Mathematics Test - 42

Poisson distribution shows the number of times an event is likely to occur within a specified time. The Poisson distribution probability formula is P(x; μ) = (e^-μ) (μ^x) / x!

Let A and B be two matrices such that AB = A and BA = B Now, consider AB = A Take Transpose on both side (AB)^T = A^T 

⇒ A^T = B^T ⋅ A^T ...(1)

Now, BA = B 

Take, Transpose on both side (BA)^T = B^T 

⇒ B^T = A^T⋅B^T…(2)

Now, from equation (1) and (2). we have A^T = (A^T . B^T)A^T 

A^T=A^T(B^TA^T)

= A^T(AB)^T(∵(AB)^T = B^T = B^TA^T)

= A^T ⋅ A^T 

Thus, A^T = (A^T)²

For any square matrix A, both (A + A^T) is symmetric and (A - A^T) is skew-symmetric. Therefore, options A and B are true.

• (A + A^T) is symmetric.
• (A - A^T) is skew-symmetric.

Step 1: List the Constraints and Objective Function

• Objective Function: Maximize Z = 5x + 6y

• Constraints:

  1. x + y ≤ 6

  2. 2x + y ≤ 10

  3. x + 2y ≤ 10

  4. x ≥ 0

  5. y ≥ 0

Step 2: Define the Feasible Region (Graphical Method)

We'll plot each inequality as an equation (treat "=" as equality to draw the line), then find the region that satisfies all constraints.

Convert each constraint into a line:

1. x + y = 6

   • When x = 0, y = 6 → (0, 6)

   • When y = 0, x = 6 → (6, 0)

2. 2x + y = 10

   • When x = 0, y = 10 → (0, 10)

   • When y = 0, x = 5 → (5, 0)

3. x + 2y = 10

   • When x = 0, y = 5 → (0, 5)

   • When y = 0, x = 10 → (10, 0)

4. x ≥ 0, y ≥ 0 → We're in the first quadrant only.

Step 3: Find the Corner Points of the Feasible Region

We find intersection points of the lines (edges of the feasible region):

1. Intersection of x + y = 6 and 2x + y = 10
   Subtract first equation from second:
   (2x + y) - (x + y) = 10 - 6 → x = 4
   Put x = 4 into x + y = 6 → y = 2
   So point is (4, 2)

2. Intersection of x + y = 6 and x + 2y = 10
   Subtract:
   (x + 2y) - (x + y) = 10 - 6 → y = 4
   Put y = 4 into x + y = 6 → x = 2
   So point is (2, 4)

3. Intersection of 2x + y = 10 and x + 2y = 10
   Multiply second equation by 2:
   2x + 4y = 20
   Subtract from first: (2x + 4y) - (2x + y) → 3y = 10 → y = 10/3 ≈ 3.33
   Put y = 10/3 into x + 2y = 10 → x = 10 - 20/3 = 10/3 ≈ 3.33
   So point is (10/3, 10/3), but check: x + y = 20/3 ≈ 6.67 > 6 → not feasible

4. Point on x-axis: x = 5, y = 0
   All constraints satisfied → Point is (5, 0)

5. Point on y-axis: x = 0, y = 5
   All constraints satisfied → Point is (0, 5)

6. Origin: (0, 0) satisfies all constraints

Step 4: Evaluate Z = 5x + 6y at Each Valid Point

• At (0, 0): Z = 0

• At (0, 5): Z = 5(0) + 6(5) = 30

• At (2, 4): Z = 5(2) + 6(4) = 10 + 24 = 34

• At (4, 2): Z = 5(4) + 6(2) = 20 + 12 = 32

• At (5, 0): Z = 5(5) + 6(0) = 25

Note: We reject (10/3, 10/3) since it doesn't satisfy x + y ≤ 6

Step 5: Find the Maximum Z

• Z = 0 at (0, 0)

• Z = 30 at (0, 5)

• Z = 34 at (2, 4) → Maximum

• Z = 32 at (4, 2)

• Z = 25 at (5, 0)

Final Answer:

The maximum value of Z is 34, and it occurs at the point (2, 4).

Correct option: a) 34

Step 1: Identify the feasible region

The feasible region is the set of points (x, y) satisfying all the constraints simultaneously.

Step 2: Convert inequalities to find boundaries

We will find intersection points of the constraints because the minimum will be at one of the vertices of the feasible region.

Step 3: Find corner points (vertices) of the feasible region

Constraint 1: x + 2y ≥ 4
Rewrite as: x = 4 - 2y

Constraint 2: 3x + y ≥ 6
Rewrite as: y = 6 - 3x

Constraint 3: x + y ≤ 8
Boundary line: x + y = 8
Rewrite as: y = 8 - x

Non-negativity: x ≥ 0, y ≥ 0

Step 4: Find intersections of pairs of boundary lines

Intersection of x + 2y = 4 and 3x + y = 6:

From the first, x = 4 - 2y.

Substitute into the second:

3(4 - 2y) + y = 6 → 12 - 6y + y = 6 → 12 - 5y = 6 → 5y = 6 → y = 6/5 = 1.2

Then,

x = 4 - 2(1.2) = 4 - 2.4 = 1.6

So, point A = (1.6, 1.2).

Check feasibility:

• x + y = 1.6 + 1.2 = 2.8 ≤ 8 ✓

• x, y ≥ 0 ✓

• x + 2y = 1.6 + 2(1.2) = 1.6 + 2.4 = 4 ≥ 4 ✓

• 3x + y = 3(1.6) + 1.2 = 4.8 + 1.2 = 6 ≥ 6 ✓

So A is feasible.

Intersection of x + 2y = 4 and x + y = 8:

From x + y = 8, x = 8 - y.

Substitute into x + 2y = 4:

(8 - y) + 2y = 4 → 8 + y = 4 → y = 4 - 8 = -4

y = -4 is not feasible since y ≥ 0.

No feasible intersection here.

Intersection of 3x + y = 6 and x + y = 8:

From x + y = 8, y = 8 - x.

Substitute into 3x + y = 6:

3x + (8 - x) = 6 → 3x + 8 - x = 6 → 2x + 8 = 6 → 2x = -2 → x = -1

x = -1 not feasible since x ≥ 0.

No feasible intersection here.

Intersection with axes:

• At x = 0:
  x + 2y ≥ 4 → y ≥ 2
  3x + y ≥ 6 → y ≥ 6
  x + y ≤ 8 → y ≤ 8
  So at x = 0, y ∈ [6, 8] to satisfy all constraints.

• At y = 0:
  x + 2(0) ≥ 4 → x ≥ 4
  3x + 0 ≥ 6 → x ≥ 2
  x + 0 ≤ 8 → x ≤ 8
  So at y = 0, x ∈ [4, 8].

Step 5: Check possible vertices on axes (extreme points)

• Point B: x = 4, y = 0
  Check all constraints:
  4 + 2(0) = 4 ≥ 4 ✓
  3(4) + 0 = 12 ≥ 6 ✓
  4 + 0 = 4 ≤ 8 ✓
  x, y ≥ 0 ✓

• Point C: x = 0, y = 6
  Check constraints:
  0 + 2(6) = 12 ≥ 4 ✓
  3(0) + 6 = 6 ≥ 6 ✓
  0 + 6 = 6 ≤ 8 ✓
  x, y ≥ 0 ✓

• Point D: x = 0, y = 8
  Check:
  0 + 2(8) = 16 ≥ 4 ✓
  3(0) + 8 = 8 ≥ 6 ✓
  0 + 8 = 8 ≤ 8 ✓
  x, y ≥ 0 ✓

Step 6: Check Z = 4x + 3y at feasible points:

Step 7: Conclusion

The minimum value of Z is 10 at the point (1.6, 1.2).

Final answer:

Minimum value of Z = 10 at (x, y) = (1.6, 1.2).

The number of arbitrary constants for a particular solution of n^th order differential equation is always zero.

The number of arbitrary constants in a general solution of a nth order differential equation is n.

Therefore, the number of arbitrary constants in the general solution of a second order D.E is 2.

f(x) = x -[x] is derivable at all x ∈ R – I , and f ‘(x) = 1 for all x ∈ R – I .

Given, y² = 3x ……….(1) and y = 2x + 4 ……….(2)

Differentiating both sides of (1) with respect to y we get,

2y = 3(dx/dy)

Or dx/dy = 2y/3

Let P (x₁, y₁) be any point on the parabola (1). Then the slope of the normal to the parabola (1) at point P is

-[dx/dy]_P = -2y₁/3

If the normal at the point P to the parabola (1) be perpendicular to the line (2) then we must have,

-2y₁/3*2 = -1

Since the slope of the line (2) is 2

Or y₁ = 3/4

Since the point P(x₁, y₁) lies on (1) hence,

y₁² = 3x₁

As, y₁ = 3/4, so, x₁ = 3/16

Therefore, the required equation of the normal is

y – y₁ = -(2y₁)/3*(x – x₁)

Putting the value of x₁ and y₁ in the above equation we get,

16x + 32y = 27

And the coordinates of the foot of the normal are (x₁, y₁) = (3/16, 3/4)

Given, x² + 3y² = 12 Or x²/12 + y²/4 = 1

Differentiating both sides of (1) with respect to y we get,

2x*(dx/dy) + 3*2y = 0

Or dx/dy = -3y/x

Suppose the normal to the ellipse (1) at the point P(√12cosθ, 2sinθ) makes an angle 60° with the major axis. Then, the slope of the normal at P is tan60°

Or -[dx/dy]_P = tan60°

Or -(-(3*2sinθ)/√12cosθ) = √3

Or √3tanθ = √3

Or tanθ = 1

Now the centre of the ellipse (1) is C(0, 0)

Therefore, the slope of the line CP is,

(2sinθ – 0)/(√12cosθ – 0) = (1/√3)tanθ = 1/√3 [as, tanθ = 1]

Therefore, the line CP is inclined at 30° to the major axis.