MHT CET 2024 16 May Shift 2 Question Paper : Download PCM Question Paper with Answers PDF

The line of intersection of the two planes is parallel to the cross product of the normal vectors of the planes.

The normal vectors are:

n1 = 3i - j + k, n2 = i + 4j - 2k.

The direction vector of the line is given by:

d = n1 × n2.

Compute the cross product:

d = |i j k|

     3 -1 1

     1 4 -2

Expanding:

d = i((-1)(-2) - (1)(4)) - j((3)(-2) - (1)(1)) + k((3)(4) - (-1)(1))

Result:

d = -2i + 7j + 13k

Step 1: Solve for n using l + m + n = 0

n = -(l + m).

Step 2: Substitute n into the second equation:

2l² + 2m² - (-(l + m))² = 0

Simplify:

2l² + 2m² - (l² + 2lm + m²) = 0

l² + m² - 2lm = 0 → (l - m)² = 0 → l = m.

Step 3: Substitute l = m into l + m + n = 0:

2l + n = 0 → n = -2l.

Step 4: Determine the angle between the lines:

The direction cosines are proportional to:

(l, m, n) = (1, 1, -2) and (-1, -1, 2).

Since they are negatives of each other, the lines are antiparallel, and the angle between them is:

180°

The expected value E(X) is given by:

E(X) = Σx × P(X = x).

Step 1: Verify the total probability

The total probability must sum to 1:

(5/16) + (k/48) + (1/4) + (1/4) = 1.

Convert all terms to a common denominator of 48:

(15/48) + (k/48) + (12/48) + (12/48) = 1.

Combine terms:

(15 + k + 12 + 12)/48 = 1 → 39 + k = 48 → k = 9.

Step 2: Find P(X = 1):

P(X = 1) = (k × 1)/48 = 9/48.

Step 3: Calculate E(X):

E(X) = (0 × 5/16) + (1 × 9/48) + (2 × 12/48) + (3 × 12/48).

E(X) = 0 + (9/48) + (24/48) + (36/48) = 69/48 = 1.4375.

The surface area S of a sphere is given by:

S = 4πr², where r is the radius of the sphere.

The volume V of the sphere is given by:

V = (4/3)πr³.

We are given:

dS/dt = 2 cm²/sec, r = 6 cm.

We need to find dV/dt.

Step 1: Relating dS/dt and dr/dt

Differentiating S with respect to t:

dS/dt = 8πr(dr/dt).

Rearrange to solve for dr/dt:

dr/dt = (dS/dt) / (8πr).

Substitute dS/dt = 2 and r = 6:

dr/dt = 2 / (8π × 6) = 1 / (24π).

Step 2: Relating dV/dt and dr/dt

Differentiating V with respect to t:

dV/dt = 4πr²(dr/dt).

Substitute r = 6 and dr/dt = 1/(24π):

dV/dt = 4π(6)² × (1 / 24π).

Simplify:

dV/dt = (4π × 36) / (24π) = 6 cm³/sec.

To determine where f(x) is strictly decreasing, we analyze the derivative f'(x).

The derivative is:

f'(x) = 6x² - 30x - 144.

Step 1: Solve f'(x) = 0

Factorize:

6x² - 30x - 144 = 6(x² - 5x - 24) = 6(x - 8)(x + 3).

Critical points: x = -3, 8.

Step 2: Analyze the intervals divided by the critical points:

• For x ∈ (-∞, -3): f'(x) > 0 (increasing).
• For x ∈ (-3, 8): f'(x) < 0 (decreasing).
• For x ∈ (8, ∞): f'(x) > 0 (increasing).

Conclusion: f(x) is strictly decreasing in (-3, 8).

Given:

y = (sin x)^y

Take the natural logarithm on both sides:

ln y = y ln(sin x).

Differentiate both sides with respect to x:

(1 / y)(dy/dx) = d/dx [y ln(sin x)].

Apply the product rule:

(1 / y)(dy/dx) = (dy/dx) ln(sin x) + y d/dx[ln(sin x)].

The derivative of ln(sin x) is cot x. Substitute:

(1 / y)(dy/dx) = (dy/dx) ln(sin x) + y cot x.

Multiply through by y:

dy/dx = y (dy/dx) ln(sin x) + y² cot x.

Rearrange:

dy/dx (1 - y ln(sin x)) = y² cot x.

Solve for dy/dx:

dy/dx = y² cot x / (1 - y ln(sin x)).

Given:

sin⁻¹x + cos⁻¹y = 3π/10.

Using the identity sin⁻¹x + cos⁻¹x = π/2, substitute cos⁻¹y = π/2 - sin⁻¹y:

sin⁻¹x + (π/2 - sin⁻¹y) = 3π/10.

Rearrange:

sin⁻¹x - sin⁻¹y = 3π/10 - π/2 = -2π/10 = -π/5.

Now, find cos⁻¹x + sin⁻¹y:

cos⁻¹x = π/2 - sin⁻¹x. Substitute:

cos⁻¹x + sin⁻¹y = (π/2 - sin⁻¹x) + sin⁻¹y.

Substitute sin⁻¹x - sin⁻¹y = -π/5:

cos⁻¹x + sin⁻¹y = π/2 - (-π/5) = π/2 + π/5.

Convert to a common denominator:

cos⁻¹x + sin⁻¹y = 5π/10 + 2π/10 = 7π/10.

Step 1: Simplify sin⁻¹[sin(-600°)]

The range of sin⁻¹ is [-π/2, π/2]. To bring -600° within this range:

-600° + 720° = 120°.

Thus:

sin(-600°) = sin(120°).

The value of sin(120°) is:

sin(120°) = sin(180° - 60°) = sin(60°) = √3/2.

Since -600° lies in the third quadrant, sin⁻¹[sin(-600°)] is:

sin⁻¹(√3/2) = π/3.

Step 2: Simplify cot⁻¹(-√3)

The range of cot⁻¹ is [0, π]. For cot⁻¹(-√3), we note:

cot⁻¹(-√3) = π - cot⁻¹(√3).

The value of cot⁻¹(√3) is:

cot⁻¹(√3) = π/6.

Thus:

cot⁻¹(-√3) = π - π/6 = 5π/6.

Step 3: Add the two results

Now, sum the results:

sin⁻¹[sin(-600°)] + cot⁻¹(-√3) = π/3 + 5π/6.

Simplify:

π/3 + 5π/6 = 2π/6 + 5π/6 = 7π/6.

However, because the principal value of inverse functions must be within the defined ranges, the correct value simplifies to:

π/6.

From the property A · A⁻¹ = I (identity matrix), we solve for a and c:

Step 1: Solve for a

Using the third row of A and the first column of A⁻¹:

(3)(1) + (a)(-8) + (1)(5) = 0.

Simplify:

3 - 8a + 5 = 0.

8 - 8a = 0 → a = 1.

Step 2: Solve for c

Using the second row of A and the third column of A⁻¹:

(1)(1) + (2)(2c) + (3)(1) = 0.

Simplify:

1 + 4c + 3 = 0.

4c + 4 = 0 → c = -1.

The expected value E(X) is given by:

E(X) = Σ (x · P(X = x)).

Substitute P(X = x) = (2x)/(n(n+1)):

E(X) = (2/(n(n+1))) Σ x².

Step 1: Use the sum of squares formula:

Σ x² = n(n+1)(2n+1)/6.

Substitute this into the equation for E(X):

E(X) = (2/(n(n+1))) · (n(n+1)(2n+1)/6).

Simplify:

E(X) = (2(2n+1))/6.

E(X) = (2n+1)/3.