Step 1: Determine the first 50 even natural numbers.

The sequence of the first 50 even natural numbers is:

2, 4, 6, ..., 100.

This sequence forms an arithmetic progression characterized by:

• First term a = 2
• Common difference d = 2
• Number of terms n = 50

Step 2: Calculate the sum and the mean of the sequence.

The sum S of an arithmetic progression can be calculated using the formula:

S = (n/2) × (2a + (n-1)d)

Substituting the known values:

S = (50/2) × (2 × 2 + (50 - 1) × 2) = 25 × (4 + 98) = 25 × 102 = 2550

The mean (μ) is then:

μ = S/n = 2550/50 = 51

Step 3: Compute the sum of squares of the sequence.

The sum of squares of the first n even natural numbers is given by:

Sum of squares = 4 × (n(n + 1)(2n + 1)) / 6

Plugging in n = 50:

Sum of squares = 4 × (50 × 51 × 101) / 6 = 4 × 257550 / 6 = 4 × 42925 = 171700

Step 4: Determine the variance of the sequence.

First, find the expected value of X^2:

E(X^2) = Sum of squares / n = 171700 / 50 = 3434

Then, calculate the variance (σ^2):

Variance = E(X^2) - (μ)^2 = 3434 - (51)^2 = 3434 - 2601 = 833

Conclusion: The variance of the first 50 even natural numbers is 833.

Step 1: Simplify the given trigonometric expression.

We utilize the identity:

(1 + sin(x)) / (1 + cos(x)) = tan(x/2)

Step 2: Substitute the identity into the integral.

Applying this substitution, the integral becomes:

I = ∫ e^x × tan(x/2) dx

Step 3: Apply substitution method.

Let u = x/2, which implies du = (1/2) dx, and therefore dx = 2 du.

Substituting these into the integral:

I = 2 ∫ e^(2u) × tan(u) du

Step 4: Integrate the expression.

This integral is standard, and its solution is:

I = e^(2u) × tan(u) + C

Step 5: Revert the substitution to the original variable.

Substituting back u = x/2 into the result:

I = e^x × tan(x/2) + C

Conclusion: The integral simplifies to:

∫ e^x × tan(x/2) dx = e^x × tan(x/2) + C

We are given the differential equation:

x cos(y) dy = (x e^x log(x) + e^x) dx

This equation can be rearranged to separate the variables:

cos(y) dy = (e^x log(x) + (e^x / x)) dx

Step 1: Integrate both sides.

Integrate the left side with respect to y and the right side with respect to x:

∫ cos(y) dy = ∫ (e^x log(x) + (e^x / x)) dx

sin(y) = ∫ e^x log(x) dx + ∫ (e^x / x) dx + C

Step 2: Evaluate the integrals on the right-hand side.

For the first integral, use integration by parts:

Let u = log(x), so du = (1 / x) dx, and dv = e^x dx, so v = e^x.

Applying the integration by parts formula ∫ u dv = uv - ∫ v du, we get:

∫ e^x log(x) dx = e^x log(x) - ∫ (e^x / x) dx

Therefore, the right side becomes:

sin(y) = e^x log(x) - ∫ (e^x / x) dx + ∫ (e^x / x) dx + C

The integrals ∫ (e^x / x) dx cancel out:

sin(y) = e^x log(x) + C

Conclusion: The solution to the differential equation is:

sin(y) = e^x log(x) + C

The expected value E(X) is given by:

E(X) = Σ (x × P(X = x)) = (-2)(0.2) + (-1)(0.3) + (0)(0.1) + (1)(0.15) + (2)(0.25)

E(X) = -0.4 - 0.3 + 0 + 0.15 + 0.5 = -0.05

The expected value of X^2 is:

E(X^2) = Σ (x^2 × P(X = x)) = (-2)^2(0.2) + (-1)^2(0.3) + (0)^2(0.1) + (1)^2(0.15) + (2)^2(0.25)

E(X^2) = 0.8 + 0.3 + 0 + 0.15 + 1 = 2.25

The variance Var(X) is given by:

Variance = E(X^2) - (E(X))^2 = 2.25 - (-0.05)^2 = 2.25 - 0.0025 = 2.2475

Thus, the variance is 2.2475

We are given the logical statement p ↔ (q → p). This biconditional expression is false only when one part is true and the other is false.

Case 1: Let p = True and q = False

• q → p evaluates to False → True = True
• Therefore, p ↔ (q → p) becomes True ↔ True = True

This case does not make the statement false.

Case 2: Let p = False and q = True

• q → p evaluates to True → False = False
• Therefore, p ↔ (q → p) becomes False ↔ False = True

This case also does not make the statement false.

Case 3: Let p = False and q = False

• q → p evaluates to False → False = True
• Therefore, p ↔ (q → p) becomes False ↔ True = False

This case satisfies the condition where the biconditional statement is false.

Next, we evaluate the provided options under the scenario p = False and q = False:

False → (False ∨ True) = False → True = True

False ∧ (True ∧ False) = False ∧ False = False

(False ∨ True) → False = True → False = False

• Option 1: p is false.
• Option 2: p → (p ∨ ~q). Substituting the values:
• Option 3: p ∧ (~p ∧ q). Substituting the values:
• Option 4: (p ∨ ~q) → p. Substituting the values:

Among these options, Option 2 is the only statement that evaluates to true when p = False and q = False.

A tautology is a logical statement that is always true, irrespective of the truth values of its individual components.

Consider the statement [(p → q) ∧ ~q] → r. For this statement to be a tautology, it must hold true under all possible truth assignments of p, q, and r.

Analyzing the Antecedent:

The antecedent of the implication is (p → q) ∧ ~q. This conjunction is true only when both p → q is true and ~q (not q) is true.

• ~q being true implies that q is false.
• For p → q to be true while q is false, p must also be false. (Recall that p → q is only false when p is true and q is false.)

Therefore, the antecedent (p → q) ∧ ~q is true exclusively when both p and q are false.

Implications for r:

The entire statement [(p → q) ∧ ~q] → r is a conditional statement that evaluates to false only when the antecedent is true and the consequent r is false. To ensure that the entire statement is always true (i.e., a tautology), r must be true in every scenario where the antecedent is true.

Since the antecedent is true only when ~q is true, r must also be true whenever ~q is true. This relationship implies that r must logically follow ~q, meaning r should be equivalent to ~q.

Conclusion: For the statement [(p → q) ∧ ~q] → r to be a tautology, r must be equivalent to ~q.

We are given:

• Total number of bulbs, N = 100
• Defective bulbs, K = 10
• Non-defective bulbs, N - K = 90
• Bulbs selected, n = 5

We need to find the probability of selecting at most one defective bulb, i.e., P(X ≤ 1).

Step 1: Probability of selecting 0 defective bulbs

Selecting all 5 bulbs from the 90 non-defective ones:

P(X = 0) = (C(90, 5)) / (C(100, 5)) ≈ 0.5837

Step 2: Probability of selecting 1 defective bulb

Selecting 1 defective bulb from 10 and 4 non-defective bulbs from 90:

P(X = 1) = (C(10, 1) × C(90, 4)) / (C(100, 5)) ≈ 0.31

Step 3: Total probability

Summing the probabilities for 0 and 1 defective bulbs:

P(X ≤ 1) = P(X = 0) + P(X = 1) = 0.5837 + 0.31 = 0.8937

Using Hypergeometric Distribution:

The hypergeometric distribution formula is:

P(X = k) = (C(K, k) × C(N - K, n - k)) / C(N, n)

Thus,

P(X ≤ 1) = (C(10, 0) × C(90, 5)) / C(100, 5) + (C(10, 1) × C(90, 4)) / C(100, 5) ≈ 0.8937

Conclusion: The probability of selecting at most one defective bulb out of five is approximately 0.8937.