MHT CET 2024 4 May Shift 2 Question Paper : Download PCM Question Paper with Answers PDF

Step 1: Identify the first 50 even natural numbers.

The sequence is: 2, 4, 6, ..., 100.

This is an arithmetic progression with:

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

Step 2: Calculate the sum and mean.

The sum of the sequence is:

S = (n/2)(2a + (n-1)d) = (50/2)(2(2) + (50-1)2) = 25(4 + 98) = 2550.

The mean is:

Mean = S/n = 2550/50 = 51.

Step 3: Calculate the sum of squares.

Sum of squares:

Σ(i²) = 4Σ(i²) = 4 × (n(n+1)(2n+1)/6).

For n = 50:

4 × (50 × 51 × 101)/6 = 171700.

Step 4: Calculate the variance.

E(X²) = Σ(i²)/n = 171700/50 = 3434.

Variance = E(X²) - (Mean)² = 3434 - 51² = 3434 - 2601 = 833.

Conclusion: The variance is 833.

Step 1: Simplify the trigonometric expression.

Using the identity:

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

Step 2: Substitute into the integral.

I = ∫ e^x · tan(x/2) dx.

Step 3: Use substitution.

Let u = x/2, so du = (1/2) dx and dx = 2 du.

The integral becomes:

I = 2 ∫ e^2u · tan(u) du.

Step 4: Solve the integral.

This is a standard integral, and the result is:

I = e^2u · tan(u) + C.

Step 5: Substitute back for u.

Substituting u = x/2:

I = e^x · tan(x/2) + C.

Step 1: Rewrite the given equation.

The given equation is:

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

Rewriting, we have:

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

Step 2: Solve each integral.

• On the left-hand side: ∫ cos(y) dy = sin(y) + C₁.
• On the right-hand side: ∫ (e^x log(x) + e^x / x) dx.

Step 3: Combine results.

After integrating both sides and simplifying:

sin(y) = x e^x + C₂.

Conclusion: The solution is \( x e^x + C \).

Step 1: Calculate the expected value \( E(X) \).

\( E(X) = Σ(x · P(X)) \).

= (-2)(0.2) + (-1)(0.3) + (0)(0.1) + (1)(0.15) + (2)(0.25).

= -0.4 - 0.3 + 0 + 0.15 + 0.5 = -0.05.

Step 2: Calculate \( E(X^2) \).

\( E(X^2) = Σ(x² · P(X)) \).

= (-2)²(0.2) + (-1)²(0.3) + (0)²(0.1) + (1)²(0.15) + (2)²(0.25).

= 0.8 + 0.3 + 0 + 0.15 + 1 = 2.25.

Step 3: Calculate the variance.

\( \text{Var}(X) = E(X^2) - (E(X))^2 \).

= 2.25 - (-0.05)² = 2.25 - 0.0025 = 2.2475.

Conclusion: Variance = 2.2475.

Step 1: Analyze the statement \( p \leftrightarrow (q \rightarrow p) \).

This biconditional statement is false only when one side is true and the other is false.

By testing different truth values of \( p \) and \( q \), we find that the statement is false when \( p \) is false and \( q \) is false.

Step 2: Evaluate the given options under these conditions:

• Option (1): \( p \) is false.
• Option (2): \( p \rightarrow (p ∨ ~q) \) becomes false → (false ∨ true), which evaluates to true.
• Option (3): \( p ∧ (~p q) \) is false.
• Option (4): \( (p ∨ ~q) → p \) becomes (false ∨ true) → false, which is false.

Step 3: Conclusion.

Option (2) is the only statement that holds true under the given conditions.

Step 1: Analyze the antecedent \( (p → q) ∧ ~q \).

This antecedent is true when \( p → q \) is true and \( ~q \) is true. If \( ~q \) is true, then \( q \) is false, and \( p → q \) is true if \( p \) is false.

Step 2: Evaluate the consequent \( r \).

To make the statement a tautology, \( r \) must always be true when the antecedent is true. Since the antecedent implies \( ~q \), \( r \) must be equivalent to \( ~q \).

Step 3: Conclusion.

The correct answer is \( r = ~q \).

Step 1: Use the hypergeometric distribution to calculate probabilities.

For selecting 0 defective bulbs:

P(X = 0) = [C(90, 5) / C(100, 5)]

For selecting 1 defective bulb:

P(X = 1) = [C(10, 1) × C(90, 4) / C(100, 5)]

Step 2: Calculate P(X ≤ 1).

Add probabilities for X = 0 and X = 1.

The total probability simplifies to (7/5)(9/10)⁴.