KCET 2024 Question Paper with Answer Key PDF for Mathematics (April 18)
Content Curator | Updated On - Dec 25, 2024
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KCET 2024 Mathematics Question Paper with Answer Key PDF
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KCET 2024 Mathematics Question Paper with Answer Key PDF
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| 1. Two finite sets have m and n elements respectively. The total number of subsets of the first set is 56 more than the total number of subsets of the second set. The values of m and n respectively are: (A) 7, 6 (B) 5, 1 (C) 6, 3 (D) 8, 7 |
(C) 6, 3 | The total number of subsets for a set with m elements is 2m, and for n elements is 2n. The difference is given as 56, so 2m - 2n = 56. Solving this for m = 6 and n = 3. |
| 2. If f(x) = [x] - 5|x| + 6 = 0, where [x] denotes the greatest integer function, then: (A) x ∈ [3, 4] (B) x ∈ (2, 4) (C) x ∈ [2, 3] (D) x ∈ (2, 3] |
(B) x ∈ (2, 4) | Evaluating the equation within the domain of x, we find that x ∈ (2, 4) satisfies the given function equation. |
| 3. If in two circles, arcs of the same length subtend angles 30° and 78° at the center, then the ratio of their radii is: (A) 5/13 (B) 13/5 (C) 13/4 (D) 4/13 |
(B) 13/5 | The length of an arc is given by l = r*theta, where r is the radius and theta is the angle subtended in radians. For the two circles: r1/r2 = theta2/theta1. Convert degrees to radians: theta1 = 30 degrees = pi/6, and theta2 = 78 degrees = 13pi/30. Therefore, r1/r2 = (13pi/30) / (pi/6) = 13/5. |
| 4. If ABC is right angled at C, then the value of tan A + tan B is: (A) a + b (B) a2 / b2 (C) c2 / ab (D) b2 / ac |
(C) c2 / ab | For a right triangle, using the tangent addition formula, tan A + tan B = c2 / ab when ABC is right at C. |
| 5. The real value of 'α' for which 1 - i sin α / 1 + 2 isin α is purely real is: (A) (n + 1) π/2, n ∈ N (B) (2n + 1) π / 2, n ∈ N (C) nπ, n ∈ N (D) (2n - 1) π / 2, n ∈ N |
(C) nπ, n ∈ N | For the expression to be purely real, α must be an integral multiple of π, where sin α = 0. |
| 6. The length of a rectangle is five times the breadth. If the minimum perimeter of the rectangle is 180 cm, then: (A) Breadth ≤ 15 cm (B) Breadth ≥ 15 cm (C) Length ≤ 15 cm (D) Length = 15 cm |
(B) Breadth ≥ 15 cm | Let the breadth be b and the length be 5b. The perimeter is 2(l + b) = 2(5b + b) = 12b. Given the minimum perimeter is 180 cm, 12b = 180 implies b = 15 cm. Thus, the breadth is b ≥ 15 cm. |
| 7. The value of 49C3 + 48C3 + 47C3 + 46C3 + 45C3 + 45C4 is: (A) 50C4 (B) 50C3 (C) 50C2 (D) 50C1 |
(A) 50C4 | Using the property of binomial coefficients, the recursive addition property yields 49C3 + 48C3 + 47C3 + 46C3 + 45C3 + 45C4 = 50C4. |
| 8. In the expansion of (1 + x)^n, the sequence nC1/nC0 + 2 * nC2/nC1 + 3 * nC3/nC2 + ... + n * nCn/nC{n-1} is equal to: (A) n(n+1)/2 (B) n/2 (C) (n+1)/2 (D) 3n(n+1) |
(A) n(n+1)/2 | The series 1 + 2 + 3 + ... + n sums to n(n+1)/2, showing the power of series and combinatorial identities in simplifying expressions. |
| 9. If S_n stands for the sum to n-terms of a G.P. with a as the first term and r as the common ratio, then S_1/S_2 is: (A) r^n + 1 (B) 1/r^n + 1 (C) r^n - 1 (D) 1/r^n - 1 |
(B) 1/r^n + 1 | The sum of the first term S_1 = a. For two terms, S_2 = a(1 + r). Then, S_1/S_2 = a/a(1 + r) = 1/(1 + r), leading to the simplified form 1/(r^n + 1). |
| 10. If A.M. and G.M. of roots of a quadratic equation are 5 and 4 respectively, then the quadratic equation is: (A) x^2 - 10x - 16 = 0 (B) x^2 + 10x + 16 = 0 (C) x^2 + 10x - 16 = 0 (D) x^2 - 10x + 16 = 0 |
(D) x^2 - 10x + 16 = 0 | Let the roots be α and β. Then, A.M. = (α + β)/2 = 5 implies α + β = 10. G.M. = sqrt(αβ) = 4 implies αβ = 16. The quadratic equation is x^2 - (α + β)x + αβ = 0, i.e., x^2 - 10x + 16 = 0. |
| 11. The angle between the line x + y = 3 and the line joining the points (1, 1) and (-3, 4) is: (A) tan-1(7) (B) tan-1(-1/7) (C) tan-1(1/7) (D) tan-1(2/7) |
(C) tan-1(1/7) | The slope of the line x + y = 3 is -1. The slope of the line joining (1, 1) and (-3, 4) is -3/4. The angle theta between the two lines is given by tan theta = |(m1 - m2)/(1 + m1m2)| = 1/7. |
| 12. The equation of the parabola whose focus is (6, 0) and directrix is x = -6 is: (A) y^2 = 24x (B) y^2 = -24x (C) x^2 = 24y (D) x^2 = -24y |
(A) y^2 = 24x | The vertex is at (0, 0), and the parabola opens to the right. The equation of the parabola is y^2 = 4ax, where a = 6, hence y^2 = 24x. |
| 13. limx → π/4 (sqrt(2)cos x - 1)/(cot x - 1) is equal to: (A) 2 (B) sqrt(2) (C) 1/2 (D) 1/sqrt(2) |
(C) 1/2 | For x approaching π/4, using approximations of cos x and cot x around π/4, the limit simplifies to 1/2. |
| 14. The negation of the statement “For every real number x, x^2 + 5 is positive” is: (A) For every real number x, x^2 + 5 is not positive (B) For every real number x, x^2 + 5 is negative (C) There exists at least one real number x such that x^2 + 5 is not positive (D) There exists at least one real number x such that x^2 + 5 is positive |
(C) There exists at least one real number x such that x^2 + 5 is not positive | The correct negation of a universal quantifier (For every x) is an existential quantifier (There exists an x) with the condition negated. |
| 15. Let a, b, c, and d be the observations with mean m and standard deviation S. The standard deviation of the observations a + k, b + k, c + k, d + k is: (A) kS (B) S + k (C) S/k (D) S |
(D) S | Adding a constant k to each observation shifts all data points by k but does not change the spread or standard deviation, which remains S. |
| 16. Let f : R to R be given by f(x) = tan x. Then f-1(1) is: (A) π/4 (B) nπ + π/4; n in Z (C) π/3 (D) nπ + π/3; n in Z |
(B) nπ + π/4; n in Z | The function tan x has a periodicity of π. The principal value of tan-1(1) is π/4, and due to periodicity, it includes all nπ + π/4, where n is an integer. |
| 17. Let f : R to R be defined by f(x) = x^2 + 1. Then the pre-images of 17 and -3 respectively are: (A) phi, {4, -4} (B) {3, -3}, phi (C) {4, -4}, phi (D) {4, -4}, {2, -2} |
(C) {4, -4}, phi | For f(x) = 17, x^2 + 1 = 17 gives x = ±4. For f(x) = -3, the equation x^2 + 1 = -3 has no real solutions, hence pre-image is phi. |
| 18. Let (g ∘ f)(x) = sin x and (f ∘ g)(x) = (sin √(x))2. Then: (A) f(x) = sin2 x, g(x) = x (B) f(x) = sin √(x), g(x) = √(x) (C) f(x) = sin2 x, g(x) = √(x) (D) f(x) = sin √(x), g(x) = x2 |
(D) f(x) = sin √(x), g(x) = x2 | Given (g ∘ f)(x) = sin x and (f ∘ g)(x) = (sin √(x))2, it implies f(x) = sin √(x) and g(x) = x2, establishing the correct functional transformations between g and f. |
| 19. Let A = {2, 3, 4, 5, ..., 16, 17, 18}. Let R be the relation on the set A of ordered pairs of positive integers defined by (a, b) R (c, d) if and only if ad = bc for all (a, b), (c, d) in A x A. Then the number of ordered pairs of the equivalence class of (3, 2) is: (A) 4 (B) 5 (C) 6 (D) 7 |
(C) 6 | For (3, 2), the equivalence class consists of all (c, d) such that 3d = 2c. Solving for integer pairs (c, d) in the set A gives six solutions. |
| 20. If cos-1 x + cos-1 y + cos-1 z = 3π, then x(y + z) + y(z + x) + z(x + y) equals to: (A) 0 (B) 1 (C) 6 (D) 12 |
(C) 6 | Given cos-1 x + cos-1 y + cos-1 z = 3π, it follows that x = y = z = -1. Substituting x = y = z = -1 into the expression x(y + z) + y(z + x) + z(x + y) results in 6. |
| 21. If 2sin-1 x - 3cos-1 x = 4x, x ∈ [-1, 1], then 2sin-1 x + 3cos-1 x is equal to: (A) (4 - 6π)/5 (B) (6π - 4)/5 (C) 3π/2 (D) 0 |
(B) (6π - 4)/5 | Using sin-1 x + cos-1 x = π/2, substitute and solve for 2sin-1 x + 3cos-1 x, which yields (6π - 4)/5. |
| 22. If A is a square matrix such that A² = A, then (I + A)³ is equal to: (A) 7A - I (B) 7A (C) 7A + I (D) I - 7A |
(B) 7A | Expanding (I + A)³ using A² = A simplifies to I + 7A. Since A² = A, this becomes 7A. |
| 23. If A = [1 1; 1 1], then A10 is equal to: (A) 28 A (B) 29 A (C) 210 A (D) 211 A |
(B) 29 A | A² = 2A leads to A10 = 29 A by induction, observing the pattern in matrix powers. |
| 24. If f(x) = determinant of matrix, then f(1) · f(3) · f(5) + f(5) · f(1) is: (A) 1 (B) 0 (C) 2 (D) None of these |
(B) 0 | Calculating f(x) at specific values and their product as stated in the expression results in 0. |
| 25. If P is the adjoint of a 3x3 matrix A and |A| = 4, then α is equal to: (A) 4 (B) 5 (C) 11 (D) 0 |
(C) 11 | Using the properties of determinants and adjoints, solve for α to find α = 11 when |P| = 16. |
| 26. If A and B are matrices, then dB/dx is: (A) 3A (B) -3B (C) 3B + 1 (D) 1 - 3A |
(A) 3A | Differentiating matrix B with respect to x and considering the elements of B, the derivative dB/dx results in 3A. |
| 27. Let f(x) be a matrix function. Then limx → 0 f(x)/x² is: (A) -1 (B) 0 (C) 3 (D) 2 |
(B) 0 | As x approaches 0, the terms in f(x) divided by x² approach 0, leading to the limit of 0. |
| 28. Which one of the following observations is correct for the features of the logarithm function to any base b > 1? (A) The domain of the logarithm function is R. (B) The range of the logarithm function is R⁺. (C) The point (1, 0) is always on the graph of the logarithm function. (D) The graph of the logarithm function is decreasing as we move from left to right. |
(C) The point (1, 0) is always on the graph of the logarithm function. | For any logarithm function with base b > 1, the point (1, 0) is always present on its graph because log_b(1) = 0. |
| 29. The function f(x) = |cos x| is: (A) Everywhere continuous and differentiable. (B) Everywhere continuous but not differentiable at odd multiples of π/2. (C) Neither continuous nor differentiable at 2n + 1, n in Z. (D) Not differentiable everywhere. |
(B) Everywhere continuous but not differentiable at odd multiples of π/2. | |cos x| is continuous everywhere but has points of non-differentiability at odd multiples of π/2 where cos x = 0. |
| 30. If y = 2x3x, then dy/dx at x = 1 is: (A) 2 (B) 6 (C) 3 (D) 1 |
(B) 6 | Using logarithmic differentiation, find dy/dx for y = 2x3x and evaluate at x = 1 to find dy/dx = 6. |
| 31. Let the function satisfy the equation f(x + y) = f(x)f(y) for all x, y ∈ ℝ, where f(0) ≠ 0. If f(5) = 3 and f'(0) = 2, then f'(5) is: (A) 6 (B) 0 (C) 3 (D) -6 |
(A) 6 | Given f'(x) = 2f(x), with f(5) = 3, we calculate f'(5) = 2 * 3 = 6. |
| 32. The value of C in (0, 2) satisfying the mean value theorem for the function f(x) = x(x - 1)², x ∈ [0, 2] is equal to: (A) 3/4 (B) 4/3 (C) 1/2 (D) 2/3 |
(B) 4/3 | Solving f'(C) = 1, we find C = 4/3, satisfying the mean value theorem. |
| 33. d/dx [ cos² ( cot-1 sqrt((2 + x)/(2 - x)) ) ] is: (A) 3/4 (B) 1/2 (C) 1 (D) 1/4 |
(D) 1/4 | Using chain rule and trigonometric identities, the derivative simplifies to 1/4. |
| 34. For the function f(x) = x³ - 6x² + 12x - 3, x = 2 is: (A) A point of minimum (B) A point of inflection (C) Not a critical point (D) A point of maximum |
(B) A point of inflection | With both first and second derivatives zero at x = 2, and f'''(2) ≠ 0, x = 2 is a point of inflection. |
| 35. The function x^x, x > 0 is strictly increasing at: (A) ∀ x ∈ ℝ (B) x < 1/e (C) x > 1/e (D) x < 0 |
(C) x > 1/e | Differentiating x^x and analyzing its monotonicity, the function is strictly increasing for x > 1/e. |
| 36. The maximum volume of the right circular cone with slant height 6 units is: (A) 4√3 π cubic units (B) 16√3 π cubic units (C) 3√3 π cubic units (D) 6√3 π cubic units |
(B) 16√3 π cubic units | Using geometry and calculus, the maximum volume calculated for the cone is 16√3 π cubic units. |
| 37. If f(x) = x e^(x(1-x)), then f(x) is: (A) Increasing in ℝ (B) Decreasing in ℝ (C) Decreasing in [1/2, 1] (D) Increasing in [-1/2, 1] |
(D) Increasing in [-1/2, 1] | Differentiating f(x), we find it is increasing in the interval [-1/2, 1]. |
| 38. ∫ (sin x / (3 + 4cos² x)) dx = (A) 1/2√3 tanminus1(2cos x/√3) + C (B) 1/√3 tan-1(cos x/3) + C (C) 1/2√3 tan-1(cos x/3) + C (D) -1/√3 tan-1(2cos x/√3) + C |
(A)minus 1/2√3 tan-1(2cos x/√3) + C | Using trigonometric substitutions, the integral simplifies to 1/2√3 tan-1(2cos x/√3) + C. |
| 39. ∫ from -π to π (1 - x²)sin x cos² x dx = (A) π/3 (B) 2π - π² (C) π³/2 (D) 0 |
(D) 0 | The integrand is an odd function over a symmetric interval, resulting in an integral value of 0. |
| 40. ∫ (1 / (x (6(log x)² + 7log x + 2))) dx = (A) 1/2 log |2log x + 1|/(3log x + 2) + C (B) log |2log x + 1|/(3log x + 2) + C (C) log |3log x + 2|/(2log x + 1) + C (D) 1/2 log |3log x + 2|/(2log x + 1) + C |
(B) log |2log x + 1|/(3log x + 2) + C | Using partial fractions and substitution, the integral simplifies to log |2log x + 1|/(3log x + 2) + C. |
| 41. ∫ (sin(5x/2) / sin(x/2)) dx = (A) 2x + sin x + 2sin 2x + C (B) x + 2sin x + 2sin 2x + C (C) x + 2sin x + sin 2x + C (D) 2x + sin x + sin 2x + C |
(C) x + 2sin x + sin 2x + C | Integrating the function simplified through trigonometric identities gives x + 2sin x + sin 2x + C. |
| 42. ∫ from 1 to 5 (|x - 3| + |1 - x|) dx = (A) 12 (B) 5/6 (C) 21 (D) 10 |
(A) 12 | The total integral computed over two intervals simplifies to 12. |
| 43. lim as n → ∞ of (n/(n² + 1²) + n/(n² + 2²) + ... + n/(n² + n²)) = (A) π/4 (B) tan-1 3 (C) tan-1 2 (D) π/2 |
(C) tan-1 2 | The limit, treated as a Riemann sum, approaches tan-1 2. |
| 44. The area of the region bounded by the line y = 3x and the curve y = x³ in square units is: (A) 10 (B) 9/2 (C) 9 (D) 5 |
(B) 9/2 | Integrating the difference between the line and the curve from their intersection points gives an area of 9/2. |
| 45. The area of the region bounded by the line y = x and the curve y = x³ is: (A) 0.2 sq. units (B) 0.3 sq. units (C) 0.4 sq. units (D) 0.5 sq. units |
(D) 0.5 sq. units | Calculating the integral between the intersections of the line and curve gives an area of 0.5 square units. |
| 46. The solution of e^(dy/dx) = x + 1, y(0) = 3 is: (A) y - 2 = xlog x (B) y - x - 3 = xlog x (C) y - x - 3 = (x + 1)log(x + 1) (D) y + x - 3 = (x + 1)log(x + 1) |
(D) y + x - 3 = (x + 1)log(x + 1) | The integration of the differential equation with initial condition yields y + x - 3 = (x + 1)log(x + 1). |
| 47. The family of curves whose x and y intercepts of a tangent at any point are respectively double the x and y coordinates of that point is: (A) xy = C (B) x² + y² = C (C) x² - y² = C (D) y/x = C |
(A) xy = C | The conditions of the problem correspond to the equation xy = C. |
| 48. The vectors AB = 3i + 4k and AC = 5i - 2j + 4k are the sides of a triangle ABC. The length of the median through A is: (A) √18 (B) √72 (C) √33 (D) √288 |
(C) √33 | The length of the median through A is calculated as √33 using vector addition and the Pythagorean theorem. |
| 49. The volume of the parallelepiped whose co-terminous edges are i + j, i + k, i + j is: (A) 6 cu. units (B) 2 cu. units (C) 4 cu. units (D) 3 cu. units |
(B) 2 cu. units | Using the scalar triple product, the volume of the parallelepiped is calculated as 2 cubic units. |
| 50. Let a and b be two unit vectors and theta is the angle between them. Then a + b is a unit vector if: (A) theta = π/4 (B) theta = π/3 (C) theta = 2π/3 (D) theta = π/2 |
(C) theta = 2π/3 | The sum of two unit vectors results in a unit vector if the angle between them is 2π/3, satisfying the condition for the magnitude. |
| 51. If vectors a, b, c are three non-coplanar vectors and vectors p, q, r are defined by: p = (a × c) / [a b c], q = (c × b) / [a b c], r = (b × a) / [a b c], then (i + b) · p + (b + c) · q + (c + a) · r is: (A) 0 (B) 1 (C) 2 (D) 3 |
(D) 3 | Each dot product evaluates to 1, summing up to 3. |
| 52. If lines (x - 1)/-3 = (y - 2)/2k = (z - 3)/2 and (x - 1)/3k = (y - 5)/1 = (z - 6)/-5 are mutually perpendicular, then k is equal to: (A) -10/7 (B) 7/10 (C) -10 (D) -7 |
(A) -10/7 | The dot product of the direction ratios gives -10/7, ensuring perpendicularity. |
| 53. The distance between the two planes 2x + 3y + 4z = 4 and 4x + 6y + 8z = 12 is: (A) 2 units (B) 8 units (C) 2/sqrt(29) units (D) 4 units |
(C) 2/sqrt(29) units | The distance formula for parallel planes gives 2/sqrt(29) units. |
| 54. The sine of the angle between the straight line (x - 2)/3 = (y - 3)/4 = (4-z)/-5 and the plane 2x - 2y + z = 5 is: (A) 1/5sqrt(2) (B) 2/5sqrt(2) (C) 3/50 (D) 3/sqrt(50) |
(A) 1/5sqrt(2) | Using the sine formula with direction ratios and the normal vector of the plane. |
| 55. The equation xy = 0 in three-dimensional space represents: (A) A pair of straight lines (B) A plane (C) A pair of planes at right angles (D) A pair of parallel planes |
(C) A pair of planes at right angles | Represents the yz-plane and xz-plane, which are perpendicular. |
| 56. The plane containing the point (3, 2, 0) and the line (x - 3)/1 = (y - 6)/5 = (z - 4)/4 is: (A) x - y + z = 1 (B) x + y + z = 5 (C) x + 2y - z = 1 (D) 2x - y + z = 5 |
(A) x - y + z = 1 | Plane equation derived from point and parallel line properties. |
| 57. Corner points of the feasible region for an LPP are (0, 2), (3, 0), (6, 0), (6, 8) and (0, 5). Let z = 4x + 6y be the objective function. The minimum value of z occurs at: (A) Only (0, 2) (B) Only (3, 0) (C) The mid-point of the line segment joining the points (0, 2) and (3, 0) (D) Any point on the line segment joining the points (0, 2) and (3, 0) |
(D) Any point on the line segment joining the points (0, 2) and (3, 0) | Minimum value of z is the same at these points and along the line segment joining them. |
| 58. A die is thrown 10 times. The probability that an odd number will come up at least once is: (A) 11/1024 (B) 1013/1024 (C) 1023/1024 (D) 1/1024 |
(C) 1023/1024 | Probability of getting at least one odd number is calculated by subtracting the probability of no odd numbers from 1. |
| 59. A random variable X has the following probability distribution: X: 0, 1, 2 P(X): 25/36, k, 1/36 Mean = 1/3, then variance is: (A) 1 (B) 5/18 (C) 7/18 (D) 11/18 |
(B) 5/18 | Variance calculated using E(X²) and E(X) values derived from solving for k. |
| 60. If a random variable X follows the binomial distribution with parameters n = 5, p, and P(X = 2) = 9P(X = 3), then p is equal to: (A) 10 (B) 1/10 (C) 5 (D) 1.2 |
(B) 1/10 | Calculated using the condition P(X = 2) = 9P(X = 3) within the binomial probability formula. |
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