formulas systematically to combine and cancel angles.

Exam Tip: Memorize the transformation formulas \(2\cos A \cos B = \cos(A+B) + \cos(A-B)\) as they are frequently used in product-simplification problems.

Question 8. Show that \(\sqrt{2 + \sqrt{2 + \sqrt{2 + 2\cos 8\theta}}} = 2\cos \theta\)
Answer: Start from the innermost expression and use the trigonometric identity \(1 + \cos 2A = 2\cos^2 A\).
Innermost term: \(2 + 2\cos 8\theta = 2(1 + \cos 8\theta) = 2(2\cos^2 4\theta) = 4\cos^2 4\theta\).
Substitute this back into the square root:
\(\sqrt{2 + \sqrt{2 + \sqrt{4\cos^2 4\theta}}} = \sqrt{2 + \sqrt{2 + 2\cos 4\theta}}\).
Now simplify the next layer using the same identity for \(2 + 2\cos 4\theta\):
\(2 + 2\cos 4\theta = 2(1 + \cos 4\theta) = 2(2\cos^2 2\theta) = 4\cos^2 2\theta\).
Substitute back into the expression:
\(\sqrt{2 + \sqrt{4\cos^2 2\theta}} = \sqrt{2 + 2\cos 2\theta}\).
Apply the identity one more time for \(2 + 2\cos 2\theta\):
\(2 + 2\cos 2\theta = 2(1 + \cos 2\theta) = 2(2\cos^2 \theta) = 4\cos^2 \theta\).
Finally, take the outermost square root:
\(\sqrt{4\cos^2 \theta} = 2\cos \theta = \text{RHS}\).
In simple words: Work outwards from the deepest square root by repeatedly applying the half-angle formula \(1 + \cos A = 2\cos^2(A/2)\).

Exam Tip: Nested root problems in trigonometry almost always rely on successive applications of the identity \(1 + \cos 2A = 2\cos^2 A\).

Question 9. Let \(U = \{1, 2, 3, 4, 5, 6, 7, 8, 9, 10\}, A = \{1, 2, 3, 4\}, B = \{2, 4, 6, 8, 10\}, C = \{3, 4, 5, 6\}\)
i) Find \((C - B)^c\)
ii) Verify that \((A \cap B)^c = A^c \cup B^c\)
Answer:
i) Find \((C - B)^c\)
Given sets:
\(C = \{3, 4, 5, 6\}\)
\(B = \{2, 4, 6, 8, 10\}\)
\(C - B\) is the set of elements in \(C\) that are not in \(B\):
\(C - B = \{3, 5\}\)
Now find the complement \((C - B)^c\) with respect to universal set \(U\):
\((C - B)^c = U - \{3, 5\} = \{1, 2, 4, 6, 7, 8, 9, 10\}\).

ii) Verify that \((A \cap B)^c = A^c \cup B^c\)
First, find \(A \cap B\):
\(A = \{1, 2, 3, 4\}\)
\(B = \{2, 4, 6, 8, 10\}\)
\(A \cap B = \{2, 4\}\)
Now, LHS = \((A \cap B)^c = U - \{2, 4\} = \{1, 3, 5, 6, 7, 8, 9, 10\}\).
Next, find complements of \(A\) and \(B\):
\(A^c = U - A = \{5, 6, 7, 8, 9, 10\}\)
\(B^c = U - B = \{1, 3, 5, 7, 9\}\)
Now, RHS = \(A^c \cup B^c = \{5, 6, 7, 8, 9, 10\} \cup \{1, 3, 5, 7, 9\} = \{1, 3, 5, 6, 7, 8, 9, 10\}\).
Since LHS = RHS, De Morgan's Law \((A \cap B)^c = A^c \cup B^c\) is verified.
In simple words: Perform set differences and intersections step by step, then apply complements relative to the universal set U.

Exam Tip: Clearly write out the universal set elements when finding set complements to avoid missing numbers.

Question 10. Using properties of sets prove that
i) \(A \cup (A \cap B) = A\)
ii) \(A \cap (A \cup B) = A\)
Answer:
i) Prove \(A \cup (A \cap B) = A\) (Absorption Law)
Using distributive law: \(P \cup (Q \cap R) = (P \cup Q) \cap (P \cup R)\)
\(A \cup (A \cap B) = (A \cup A) \cap (A \cup B)\)
Since \(A \cup A = A\) (Idempotent Law):
\(= A \cap (A \cup B)\)
By the Absorption Law of sets, \(A \cap (A \cup B) = A\).

ii) Prove \(A \cap (A \cup B) = A\) (Absorption Law)
Using set properties and logical equivalence:
Let \(x\) be an arbitrary element such that \(x \in A \cap (A \cup B)\).
\(\Rightarrow x \in A\) and \(x \in (A \cup B)\)
\(\Rightarrow x \in A\) and (\(x \in A\) or \(x \in B\))
Since \(x \in A\) is true, the statement (\(x \in A\) or \(x \in B\)) is automatically true whenever \(x \in A\).
Therefore, \(x \in A\).
Thus, \(A \cap (A \cup B) \subseteq A\)  --- (1)
Conversely, if \(x \in A\), then \(x \in A\) and (\(x \in A\) or \(x \in B\)), which means \(x \in A \cap (A \cup B)\).
Thus, \(A \subseteq A \cap (A \cup B)\)  --- (2)
From (1) and (2), \(A \cap (A \cup B) = A\).
In simple words: Prove set equality either by using standard set algebra laws or by showing mutual subset containment using element arguments.

Exam Tip: Memorize fundamental set laws like Idempotent, Distributive, and Absorption laws to quickly prove set identities.

Question 11. The cost and revenue functions of the product are given by \(C(x) = 2x + 400\) and \(R(x) = 6x + 20\) respectively, where \(x\) is the number of items produced by the manufacturer. How many items the manufacturer must sell to realize some profit?
Answer: Profit \(P(x)\) is given by the difference between total revenue and total cost:
\(P(x) = R(x) - C(x)\)
Substitute the given functions:
\(P(x) = (6x + 20) - (2x + 400)\)
\(P(x) = 6x + 20 - 2x - 400\)
\(P(x) = 4x - 380\)
To realize some profit, the profit must be strictly greater than zero:
\(P(x) > 0\)
\(4x - 380 > 0\)
\(4x > 380\)
\(x > \frac{380}{4}\)
\(x > 95\)
Since the number of items must be an integer, the manufacturer must sell at least 96 items to realize a profit.
In simple words: Set up the profit function as Revenue minus Cost, then solve the inequality for profit greater than zero.

Exam Tip: Always remember that profit equals Revenue minus Cost (\(P = R - C\)). For "some profit", set \(P > 0\).

SECTION C

Question 12. In a town of 10,000 families, it was found that 40% families buys newspaper A, 20% families buy newspaper B, and 10% families buy newspaper C, 5% buy newspaper A & B, 3% buy newspaper B & C and 4% buy newspaper A & C. If 2% families buy all three newspapers, then find the number of families which buy newspaper
i) A only
ii) at least one of the three papers
iii) none of A, B & C
Answer: Let \(U\) be the universal set of all families in the town, so \(n(U) = 10,000\).
Let \(A, B, C\) be the sets of families buying newspapers A, B, and C respectively.
Given percentages converted to numbers of families:
\(n(A) = 40\% \text{ of } 10,000 = 4,000\)
\(n(B) = 20\% \text{ of } 10,000 = 2,000\)
\(n(C) = 10\% \text{ of } 10,000 = 1,000\)
\(n(A \cap B) = 5\% \text{ of } 10,000 = 500\)
\(n(B \cap C) = 3\% \text{ of } 10,000 = 300\)
\(n(A \cap C) = 4\% \text{ of } 10,000 = 400\)
\(n(A \cap B \cap C) = 2\% \text{ of } 10,000 = 200\)

i) Find the number of families which buy newspaper A only:
Number of families buying A only is given by:
\(n(A \text{ only}) = n(A) - n(A \cap B) - n(A \cap C) + n(A \cap B \cap C)\)
\(n(A \text{ only}) = 4000 - 500 - 400 + 200 = 3300\).

ii) Find the number of families which buy at least one of the three papers:
This is given by \(n(A \cup B \cup C)\):
\(n(A \cup B \cup C) = n(A) + n(B) + n(C) - n(A \cap B) - n(B \cap C) - n(A \cap C) + n(A \cap B \cap C)\)
\(n(A \cup B \cup C) = 4000 + 2000 + 1000 - 500 - 300 - 400 + 200 = 6000\).

iii) Find the number of families which buy none of A, B & C:
This is given by the complement \(n((A \cup B \cup C)^c)\):
\(n(\text{none}) = n(U) - n(A \cup B \cup C) = 10,000 - 6,000 = 4,000\).
In simple words: Use standard set cardinality formulas and Venn diagram principles to find individual intersections and unions.

Exam Tip: For "only A", remember to subtract both pairwise intersections and then add back the triple intersection \(n(A \cap B \cap C)\).

Question 13. Solve: \(3x + 2y \le 12, 3x + 4y \ge 12, x \ge 0, y \ge 1\) graphically.
Answer: Step 1: Convert inequalities into equations to find boundary lines:
1. Line 1: \(3x + 2y = 12\)
- When \(x = 0\), \(y = 6 \rightarrow (0, 6)\)
- When \(y = 0\), \(x = 4 \rightarrow (4, 0)\)
2. Line 2: \(3x + 4y = 12\)
- When \(x = 0\), \(y = 3 \rightarrow (0, 3)\)
- When \(y = 0\), \(x = 4 \rightarrow (4, 0)\)
3. Line 3: \(x = 0\) (Y-axis)
4. Line 4: \(y = 1\) (Horizontal line)

Step 2: Determine half-plane regions using test point \((0,0)\):
- For \(3x + 2y \le 12\): Test \((0,0) \rightarrow 0 \le 12\) (True, region contains origin).
- For \(3x + 4y \ge 12\): Test \((0,0) \rightarrow 0 \ge 12\) (False, region away from origin).
- For \(x \ge 0\) and \(y \ge 1\): Region lies to the right of Y-axis and above the line \(y = 1\).

Step 3: Intersection points of the feasible region:
- Intersection of \(3x + 2y = 12\) and \(y = 1\):
\(3x + 2(1) = 12 \Rightarrow 3x = 10 \Rightarrow x = \frac{10}{3} \rightarrow \left(\frac{10}{3}, 1\right)\)
- Intersection of \(3x + 4y = 12\) and \(y = 1\):
\(3x + 4(1) = 12 \Rightarrow 3x = 8 \Rightarrow x = \frac{8}{3} \rightarrow \left(\frac{8}{3}, 1\right)\)
- Intersection of \(3x + 2y = 12\) and \(3x + 4y = 12\):
Subtracting gives \(2y = 0 \Rightarrow y = 0\), which is outside the region \(y \ge 1\). Checking boundary constraints gives the bounded polygon vertices.
In simple words: Plot the boundary lines, determine the shaded region for each inequality, and find the overlapping feasible polygon.

Exam Tip: Always test the origin \((0,0)\) to determine which side of a non-origin passing line should be shaded.

Question 14. In any triangle \(\Delta ABC\), prove that: \((b^2 - c^2)\cot A + (c^2 - a^2)\cot B + (a^2 - b^2)\cot C = 0\)
Answer: Recall the Sine Rule and Cosine Rule in any triangle \(\Delta ABC\):
By Sine Rule: \(\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C} = 2R\)
\(\Rightarrow a = 2R\sin A, \, b = 2R\sin B, \, c = 2R\sin C\)
By Cosine Rule:
\(\cos A = \frac{b^2 + c^2 - a^2}{2bc}\)
We know that \(\cot A = \frac{\cos A}{\sin A}\).
Consider the first term of the expression: \((b^2 - c^2)\cot A\)
\((b^2 - c^2) \frac{\cos A}{\sin A} = (b^2 - c^2) \frac{b^2 + c^2 - a^2}{2bc \sin A}\)
Using \(a = 2R\sin A \Rightarrow \sin A = \frac{a}{2R}\):
\(= (b^2 - c^2) \frac{b^2 + c^2 - a^2}{2bc \left(\frac{a}{2R}\right)} = R \cdot \frac{(b^2 - c^2)(b^2 + c^2 - a^2)}{abc}\)
Similarly, write the second and third terms:
Second term: \(R \cdot \frac{(c^2 - a^2)(c^2 + a^2 - b^2)}{abc}\)
Third term: \(R \cdot \frac{(a^2 - b^2)(a^2 + b^2 - c^2)}{abc}\)
Summing all three terms:
\(\text{LHS} = \frac{R}{abc} [ (b^2 - c^2)(b^2 + c^2 - a^2) + (c^2 - a^2)(c^2 + a^2 - b^2) + (a^2 - b^2)(a^2 + b^2 - c^2) ]\)
Expanding each term in the numerator:
\((b^4 - c^4) - a^2(b^2 - c^2) + (c^4 - a^4) - b^2(c^2 - a^2) + (a^4 - b^4) - c^2(a^2 - b^2)\)
Notice that the terms \(b^4, c^4, a^4\) cancel out completely:
\(= -a^2b^2 + a^2c^2 - b^2c^2 + a^2b^2 - c^2a^2 + b^2c^2 = 0\)
\(= \frac{R}{abc} [0] = 0 = \text{RHS}\).
In simple words: Convert cotangent into cosine and sine, apply the cosine rule, and expand algebraic terms so they completely cancel out to zero.

Exam Tip: Substitute trigonometric ratios into side lengths using sine and cosine rules, then rely on algebraic cancellation of cyclic terms.