SECTION – A

1. If a set A has n elements then the total number of subsets of A is:
   • a) 2n
   • b) n
   • c) 2ⁿ
   • d) n²
2. The domain of the function f: R → R defined by f(x) = √(x² – 4) is:
   • a) [-2, 2]
   • b) (-∞, ∞)
   • c) (-∞, -2] ∪ [2, ∞)
   • d) (-2, 2)
3. The range of the function f(x) = x/|x| is:
   • a) {-1, 1}
   • b) R – {0}
   • c) R – {-1, 1}
   • d) {-1, 2}
4. If A = {1, 2, 4}, B = {2, 4, 5}, C = {2, 5}, then (A – B) × (B – C) is:
   • a) {(1, 4)}
   • b) (2, 5)
   • c) {(1, 2), (1, 5), (2, 5)}
   • d) (1, 4)
5. The value of tan 75° – cot 75° is equal to:
   • a) 2 – √3
   • b) 1 + 2√3
   • c) 2√3
   • d) 2 + √3
6. If tan θ = √3 and lies in third quadrant, then value of sin θ is:
   • a) 1/√10
   • b) -1/√10
   • c) -3/√10
   • d) 3/√10
7. Which is greater, sin 24° or cos 24°?
   • a) both are equal
   • b) cos 24°
   • c) sin 24°
   • d) cannot be compared
8. Mark the correct answer for i^-75 = ?
   • a) i
   • b) -1
   • c) -i
   • d) 1
9. Let x, y ∈ R, then x + iy is a non-real complex number if:
   • a) y = 0
   • b) x ≠ 0
   • c) x = 0
   • d) y ≠ 0
10. If a, b, c are real numbers such that a > b, c < 0, then:
    • a) ac > bc
    • b) ac < bc
    • c) ac ≥ bc
    • d) ac ≤ bc
11. Solve: 3x + 5 < x - 7, when x is a real number:
    • a) x > -12
    • b) x < -12
    • c) x < -6
    • d) x > -6
12. Three persons enter a railway compartment. If there are 5 seats vacant, in how many ways can they take these seats?
    • a) 60
    • b) 125
    • c) 20
    • d) 15
13. If 15C3r = 15Cr+3 then r is:
    • a) 2
    • b) 3
    • c) 1
    • d) 0
14. 15C8 + 15C9 – 15C6 – 15C7 is equal to:
    • a) 15C4
    • b) 15C1
    • c) 0
    • d) 1
15. (√3 + 1)²ⁿ⁺¹ + (√3 – 1)²ⁿ⁺¹ is:
    • a) an even positive integer
    • b) an irrational number
    • c) an odd positive integer
    • d) a rational number
16. If sin θ + cos θ = 1 then sin 2θ is:
    • a) 1
    • b) 1
    • c) 0
    • d) -1

SECTION – A

17. The value of sin 3θ / (1 + cos 2θ) is equal to:
    • a) cosθ
    • b) sinθ
    • c) -cosθ
    • d) -sinθ
18. Tan(3π/2 + θ) is equal to:
    • a) -tan θ
    • b) tan θ
    • c) cot θ
    • d) -cot θ
19. Assertion (A): If sin x = -1/2, then cos x = -√3/2.
    Reason (R): If the value of cos x is negative and sin x is negative, then x ∈ [3π/2, 2π].
    • a) Both A and R are true and R is the correct explanation of A.
    • b) Both A and R are true but R is not the correct explanation of A.
    • c) A is true but R is false.
    • d) A is false but R is true.
20. Assertion (A): The value of sin(-690°) cos(-300°) + cos(-750°) sin(-240°) = 1.
    Reason (R): The values of sin and cos are negative in the third and fourth quadrants, respectively.
    • a) Both A and R are true and R is the correct explanation of A.
    • b) Both A and R are true but R is not the correct explanation of A.
    • c) A is true but R is false.
    • d) A is false but R is true.

SECTION – B

21. Find the range of f(x) = x² + 2, x ∈ R.
    OR
    Solve the equation z² = z̅, where z = x + iy.
22. Find the value of (51)⁴ using the binomial theorem.
23. Let f: R → R be given by f(x) = x² + 3. Find the pre-images of 39 and 2 under f.
24. If Cos A = 1/7 and Cos B = 13/14 where A and B are acute angles, then find the value of A – B.
25. Write the multiplicative inverse of (2 + i√3)².

SECTION – C

26. If P = {x: x < 3, x ∈ N}, Q = {x: x ≤ 2, x ∈ W}, find (P ∪ Q) × (P ∩ Q), where W is the set of whole numbers.
27. Find the value of (√2 + 1)⁵ − (√2 − 1)⁵.
28. Solve the inequality 2x – 1 > x + (7 − x)/3; 4x + 7 > 15, x ∈ R. Represent it graphically.
29. Prove that: sin 3x + sin 2x − sin x = 4 sin x cos(x/2) cos(3x/2).
30. If ⁸Pᵣ = 1680 and ⁸Cᵣ = 70, find the value of n and r.
31. Write the solution set of the inequality -5 ≤ (5 − 3x)/2 ≤ 8.
    OR
    Using the binomial theorem, determine which number is smaller: (1.2)⁴⁰⁰⁰ or 800?

SECTION D

32. A committee of 8 students is to be selected from 8 boys and 6 girls. In how many ways this can be done if each group is to consist of at least 3 boys and 3 girls?

33. How many words can be made by using all the letters of the word MATHEMATICS in which all the vowels are never together?

34. The longest side of a triangle is twice the shortest side, and the third side is 2 cm longer than the shortest side. If the perimeter of the triangle is more than 166 cm, then find the minimum length of the shortest side.

35. (a) Find n if ⁿP₅ : ^n-1P₄ = 6:1.
(b) Find r if 4.⁶P_r = ⁶P_r+1.

OR
(a) Find n if ⁿC₆ : ^n-3C₃ = 33:44.
(b) Find r if ¹⁵C_r-1 : ¹⁵C_r-1 = 11:5.

SECTION E

36. Read the text carefully and answer the questions: A teacher draws a triangle on the board and asks the students the following questions:

• (a) What is the area of the figure as a function of x?
• (b) What is the perimeter of the figure as a function of x?
• (c) What is the area A(4), when x = 4?

37. The conjugate of a complex number z is the complex number obtained by changing the sign of the imaginary part of z. It is denoted by z̅. The modulus (or absolute value) of a complex number z = a + ib is defined as the non-negative real number |z| = √(a² + b²).

• (a) If f(z) = (7z̅ – z)/(1 – z̅²), where z = 1 + 2i, then find |f(z)|.
• (b) Find the value of (z + 3)(z̅ + 3).
• (c) If (x – iy)(3 + 5i) is the conjugate of -6 – 24i, then find the value of x + y.
• (d) If z = 3 + 4i, then find z̅.

38. Read the text carefully and answer the questions: A state cricket authority has to choose a team of 11 members, so the authority asks 2 coaches of a government academy to select the team members that have experience as well as the best performers in the last 15 matches. They can make up a team of 11 cricketers amongst 15 possible candidates. In how many ways can the final eleven be selected from 15 cricket players if . (a) Two of them bein leg spinners, in how manay ways can be the final elevenbe selected from 15 cricket players if one and only one leg spinner must be included? (b) If there are 6 bowlers, 3 wicketkeepers, and 6 batsmen in all. In how many ways can be te final eleven be selected from 15 cricket players if 4 bowlers, 2 wicketkeepers 5 batsmen are included. ?

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SECTION – A SOLUTIONS

1. If a set A has n elements, then the total number of subsets of A is:

• a) 2n
• b) n
• c) 2ⁿ (Correct Answer)
• d) n²

Explanation:

Let A be a set with n elements. The total number of subsets of a set is given by the formula 2ⁿ.

This is because for each element in the set, there are two choices: either the element is included in the subset or it is not included. Thus, for n elements, the number of different subsets is 2ⁿ.

For example, if a set A = {1, 2}, which has n = 2 elements, the subsets are:

• ∅ (empty set)
• {1}
• {2}
• {1, 2}

Thus, for n = 2, the total number of subsets is 2² = 4.

Therefore, the correct answer is 2ⁿ.

SECTION – A SOLUTIONS

1. If a set A has n elements, then the total number of subsets of A is:

• a) 2n
• b) n
• c) 2ⁿ (Correct Answer)
• d) n²

Explanation:

The total number of subsets of a set A having n elements is given by the formula 2ⁿ. This is because for each element, there are two choices: either include it in the subset or exclude it. Therefore, the number of subsets is 2ⁿ.

Example: If A = {1, 2}, the subsets are ∅, {1}, {2}, and {1, 2}. Thus, for n = 2, the total number of subsets is 2² = 4.

2. The domain of the function f(x) = √(x² – 4) is:

• a) [-2, 2]
• b) (-∞, ∞)
• c) (-∞, -2] ∪ [2, ∞) (Correct Answer)
• d) (-2, 2)

Explanation:

For f(x) = √(x² – 4) to be defined, the expression inside the square root must be non-negative, i.e., x² – 4 ≥ 0. This gives x ≤ -2 or x ≥ 2, which corresponds to the domain (-∞, -2] ∪ [2, ∞).

3. The range of the function f(x) = |x| / x is:

• a) {-1, 1} (Correct Answer)
• b) R – {0}
• c) R – {-1, 1}
• d) {-1, 2}

Explanation:

The function f(x) = |x| / x equals 1 when x > 0 and -1 when x < 0. Therefore, the range of the function is {-1, 1}. It is undefined at x = 0.

4. If A = {1, 2, 4}, B = {2, 4, 5}, C = {2, 5}, then (A – B) × (B – C) is:

• a) {(1, 4)}
• b) (2, 5)
• c) {(1, 2), (1, 5), (2, 5)} (Correct Answer)
• d) (1, 4)

Explanation:

First, A – B = {1}, B – C = {4, 5}. Now, the Cartesian product (A – B) × (B – C) is {(1, 4), (1, 5)}.

5. The value of tan 75° – cot 75° is equal to:

• a) 2 – √3
• b) 1 + 2√3
• c) 2√3
• d) 2 + √3 (Correct Answer)

Explanation:

We know that tan 75° = 2 + √3 and cot 75° = 2 – √3. Therefore, tan 75° – cot 75° = (2 + √3) – (2 – √3) = 2√3.

6. If tan θ = √3 and θ lies in the third quadrant, then the value of sin θ is:

• a) 1/√10
• b) -1/√10 (Correct Answer)
• c) -3/√10
• d) 3/√10

Explanation:

In the third quadrant, both sine and tangent are negative. Given tan θ = √3, we can use the identity sin²θ + cos²θ = 1 to find sin θ. We find that sin θ = -1/√10.

7. Which is greater, sin 24° or cos 24°?

• a) both are equal
• b) cos 24° (Correct Answer)
• c) sin 24°
• d) cannot be compared

Explanation:

Since cosine decreases slower than sine as the angle increases from 0° to 90°, cos 24° is greater than sin 24°.

8. Mark the correct answer for i^-75 = ?

• a) i
• b) -1
• c) -i (Correct Answer)
• d) 1

Explanation:

Using the periodicity of i: i⁴ = 1, we reduce -75 mod 4, which gives i^-75 = i¹ = -i.

9. Let x, y ∈ R, then x + iy is a non-real complex number if:

• a) y = 0
• b) x ≠ 0
• c) x = 0
• d) y ≠ 0 (Correct Answer)

Explanation:

A complex number x + iy is non-real when its imaginary part y is non-zero, i.e., y ≠ 0.

10. If \(a\), \(b\), \(c\) are real numbers such that \(a > b\), \(c < 0\), then:

• a) \(ac > bc\)
• b) \(ac < bc\)
• c) \(ac \geq bc\)
• d) \(ac \leq bc\)

Correct Answer: b) \(ac < bc\)

Solution: Since \(c\) is negative, multiplying both sides of the inequality \(a > b\) by \(c\) reverses the inequality sign, giving \(ac < bc\).

11. Solve: \(3x + 5 < x - 7\), when \(x\) is a real number:

• a) \(x > -12\)
• b) \(x < -12\)
• c) \(x < -6\)
• d) \(x > -6\)

Correct Answer: c) \(x < -6\)

Solution: Simplifying the inequality:
Step 1: Subtract \(x\) from both sides: \(3x + 5 – x < x - 7 - x\)
Step 2: Simplify: \(2x + 5 < -7\)
Step 3: Subtract 5 from both sides: \(2x < -12\)
Step 4: Divide by 2: \(x < -6\)

12. Three persons enter a railway compartment. If there are 5 seats vacant, in how many ways can they take these seats?

• a) 60
• b) 125
• c) 20
• d) 15

Correct Answer: a) 60

Solution: The number of ways to arrange 3 people in 5 seats is given by permutations: \(5P3 = 5 \times 4 \times 3 = 60\).

13. If \(15C_{3r} = 15C_{r+3}\), then \(r\) is:

• a) 2
• b) 3
• c) 1
• d) 0

Correct Answer: b) 3

Solution: Using the property of combinations: \(15C_r = 15C_{15-r}\). Setting \(3r = r+3\), we solve for \(r = 3\).

14. \(15C_8 + 15C_9 – 15C_6 – 15C_7\) is equal to:

• a) \(15C_4\)
• b) \(15C_1\)
• c) 0
• d) 1

Correct Answer: d) 1

Solution: Simplifying the combinations using Pascal’s identity yields a result of 1.

15. \((\sqrt{3} + 1)^{2n+1} + (\sqrt{3} – 1)^{2n+1}\) is:

• a) an even positive integer
• b) an irrational number
• c) an odd positive integer
• d) a rational number

Correct Answer: c) an odd positive integer

Solution: Expanding and simplifying, the irrational terms cancel, leaving an odd positive integer.

16. If \(\sin \theta + \cos \theta = 1\), then \(\sin 2\theta\) is:

• a) 1
• b) 1
• c) 0
• d) -1

Correct Answer: c) 0

Solution: Squaring both sides gives: \(\sin^2 \theta + \cos^2 \theta + 2\sin \theta \cos \theta = 1\). Since \(\sin^2 \theta + \cos^2 \theta = 1\), we get \(\sin 2\theta = 0\).

SECTION – A Solutions

17. The value of sin 3θ / (1 + cos 2θ) is equal to:

    • a) cosθ
    • b) sinθ (Correct Answer)
    • c) -cosθ
    • d) -sinθ

    Explanation:

    We use the following trigonometric identity:

    sin 3θ = 3 sinθ – 4 sin³θ, and cos 2θ = cos²θ – sin²θ.

    Thus, sin 3θ / (1 + cos 2θ) simplifies to sinθ.

18. Tan(3π/2 + θ) is equal to:

    • a) -tan θ (Correct Answer)
    • b) tan θ
    • c) cot θ
    • d) -cot θ

    Explanation:

    The tangent function has a period of π, so tan(3π/2 + θ) is equivalent to -tanθ.

19. Assertion (A): If sin x = -1/2, then cos x = -√3/2.

    Reason (R): If the value of cos x is negative and sin x is negative, then x ∈ [3π/2, 2π].

    • a) Both A and R are true and R is the correct explanation of A.
    • b) Both A and R are true but R is not the correct explanation of A (Correct Answer)
    • c) A is true but R is false.
    • d) A is false but R is true.

    Explanation:

    In this case, both the assertion and the reason are correct. However, R does not correctly explain why cos x = -√3/2 when sin x = -1/2. The correct reasoning is based on the unit circle and specific angle values where sin and cos are negative.

20. Assertion (A): The value of sin(-690°) cos(-300°) + cos(-750°) sin(-240°) ≠ 1.

    Reason (R): The values of sin and cos are negative in the third and fourth quadrants, respectively.

    • a) Both A and R are true and R is the correct explanation of A (Correct Answer)
    • b) Both A and R are true but R is not the correct explanation of A.
    • c) A is true but R is false.
    • d) A is false but R is true.

    Explanation:

    By reducing angles to their corresponding values within [0, 360°], we get:

    • sin(-690°) = sin(30°), cos(-300°) = cos(60°)
    • cos(-750°) = cos(30°), sin(-240°) = sin(120°)

    The expression does not simplify to 1 due to the negative signs involved, and the assertion is true. The reasoning is also correct because the angles are in the third and fourth quadrants.

SECTION – B Solutions

21. Find the range of f(x) = x² + 2, x ∈ R.

    • (Correct Answer): Range: [2, ∞)

    Explanation:

    The given function f(x) = x² + 2 is a quadratic function. The term x² is always non-negative for any real value of x. Therefore, the smallest value of f(x) occurs when x = 0, which gives f(0) = 2.

    Thus, the range of the function is all real numbers greater than or equal to 2, i.e., [2, ∞).

    OR

    Solve the equation z² = z̅, where z = x + iy.

    Explanation:

    Let z = x + iy, where z̅ is the conjugate of z, so z̅ = x – iy.

    We are solving for z such that z² = z̅. Expanding both sides:

    • z² = (x + iy)² = x² – y² + 2ixy
    • z̅ = x – iy

    Equating the real and imaginary parts:

    • x² – y² = x
    • 2xy = -y

    From 2xy = -y, for y ≠ 0, we have 2x = -1, so x = -1/2.

    Substitute x = -1/2 into the real part equation:

    (-1/2)² – y² = -1/2 → 1/4 – y² = -1/2 → y² = 3/4 → y = ±√(3)/2.

    Thus, the solutions are z = -1/2 ± i√(3)/2.

SECTION – B Solutions

22. Find the value of (51)⁴ using the binomial theorem.

    • (Correct Answer): (51)⁴ = 6765201

    Explanation:

    We can express 51 as (50 + 1) and use the binomial theorem to expand (50 + 1)⁴:

    (50 + 1)⁴ = ⁴C₀50⁴ + ⁴C₁50³(1) + ⁴C₂50²(1²) + ⁴C₃50(1³) + ⁴C₄(1⁴)

    Expanding each term:

    • ⁴C₀50⁴ = 1 × 6250000 = 6250000
    • ⁴C₁50³(1) = 4 × 125000 = 500000
    • ⁴C₂50²(1²) = 6 × 2500 = 15000
    • ⁴C₃50(1³) = 4 × 50 = 200
    • ⁴C₄(1⁴) = 1 × 1 = 1

    Adding them up: 6250000 + 500000 + 15000 + 200 + 1 = 6765201.

23. Let f: R → R be given by f(x) = x² + 3. Find the pre-images of 39 and 2 under f.

    • (Correct Answer): Pre-image of 39: ±6, Pre-image of 2: No pre-image exists.

    Explanation:

    For pre-image of 39, we solve f(x) = 39:

    x² + 3 = 39 → x² = 36 → x = ±6.

    For pre-image of 2, solve f(x) = 2:

    x² + 3 = 2 → x² = -1 (no real solution exists since x² cannot be negative).

24. If Cos A = 1/7 and Cos B = 13/14 where A and B are acute angles, then find the value of A – B.

    • (Correct Answer): A – B ≈ 0.477 radians or 27.33°

    Explanation:

    Using the inverse cosine function (cos⁻¹), we find A and B:

    • A = cos⁻¹(1/7) ≈ 1.428 radians or 81.87°
    • B = cos⁻¹(13/14) ≈ 0.951 radians or 54.54°

    Therefore, A – B ≈ 1.428 – 0.951 = 0.477 radians or 27.33°.

25. Write the multiplicative inverse of (2 + i√3)².

    • (Correct Answer): Multiplicative inverse = 1/[(2 + i√3)²] = 1/(1 + 4i√3).

    Explanation:

    First, calculate (2 + i√3)²:

    (2 + i√3)² = (2² + 2 × 2 × i√3 + (i√3)²) = 4 + 4i√3 – 3 = 1 + 4i√3.

    To find the multiplicative inverse, we divide 1 by the square of the complex number:

    Multiplicative inverse = 1/(1 + 4i√3).

SECTION – C

26. If P = {x: x < 3, x ∈ N}, Q = {x: x ≤ 2, x ∈ W}, find (P ∪ Q) × (P ∩ Q), where W is the set of whole numbers.
    Solution:
    P = {1, 2} (natural numbers less than 3) and Q = {0, 1, 2} (whole numbers less than or equal to 2).
    P ∪ Q = {0, 1, 2} and P ∩ Q = {1, 2}.
    Thus, (P ∪ Q) × (P ∩ Q) = {(0, 1), (0, 2), (1, 1), (1, 2), (2, 1), (2, 2)}.
27. Find the value of (√2 + 1)⁵ − (√2 − 1)⁵.
    Solution:
    Let a = (√2 + 1) and b = (√2 – 1).
    Using the Binomial Theorem:
    (a + b)⁵ = a⁵ + b⁵ + 5a²b²(a + b).
    Calculating gives us:
    a + b = 2√2, a – b = 2, and substituting yields:
    (√2 + 1)⁵ − (√2 − 1)⁵ = 10.
28. Solve the inequality 2x – 1 > x + (7 − x)/3; 4x + 7 > 15, x ∈ R. Represent it graphically.
    Solution:
    1st Inequality: 2x – 1 > x + (7 − x)/3
    Multiply through by 3: 6x – 3 > 3x + 7 – x
    So, 6x – 3 > 2x + 7
    Result: 4x > 10 ⟹ x > 2.5.
    
    2nd Inequality: 4x + 7 > 15 ⟹ 4x > 8 ⟹ x > 2.
    
    The solution set is x > 2.5. Graphically, this can be represented on the number line.
29. Prove that: sin 3x + sin 2x − sin x = 4 sin x cos(x/2) cos(3x/2).
    Solution:
    Using the angle addition formulas, we rewrite:
    sin 3x = sin(2x + x) = sin 2x cos x + cos 2x sin x.
    After applying relevant identities and simplifications, we derive:
    sin 3x + sin 2x – sin x = 4 sin x cos(x/2) cos(3x/2).
30. If ⁸Pᵣ = 1680 and ⁸Cᵣ = 70, find the value of n and r.
    Solution:
    Using the formulas for permutations and combinations:
    ⁸Pᵣ = n!/(n-r)! and ⁸Cᵣ = n!/(r!(n-r)!)
    Setting up the equations, we can derive n = 8 and r = 5.
31. Write the solution set of the inequality -5 ≤ (5 − 3x)/2 ≤ 8.
    OR
    Using the binomial theorem, determine which number is smaller: (1.2)⁴⁰⁰⁰ or 800?
    Solution:
    For the first inequality:
    -5 ≤ (5 − 3x)/2 leads to:
    -10 ≤ 5 – 3x ⟹ 3x ≤ 15 ⟹ x ≤ 5.
    And (5 − 3x)/2 ≤ 8 leads to:
    5 − 3x ≤ 16 ⟹ -3x ≤ 11 ⟹ x ≥ -11/3.
    Thus, the solution set is x ∈ [-11/3, 5].
    
    For the OR option:
    (1.2)⁴⁰ is greater than 800 due to exponential growth, so 800 is smaller.