KCET 2024 Mathematics Question Paper: Download Question Paper with Solution PDF

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Updated on, Dec 25, 2024

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KCET 2024 Mathematics Question Paper is available here for download. KCET Mathematics question paper was conducted on April 18, 2024 by Karnataka Examination Authority (KEA). KCET 2024 Mathematics question paper consists of 60 questions to be attempted in 80 minutes for a total of 60 marks.

KCET 2024 Mathematics Question Paper with Answer Key PDF

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KCET 2024 Mathematics Question Paper with Solution

Question Answer  Detailed Solution
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)

Analyzing the equation based on the range of x values, it is determined that x values between 2 and 4 are solutions to the function's 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 formula for the length of an arc is l = r*theta, where r represents the radius and theta is the angle in radians. For two circles, the relationship between their radii, r1 and r2, and the angles they subtend, theta1 and theta2, is given as r1/r2 = theta2/theta1. To apply this, convert angles from degrees to radians: theta1 is 30 degrees, equivalent to pi/6 radians, and theta2 is 78 degrees, which converts to 13pi/30 radians. Substituting these values in, the ratio of the radii, r1/r2, becomes (13pi/30) / (pi/6), which simplifies to 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

In a right triangle where the right angle is at vertex C, the tangent addition formula states that the sum of the tangents of angles A and B, tan A + tan B, equals c2/ab. This relationship holds because angle C, being a right angle, means that angles A and B are complementary (A + B = 90°).

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 entirely real, the value of α must be a multiple of π, which ensures that sin α equals zero.

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

Assume the breadth of a rectangle is b and its length is 5 times the breadth, represented as 5b. The formula for the perimeter of a rectangle is 2(l + b), which in this case becomes 2(5b + b) = 12b. If the smallest possible perimeter is 180 cm, then setting 12b equal to 180 allows us to solve for b, giving b = 15 cm. Therefore, the minimum breadth of the rectangle is 15 cm or more.

7. The value of 49C3 + 48C3 + 47C3 + 46C3 + 45C3 + 45C4 is:
(A) 50C4
(B) 50C3
(C) 50C2
(D) 50C1
(A) 50C4

The equation 49C3 + 48C3 + 47C3 + 46C3 + 45C3 + 45C4 = 50C4 demonstrates a special relationship between binomial coefficients. It shows that the sum of several consecutive binomial coefficients with specific parameters (in this case, combinations of 49, 48, etc. choosing 3 items, and one combination choosing 4 items) is equal to a single binomial coefficient with different parameters (50 choose 4). This relationship is a consequence of the recursive addition property of binomial coefficients, which allows for the calculation of one coefficient based on the values of others.

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 sum of the series 1 + 2 + 3 + ... + n is expressed by the formula n(n+1)2\frac{n(n+1)}{2}. This formula demonstrates how series and combinatorial identities can be effectively used to simplify and calculate the sum of sequential integers.

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 original statement describes a geometric series. The ratio of the first term's sum (S1) to the sum of the first two terms (S2) simplifies to 1/(1+r), not 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

We have two numbers. We know their average and their geometric mean. Using this, we can find the equation that represents these two numbers.

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)| = 0.14285714285714285.

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 parabola has its vertex at the origin, represented by the coordinates (0, 0), and it opens towards the right direction. The standard form of the equation for such a parabola is y2=4axy^2 = 4ax. For this specific parabola, the value of aa is given as 6. Substituting 6 for aa in the equation, we get y2=24xy^2 = 24x. This equation defines how the shape of the parabola is formed and its orientation in the coordinate plane.

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

The limit of [(cos(x) - cot(x)) / (x - π/4)] as x approaches π/4, when approximated using Taylor series for cos(x) and cot(x) around π/4, simplifies to (√2 + 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 negation of "For every x" is "There exists an x" where the following condition is 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

When a constant (k) is added to every data point in a dataset, the entire dataset is shifted by that constant value. However, the spread or variability within the data, as measured by the standard deviation (S), remains unchanged.

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 tangent function (tan(x)) has a period of π. The principal value of its inverse, tan⁻¹(1), is π/4. Due to the periodicity of tan(x), all solutions for tan⁻¹(1) can be expressed as nπ + π/4, where n is any 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 the function f(x) = x² + 1, x = ±4 when f(x) = 17 and there is no pre-image when f(x) = -3.

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

If you plug a value into function 'f' and then plug the result into function 'g', you get sin(x). But if you plug a value into 'g' first and then plug the result into 'f', you get (sin(√x))². This tells us that 'f' must be sin(√x) and 'g' must be x².

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 the pair (3, 2), the equivalence class includes all pairs (c, d) such that the product of 3 and d equals the product of 2 and c. Finding integer solutions for (c, d) within the set A results in six pairs.

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 that the sum of cos-1(x) + cos-1(y) + cos-1(z) equals 3π, it implies that x, y, and z must each be -1. Plugging x = y = z = -1 into the formula x(y + z) + y(z + x) + z(x + y) yields a result of 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

we can solve it Using sin-1 x + cos-1 x = π/2, substituting and solve for 2sin-1 x + 3cos-1 x, which gives (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(I + A)^3 using A2=AA^2 = A simplifies to I+3AI + 3A. Since A2=AA^2 = A, this becomes 3A3A.

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 then by induction and observing the pattern in matrix powers, A10 = 29
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)f(x) at specific values and taking their product as stated in the expression yields a product of 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, solving for α\alpha yields 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

Upon differentiating matrix BB with respect to xx and considering the elements within BB, the derivative dBdx\frac{dB}{dx} is found to be 3A3A.

.
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 tend to 0, resulting in the limit being 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 logarithmic function with a base bb greater than 1, the point (1, 0) is consistently found on its graph because log⁡b(1)=0\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.

The function ∣cos⁡x∣ is continuous everywhere, yet it has points of non-differentiability at odd multiples of π2, where cos⁡x=0 = 0.

30. If y = 2x3x, then dy/dx at x = 1 is:
(A) 2
(B) 6
(C) 3
(D) 1
(B) 6

By employing logarithmic differentiation, calculate dy/dx for y = 2x3x and evaluate at x = 1 to determine 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), and knowing  f(5) = 3, we compute 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 By solving f'(C) = 1, we find C = 4/3, which satisfies 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

Applying the chain rule and trigonometric identities simplifies the derivative 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 the first and second derivatives equal to zero at a specific point,

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

Upon Differentiating x^x and analyzing its monotonicity, it is found that 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

By applying principles from 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]

Upon differentiating f(x), we determine that it is increasing over 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

By employing 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

When integrating an odd function across an interval that is symmetric about the origin, the result of the integral is zero. This occurs because the negative and positive areas under the curve cancel each other out.

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 By partial fractions and substitution, the integral can be simplified to the expression 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

When the function is simplified using trigonometric identities and then integrated, the result is given by the expression x+2sin⁡x+sin⁡2x+C. This outcome reflects the integral's calculation that incorporates the simplifications made from trigonometric transformations.

42. ∫ from 1 to 5 (|x - 3| + |1 - x|) dx =
(A) 12
(B) 5/6
(C) 21
(D) 10
(A) 12

When the integral is computed across two specified intervals and then simplified, the total value obtained is 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

When viewed as a Riemann sum and evaluated as a limit, the expression 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 over the interval defined by their points of intersection results in 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 intersection points of the line and the curve results in 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)

Solving the differential equation with the specified initial condition leads to the solution 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, where C is a constant.

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

Using vector addition and the Pythagorean theorem, the length of the median through vertex A is calculated to be √33 

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

By applying the scalar triple product, the volume of the parallelepiped is calculated to be 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, This specific angle satisfies the condition necessary for the resulting vector's magnitude to be one.

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, resulting in a total sum of 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 calculates to -10/7, confirming that the lines are perpendicular.

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

Using the distance formula for parallel planes, the distance between them is calculated as 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)

By applying the sine formula along with the direction ratios and the normal vector of the plane, one can determine specific geometric properties such as angles or distances related to 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

The yz-plane and the xz-plane represent two of the coordinate planes that are perpendicular to each other. The yz-plane is defined by all points where x=0, and the xz-plane is defined by all points where y=0.

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

The equation of a plane can be derived using a given point on the plane and properties of a line parallel to the plane. By utilizing these elements, the normal vector to the plane and a point through which the plane passes are identified, enabling the formulation of the plane's equation.

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)

The minimum value of z is consistent at specific points and remains the same along the line segment that connects these points. This indicates that the function representing z reaches its lowest value across this entire segment.

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

The probability of obtaining at least one odd number is calculated by subtracting the probability of getting no odd numbers from 1. This approach utilizes the complement rule in probability, which states that the probability of an event occurring is 1 minus the probability of the event not occurring.

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

The variance is calculated using the values of E(X²) and E(X) which are derived by solving for the constant k. This involves determining kk such that the expected values fit the requirements of the probability distribution, allowing for the computation of variance as Var(X)=E(X2)−[E(X)]2.

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

The calculation is based on the condition P(X=2)= 9P(X=3) within the framework of the binomial probability formula. This relationship is used to determine the parameters of the binomial distribution, such as the number of trials nn and the probability of success pp, by setting up equations that relate these probabilities according to the binomial formula.



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