SECTION – A

1. If a set A has n elements then the total number of subsets of A is
   1. 2n
   2. n
   3. 2ⁿ
   4. n²
2. The domain of the function f: R → R defined by f(x) = √(x² – 4) is
   1. [-2, 2]
   2. (-∞, ∞)
   3. (-∞, -2] ∪ [2, ∞)
   4. (-2, 2)
3. The range of the function f(x) = x / |x| is
   1. (-1, 1)
   2. R – {0}
   3. R – {-1, 1}
   4. {-1, 2}
4. If A = {1, 2, 4}, B = {2, 4, 5}, C = {2, 5}, then (A – B) × (B – C) is
   1. {(1, 4)}
   2. {(2, 5)}
   3. {(1, 2), (1, 5), (2, 5)}
   4. {(1, 4)}
5. The value of tan 75° – cot 75° is equal to
   1. 2 – √3
   2. 1 + 2√3
   3. 2√3
   4. 2 + √3
6. If tan θ = √3 and lies in third quadrant, then value of sin θ is
   1. 1/√10
   2. -1/√10
   3. -3/√10
   4. 3/√10
7. Which is greater, sin 24° or cos 24°?
   1. both are equal
   2. cos 24°
   3. sin 24°
   4. cannot be compared
8. Mark the correct answer for i ^-75 = ?
   1. i
   2. -1
   3. -i
   4. 1
9. Let x, y ∈ R, then x + iy is a non-real complex number if
   1. y = 0
   2. x ≠ 0
   3. x = 0
   4. y ≠ 0
10. If a, b, c are real numbers such that a > b, c < 0
    1. ac > bc
    2. ac < bc
    3. ac ≥ bc
    4. ac ≠ bc
11. Solve: 3x + 5 < x - 7, when x is a real number
    1. x > -12
    2. x > -12
    3. x < -6
    4. x > -6
12. Three persons enter a railway compartment. If there are 5 seats vacant, in how many ways can they take these seats?
    1. 60
    2. 125
    3. 20
    4. 15
13. If ¹⁵C₃ = ¹⁵C_r+3, then r is
    1. 2
    2. 3
    3. 1
    4. 0
14. ¹⁵C₈ + ¹⁵C₉ – ¹⁵C₆ – ¹⁵C₇
    1. 15C₄
    2. 15C₁
    3. 0
    4. 1
15. (√3 + 1)²ⁿ⁺¹ + (√3 – 1)²ⁿ⁺¹ is
    1. an even positive integer
    2. an irrational number
    3. an odd positive integer
    4. a rational number
16. If sin θ + cos θ = 1 then sin 2θ is
    1. 1
    2. 0
    3. 0
    4. -1

SECTION – A: Question Paper with Solutions

1. If a set A has n elements, then the total number of subsets of A is:
   1. 2n
   2. n
   3. 2ⁿ (Correct)
   4. n²

   Solution: The total number of subsets of a set with n elements is 2ⁿ.

2. The domain of the function f: R → R defined by f(x) = √(x² – 4) is:
   1. [-2, 2]
   2. (-∞, ∞)
   3. (-∞, -2] ∪ [2, ∞) (Correct)
   4. (-2, 2)

   Solution: The expression under the square root must be non-negative, i.e., x² – 4 ≥ 0. Solving this gives x ≤ -2 or x ≥ 2, hence the domain is (-∞, -2] ∪ [2, ∞).

3. The range of the function f(x) = x / |x| is:
   1. (-1, 1)
   2. R – {0} (Correct)
   3. R – {-1, 1}
   4. {-1, 2}

   Solution: The function f(x) = x / |x| takes the value -1 when x is negative, 1 when x is positive, and is undefined at x = 0. Hence, the range is R – {0}.

4. If A = {1, 2, 4}, B = {2, 4, 5}, C = {2, 5}, then (A – B) × (B – C) is:
   1. {(1, 4)}
   2. {(2, 5)}
   3. {(1, 2), (1, 5), (2, 5)}
   4. {(1, 4)} (Correct)

   Solution: A – B = {1}, B – C = {4}, so (A – B) × (B – C) = {(1, 4)}.

5. The value of tan 75° – cot 75° is equal to:
   1. 2 – √3 (Correct)
   2. 1 + 2√3
   3. 2√3
   4. 2 + √3

   Solution: Use the values of tan 75° and cot 75°. tan 75° = 2 + √3, cot 75° = 2 – √3, so tan 75° – cot 75° = 2 – √3.

6. If tan θ = √3 and lies in third quadrant, then value of sin θ is:
   1. 1/√10
   2. -1/√10 (Correct)
   3. -3/√10
   4. 3/√10

   Solution: In the third quadrant, sin θ is negative. Given tan θ = √3, and using the identity sin²θ + cos²θ = 1, we get sin θ = -1/√10.

7. Which is greater, sin 24° or cos 24°?
   1. both are equal
   2. cos 24° (Correct)
   3. sin 24°
   4. cannot be compared

   Solution: cos 24° is greater than sin 24° because cos θ > sin θ for angles less than 45°.

8. Mark the correct answer for i^-75 = ?
   1. i
   2. -1
   3. -i (Correct)
   4. 1

   Solution: i^-75 = i^{-75 mod 4} = i¹ = -i.

9. Let x, y ∈ R, then x + iy is a non-real complex number if:
   1. y = 0
   2. x ≠ 0
   3. x = 0
   4. y ≠ 0 (Correct)

   Solution: For a complex number to be non-real, its imaginary part (y) must be non-zero. Hence, y ≠ 0.

10. If a, b, c are real numbers such that a > b, c < 0, then:
    1. ac > bc
    2. ac < bc (Correct)
    3. ac ≥ bc
    4. ac ≠ bc

    Solution: When multiplying inequalities by a negative number, the inequality sign reverses. Therefore, ac < bc.

11. Solve: 3x + 5 < x - 7, when x is a real number:
    1. x > -12
    2. x > -12
    3. x < -6 (Correct)
    4. x > -6

    Solution: Subtracting x from both sides and simplifying, we get 2x < -12, so x < -6.

12. Three persons enter a railway compartment. If there are 5 seats vacant, in how many ways can they take these seats?
    1. 60 (Correct)
    2. 125
    3. 20
    4. 15

    Solution: The number of ways to arrange 3 people in 5 seats is P(5, 3) = 5 × 4 × 3 = 60.

13. If ¹⁵C₃ = ¹⁵C_r+3, then r is:
    1. 2 (Correct)
    2. 3
    3. 1
    4. 0

    Solution: From the symmetry property of binomial coefficients, we know that ¹⁵C₃ = ¹⁵C₁₂. Thus, r = 2.

14. ¹⁵C₈ + ¹⁵C₉ – ¹⁵C₇ + ¹⁵C₆ is equal to:
    1. ¹⁶C₈ (Correct)
    2. ¹⁷C₈
    3. ¹⁵C₇
    4. none

    Solution: Using Pascal’s identity, we know that ⁿC_k + ⁿC_k+1 = ⁿ⁺¹C_k+1. So, ¹⁵C₈ + ¹⁵C₉ = ¹⁶C₉. Similarly, the rest gives us ¹⁶C₈.

15. The remainder obtained when 7 × 8 × 9 × 10 × 11 × 12 is divided by 100 is:
    1. 20
    2. 0 (Correct)
    3. 50
    4. 90

    Solution: As the product contains both 10 and 5, which multiply to give a factor of 100, the remainder when divided by 100 is 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.