CBSE Class 12 Mathematics Question Paper 2024 (Set 2- 65/3/2) Available - Download Solution PDF with Answer Key

Solution:

Step 1: Express the derivative of 5^x.

The function 5^x can be written in exponential form as:

5^x = e^{x log 5}.

Differentiating 5^x with respect to x:

d/dx(5^x) = d/dx(e^{x log 5}) = e^{x log 5} ⋅ log 5 = 5^x ⋅ log 5.

Step 2: Express the derivative of e^x.

The derivative of e^x with respect to x is:

d/dx(e^x) = e^x.

Step 3: Find the derivative of 5^x with respect to e^x.

Using the chain rule:

d/d(e^x)(5^x) = (d/dx(5^x)) / (d/dx(e^x)) = (5^x * log 5)/e^x

Step 4: Simplify the result.
The expression can be rewritten as:

(5^x/e^x)log5 = 5^x/exlog5=5^1/exlog5

Thus, the final answer is: 5^(1/ex)log 5

Since |a| = 2 and |k| ∈ [0, 3] (from k ∈ [−3, 2]), the range of |a||k| is:

|a||k| ∈ [2 · 0, 2 · 3] = [0, 6].

Final Answer: (D) [0, 6]

The angles α, β, γ made by the line with the x-axis, y-axis, and z-axis respectively, satisfy the equation for direction cosines:

cos² α + cos² β + cos² γ = 1.

Given that α = π/4 and γ = π/4, we calculate:

cos α = cos γ = √2/2.

Substitute these values into the equation:

(√2/2)² + cos² β + (√2/2)² = 1.

Simplify:

1/2 + cos² β + 1/2 = 1 ⇒ cos² β = 0.

This implies:

cos β = 0 ⇒ β = π/2.

Final Answer: π/2

To determine the correct constraints, analyze the feasible region depicted in the graph:

• Line 1: x + 2y = 76 The region above this line is shaded, indicating the constraint: x + 2y ≥ 76.
• Line 2: 2x + y = 104 The region below this line is shaded, indicating the constraint: 2x + y ≤ 104.
• Non-negativity constraints: Since the shaded region is in the first quadrant: x ≥ 0 and y ≥ 0.

Thus, the group of constraints representing the feasible region is: x + 2y ≥ 76, 2x + y ≤ 104, x ≥ 0, y ≥ 0.

Final Answer: (C) x + 2y ≥ 76, 2x + y ≤ 104, x ≥ 0, y ≥ 0

The inverse of a diagonal matrix is obtained by taking the reciprocal of the diagonal elements.

For A =

[ 2 0 0 ]

[ 0 3 0 ]

[ 0 0 5 ], the diagonal elements are 2, 3, and 5. Thus:

A⁻¹ =

[ 1/2 0 0 ]

[ 0 1/3 0 ]

[ 0 0 1/5 ].

This matches option (A).

Final Answer: (A)

Step 1: Definition of a transpose.
For a square matrix A, the transpose A′ is obtained by interchanging the rows and columns of A.

Step 2: Subtracting A′ from A.
The matrix A − A′ satisfies the property:

(A − A′)′ = A′ − A = −(A − A′).

Thus, A − A′ is equal to the negative of its transpose, which is the definition of a skew symmetric matrix.

Step 3: Conclusion.
For any square matrix A, A − A′ is always a skew symmetric matrix.

Final Answer: (C) a skew symmetric matrix

Step 1: Understanding the function.
The function is defined as f(x) = x² + 1, where x ∈ R. The term x² is always non-negative, and adding 1 shifts the range of x² to start from 1.

Step 2: Range of f(x).
The minimum value of x² is 0, so the minimum value of f(x) = x² + 1 is:

f(x) = 0 + 1 = 1.

Thus, the range of f(x) is [1,∞).

Step 3: Conditions for f(x) to be onto.
For f(x) to be onto, the codomain A must include the entire range of f(x). Therefore, A = [1,∞).

Step 4: Conclusion.
The function f(x) = x² + 1 is onto if A = [1,∞).

Final Answer: (C) [1,∞)

For a 2 × 2 matrix A =

[ a b ]

[ c d ], the adjugate matrix is:

adj A =

[ d −b ]

[ −c a ].

If adj A = A, then:

[ d −b ] = [ a b ]

[ −c a ] [ c d ].

Equating elements:

d = a, −b = b, −c = c, a = d.

From −b = b, we get b = 0, and from −c = c, we get c = 0. Thus:

A =

[ a 0 ]

[ 0 a ].

The sum of the elements is:

a + b + c + d = a + 0 + 0 + a = 2a.

Final Answer: 2a

Step 1: Analyze the given function.
1. Case 1: For x ≥ 0, |x| = x. Then:
f(x) = |1 − x + x| = |1| = 1.
2. Case 2: For x < 0, |x| = −x. Then:
f(x) = |1 − x − x| = |1 − 2x|.

Step 2: Check continuity.
For x ≥ 0, f(x) = 1. For x < 0, f(x) = |1 − 2x|. At the transition point x = 0:
f(0+) = 1, f(0−) = |1 − 2(0)| = 1.
Similarly, at x = 1, f(1+) = 1 and f(1−) = 1. Thus, f(x) is continuous everywhere.

Final Answer: (D) continuous everywhere

Step 1: Definition of a point of inflection.
A point of inflection is a point where the concavity of the function changes. This means f′′(x) changes its sign at that point.

Step 2: Conditions for inflection points.
At a point x = c, the function f(x) has a point of inflection if: - f′′(x) changes its sign around x = c. - f′(x) = 0 but does not change its sign.

Step 3: Analyzing the options.
Option (C) correctly states that f′(x) = 0 and f′(x) does not change its sign, which is consistent with the definition of a point of inflection.

Step 4: Conclusion.
The correct answer is (C).

Step 1: Understanding the property of g(x).
The function g(x) satisfies g(−x) = −g(x), which means g(x) is an odd function.

Step 2: Integral over an asymmetric interval.
The integral ∫₀^2a g(x) dx can be related to the integral over the negative interval using the property of odd functions:

∫₀^2a g(x) dx = − ∫_−2a⁰ g(x) dx.

Step 3: Analyzing the given options.
Option (D) correctly expresses the relationship between the integral over [0, 2a] and [−2a, 0] for odd functions.

Step 4: Conclusion.
The integral ∫₀^2a g(x) dx equals − ∫_−2a⁰ g(x) dx. Thus, the correct answer is (D).

Rewriting the equation:

x log x (dy/dx) + y = 2 log x ⇒ (dy/dx) + y/(x log x) = 2/(x log x).

This is a first-order linear differential equation of the form:

(dy/dx) + P(x)y = Q(x).

Final Answer: First-order linear differential equation

To check if ⃗a and ⃗b are perpendicular, compute their dot product:

⃗a · ⃗b = (2)(1) + (−1)(−2) + (1)(1) = 2 + 2 + 1 = 0.

Since ⃗a · ⃗b = 0, the vectors are perpendicular.

Final Answer: Perpendicular vectors

For a line making angles α, β, γ with the coordinate axes, the equation:

cos² α + cos² β + cos² γ = 1 is always true because it represents the property of direction cosines. The statement cos α + cos β + cos γ = 1 is not valid since it assumes specific alignment which is not general for direction cosines.

Final Answer: (D) cos α + cos β + cos γ = 1

The restrictions on decision variables in a linear programming problem are referred to as constraints. These constraints define the feasible region within which the solution lies.

Final Answer: (B) constraints

The probability of the union is given by:

P(E ∪ F) = P(E) + P(F) − P(E ∩ F).

Substitute the values:

0.4 = 0.1 + 0.3 − P(E ∩ F) ⇒ P(E ∩ F) = 0.

The conditional probability is:

P(F | E) = P(E ∩ F) / P(E) = 0 / 0.1 = 0.

Final Answer: 0

Step 1: Definition of an identity matrix.
An identity matrix A = [aij] is a square matrix in which all the diagonal elements are 1, and all off-diagonal elements are 0. Mathematically:
aij = 1, if i = j, 0, if i ≠ j.

Step 2: Analyze each option.
- (A) aij = 0 if i = j and aij = 1 if i ≠ j: This is incorrect because it contradicts the definition of an identity matrix.
- (B) aij = 1, ∀i, j: This is incorrect because an identity matrix has 0 for all off-diagonal elements.
- (C) aij = 0, ∀i, j: This is incorrect because it implies all elements are 0, which is not an identity matrix.
- (D) aij = 0 if i ≠ j and aij = 1 if i = j: This is correct, as it matches the definition of an identity matrix.

Final Answer: (D)

The projection of ⃗a on ⃗b is:

Proj_⃗b(⃗a) = (⃗a · ⃗b) / ||⃗b||² ⃗b.

The projection of ⃗b on ⃗a is:

Proj_⃗a(⃗b) = (⃗a · ⃗b) / ||⃗a||² ⃗a.

Clearly, the two projections are not the same unless ||⃗a|| = ||⃗b||.

The angle between ⃗a and ⃗b is the same as the angle between ⃗b and ⃗a, as the cosine function is symmetric. Thus, the reason is true.

Final Answer: (D)

A scalar matrix is a special type of diagonal matrix where all diagonal elements are equal. For example:

A = [ 2 0 0 ]
[ 0 2 0 ]
[ 0 0 2 ]

is a scalar matrix and also a diagonal matrix. However, the reason given, "In a diagonal matrix, all the diagonal elements are 0," is incorrect because diagonal matrices can have any value along their diagonal elements, not necessarily 0.

Final Answer: (C)

• Step 1: Use the trigonometric identity.
  Using the trigonometric identity: sin A cos B = 1/2 [sin(A + B) + sin(A - B)], we rewrite the integral as: ∫₀^π/2 sin 2x cos 3x dx = 1/2 ∫₀^π/2 [sin(5x) + sin(-x)] dx.
• Step 2: Simplify.
  Simplify: sin(-x) = -sin(x), so the integral becomes: 1/2 ∫₀^π/2 [sin(5x) - sin(x)] dx.
• Step 3: Evaluate ∫ sin(5x) dx.
  ∫ sin(5x) dx = -1/5 cos(5x). At the limits: ∫₀^π/2 sin(5x) dx = -1/5 [cos(5π/2) - cos(0)] = -1/5 [0 - 1] = 1/5.
• Step 4: Evaluate ∫ sin(x) dx.
  ∫ sin(x) dx = -cos(x). At the limits: ∫₀^π/2 sin(x) dx = -[cos(π/2) - cos(0)] = -[0 - 1] = 1.
• Step 5: Substitute back.
  ∫₀^π/2 sin 2x cos 3x dx = 1/2 [1/5 - 1] = 1/2 [-4/5] = -2/5.

1. Substitute u = 2x − x²:
Let u = 2x − x² ⇒ du/dx = 2 − 2x.

2. Rewrite the integral for F(x):
F(x) = ∫√(2x − x²) dx = ∫√u * (1/(2 − 2x)) du.

3. Solve the integral:
After substitution and simplification, compute the antiderivative and apply the boundary condition F(1) = 0.

Final Answer: F(x) = Function involving u with F(1) = 0.

1. Position vector of C:
The formula for the position vector of a point dividing a line segment externally in the ratio m : n is:
r_C = (m r_B − n r_A) / (m − n).
Here, r_A = ˆi + 2ˆj − ˆk, r_B = −ˆi + ˆj + ˆk, m = 4, and n = 1.
Substitute:
r_C = (4(−ˆi + ˆj + ˆk) − 1(ˆi + 2ˆj − ˆk)) / (4 − 1).
Simplify:
r_C = (−4ˆi + 4ˆj + 4ˆk − ˆi − 2ˆj + ˆk) / 3 = (−5ˆi + 2ˆj + 5ˆk) / 3.
Thus:
r_C = −(5/3)ˆi + (2/3)ˆj + (5/3)ˆk.

2. Find |−→AB| : |−→BC|:
First, calculate −→AB = r_B − r_A:
−→AB = (−ˆi + ˆj + ˆk) − (ˆi + 2ˆj − ˆk) = −2ˆi − ˆj + 2ˆk.
Magnitude:
|−→AB| = √((-2)² + (−1)² + 2²) = √(4 + 1 + 4) = √9 = 3.
Now calculate −−→BC = r_C − r_B:
−−→BC = (−(5/3)ˆi + (2/3)ˆj + (5/3)ˆk) − (−ˆi + ˆj + ˆk).
Simplify:
−−→BC = (−(5/3) + 1)ˆi + ((2/3) − 1)ˆj + ((5/3) − 1)ˆk = (−2/3)ˆi − (1/3)ˆj + (2/3)ˆk.
Magnitude:
|−−→BC| = √((-2/3)² + (−1/3)² + (2/3)²) = √(4/9 + 1/9 + 4/9) = √9/9 = 1.
Thus:
|−→AB| : |−→BC| = 3 : 1.

Final Answer: r_C = −(5/3)ˆi + (2/3)ˆj + (5/3)ˆk, |−→AB| : |−→BC| = 3 : 1.

Step 1: Using the given condition.
The condition ⃗a + ⃗b − ⃗c = ⃗0 implies:
⃗a + ⃗b = ⃗c.

Taking the magnitude on both sides:
|⃗c|² = |⃗a + ⃗b|².

Step 2: Expanding the magnitude.
Expanding |⃗a + ⃗b|²:
|⃗a + ⃗b|² = |⃗a|² + |⃗b|² + 2⃗a · ⃗b.
Since |⃗a| = |⃗b| = |⃗c| = 1, this becomes:
1 = 1 + 1 + 2(⃗a · ⃗b).
Simplify:
2(⃗a · ⃗b) = −1 ⇒ ⃗a · ⃗b = −1/2.

Step 3: Relation between ⃗a and ⃗c.
Since ⃗c = ⃗a + ⃗b, the angle between ⃗a and ⃗c can be found using:
cos θ = ⃗a · ⃗c / |⃗a||⃗c|.
Substituting ⃗c = ⃗a + ⃗b:
⃗a · ⃗c = ⃗a · (⃗a + ⃗b) = |⃗a|² + ⃗a · ⃗b.
Using |⃗a|² = 1 and ⃗a · ⃗b = −1/2:
⃗a · ⃗c = 1 − 1/2 = 1/2.

Step 4: Angle between ⃗a and ⃗c.
The cosine of the angle is:
cos θ = 1/2 / 1 = 1/2.
Thus, the angle is:
θ = cos⁻¹(−1/2).

Final Answer: θ = cos⁻¹(−1/2).

Step 1: Simplify sin²(cos⁻¹(3/5)).
Let θ = cos⁻¹(3/5). Then:
cos θ = 3/5.
Using the Pythagorean identity:
sin² θ = 1 − cos² θ = 1 − (3/5)² = 1 − 9/25 = 16/25.

Step 2: Simplify tan²(sec⁻¹(3)).
Let φ = sec⁻¹(3). Then:
sec φ = 3 ⇒ cos φ = 1/3.
Using tan² φ = sec² φ − 1:
tan² φ = 3² − 1 = 9 − 1 = 8.

Step 3: Combine the results.
The expression becomes:
sin²(cos⁻¹(3/5)) + tan²(sec⁻¹(3)) = 16/25 + 8.
Simplify:
16/25 + 8 = 16/25 + 200/25 = 216/25.

Final Answer: 10.

1. Start with the given equation:
x = e^(y²).
Take the natural logarithm on both sides:
log x = y².

2. Differentiate both sides with respect to x:
1/x · dx/dx = 2y · dy/dx.

3. Rearrange for dy/dx:
dy/dx = 1 / (2y) · 1/x.

4. Substitute y² = log x into y = √(log x):
dy/dx = 1 / (2√(log x)) · 1/x.

5. Express dy/dx in terms of log x:
dy/dx = (log x − 1) / (log x)².

Proved.

1. Continuity at x = 1:
f(1−) = lim_x→1− f(x) = (1)² + 1 = 2,
f(1+) = lim_x→1+ f(x) = 3 − 1 = 2.
Thus, f(1−) = f(1+) = f(1) = 2, so f(x) is continuous at x = 1.

2. Differentiability at x = 1:
Find the left-hand derivative:
f'(x) = d/dx(x² + 1) = 2x, f'(1−) = 2(1) = 2.
Find the right-hand derivative:
f'(x) = d/dx(3 − x) = −1, f'(1+) = −1.
Since f'(1−) ≠ f'(1+), the function is not differentiable at x = 1.

Final Answer: Not differentiable at x = 1.

Step 1: Substitution. Let t = log x, so dt = (1/x) dx. This transforms the integral as:

∫[(log x)² − 5 log x + 4] dx = ∫(t² − 5t + 4) dt.

Step 2: Factorize the denominator. The quadratic expression in the denominator can be factorized as:

t² − 5t + 4 = (t − 1)(t − 4).

Thus, the integral becomes: ∫(1/(t − 1)(t − 4)) dt.

Step 3: Partial fraction decomposition. Write:

1/(t − 1)(t − 4) = A/(t − 1) + B/(t − 4),

where A and B are constants. Solving for A and B, we get:

1 = A(t − 4) + B(t − 1).

Substituting t = 1, A = 1/3. Substituting t = 4, B = −1/3.

Thus: 1/(t − 1)(t − 4) = 1/3(t − 1) − 1/3(t − 4).

Step 4: Integrate. The integral becomes:

∫(1/(t − 1)(t − 4)) dt = 1/3 ∫(1/(t − 1)) dt − 1/3 ∫(1/(t − 4)) dt.

The integrals are:

∫(1/(t − 1)) dt = log |t − 1|,

∫(1/(t − 4)) dt = log |t − 4|.

Thus:

∫(1/(t − 1)(t − 4)) dt = 1/3 log |t − 1| − 1/3 log |t − 4|.

Step 5: Back substitution. Substitute t = log x back into the result:

∫[(log x)² − 5 log x + 4] dx = 1/3 log |log x − 1| − 1/3 log |log x − 4| + C,

where C is the constant of integration.

Final Answer: 1/3 log |log x − 1| − 1/3 log |log x − 4| + C.

1. Simplify the trigonometric expression: Using the identity 1 + cos 2x = 2 cos² x and sin 2x = 2 sin x cos x, rewrite the numerator:

(2 + sin 2x) / (1 + cos 2x) = (2 + 2 sin x cos x) / (2 cos² x) = (1 / cos² x) + (sin x / cos x) = sec² x + tan x.

2. Rewrite the integral:

∫[(sec² x + tan x) e^x] dx.

3. Separate the terms:

∫[(sec² x e^x) + (tan x e^x)] dx = ∫(sec² x e^x) dx + ∫(tan x e^x) dx.

4. Solve the integrals: The first integral:

∫(sec² x e^x) dx = e^x tan x + C1.

The second integral:

∫(tan x e^x) dx = e^x ln |sec x| + C2.

5. Combine the results:

∫[(2 + sin 2x) / (1 + cos 2x)] e^x dx = e^x (tan x + ln |sec x|) + C.

Final Answer: e^x (tan x + ln |sec x|) + C.

1. Simplify the denominator: Use the identity:
sin x + cos x = √2 sin(x + π/4).
Thus:

1 / (sin x + cos x) = 1 / (√2 sin(x + π/4)).

2. Rewrite the integral:

∫(1 / (sin x + cos x)) dx = 1 / √2 ∫(csc(x + π/4)) dx from 0 to π/4.

3. Substitute: Let u = x + π/4, so du = dx. Change the limits:
When x = 0, u = π/4; when x = π/4, u = π/2.
The integral becomes:

1 / √2 ∫(csc u) du from π/4 to π/2.

4. Integrate csc u:

∫(csc u) du = ln |csc u − cot u|.

5. Evaluate at the limits:

1 / √2 [ln |csc(π/2) − cot(π/2)| − ln |csc(π/4) − cot(π/4)|].

Substitute values:
csc(π/2) = 1, cot(π/2) = 0, csc(π/4) = √2, cot(π/4) = 1.

Simplify:

ln |csc(π/2) − cot(π/2)| = ln(1), ln |csc(π/4) − cot(π/4)| = ln(√2 − 1).

Final result:

1 / √2 [0 − ln(√2 − 1)] = −ln(√2 − 1) / √2.

Final Answer: −ln(√2 − 1) / √2.

Step 1: Graph the constraints. Convert each inequality into an equation to find the boundary lines:

• x + y = 8,
• x + 2y = 16,
• 4x + y = 29.

Find the points of intersection for these lines and determine the feasible region. The feasible region is the intersection of the half-planes satisfying all constraints.

Step 2: Determine corner points. By solving the equations pairwise, find the intersection points:

• Intersection of x + y = 8 and x + 2y = 16: (0, 8),
• Intersection of x + 2y = 16 and 4x + y = 29: (5, 5.5),
• Intersection of x + y = 8 and 4x + y = 29: (7.67, 0.33),
• Intersection of x + y = 8 and y = 0: (8, 0).

Step 3: Evaluate z = 600x + 400y at the corner points:

• At (0, 8): z = 600(0) + 400(8) = 3200,
• At (5, 5.5): z = 600(5) + 400(5.5) = 5300,
• At (7.67, 0.33): z = 600(7.67) + 400(0.33) ≈ 4720,
• At (8, 0): z = 600(8) + 400(0) = 4800.

Step 4: Conclusion. The minimum value of z = 600x + 400y is 3200, which occurs at the point (0, 8).

Final Answer: The minimum value of z is 3200 at the point (0, 8).

1. Let the events be:
- P1, P2, P3: Selection of P, Q, and R as CEO.
- E: Company increases profits.

2. Use Bayes’ theorem: The required probability is:
P(P3 | E) = P(P3) · P(E | P3) / P(E).

3. Calculate the prior probabilities: From the given ratio 4:1:2:
P(P1) = 4/7, P(P2) = 1/7, P(P3) = 2/7.

4. Calculate the total probability P(E):

P(E) = P(P1) · P(E | P1) + P(P2) · P(E | P2) + P(P3) · P(E | P3).

Substitute the given probabilities:
P(E) = (4/7) · 0.3 + (1/7) · 0.8 + (2/7) · 0.5.
Simplify:
P(E) = 1.2/7 + 0.8/7 + 1.0/7 = 3.0/7.

5. Calculate P(P3 | E):

P(P3 | E) = (2/7) · 0.5 / (3.0/7) = 1.0 / 3.0 = 1/3.

Final Answer: The probability that the increase in profits is due to R’s appointment as CEO is 1/3.

Start with the given equation:
x cos(p + y) + cos p sin(p + y) = 0.

1. Rearrange the equation:
cos p sin(p + y) = −x cos(p + y).

2. Differentiate both sides with respect to x:
cos p (d/dx)sin(p + y) = −(d/dx)[x cos(p + y)].

3. Use the chain rule on both sides:
cos p cos(p + y) (dy/dx) = −cos(p + y) + x(−sin(p + y)) (dy/dx).

4. Simplify:
cos p cos(p + y) (dy/dx) = − cos(p + y) + x sin(p + y) (dy/dx).

5. Factor out (dy/dx) terms:
(dy/dx) (cos p cos(p + y) − x sin(p + y)) = − cos(p + y).

6. Divide through by the coefficient of (dy/dx):
(dy/dx) = − cos(p + y) / (cos p cos(p + y)).

Simplify:
cos p (dy/dx) = − cos²(p + y).

Proved.

For f(x) to be continuous at x = 2, the left-hand limit (LHL), right-hand limit (RHL), and the value of f(2) must all be equal.

1. Left-hand limit (LHL): For x < 2:
f(x) = (x − 2) / |x − 2| + a = −1 + a.
As x → 2−:
LHL = −1 + a.

2. Right-hand limit (RHL): For x > 2:
f(x) = (x − 2) / |x − 2| + b = 1 + b.
As x → 2+:
RHL = 1 + b.

3. Value at x = 2:
f(2) = a + b.

4. Continuity condition:
LHL = RHL = f(2).
Substitute:
−1 + a = 1 + b = a + b.
From −1 + a = a + b:
b = −1.
From 1 + b = a + b:
a = 1.

Final Answer: a = 1, b = −1.

1. Find the derivative:
f(x) = log(x) / x, f′(x) = (x · 1/x − log(x) · 1) / x² = (1 − log(x)) / x².

2. Critical points: The critical points are obtained by solving f′(x) = 0:
1 − log(x) = 0 ⇒ log(x) = 1 ⇒ x = e.

3. Sign of f′(x):
- For x ∈ (0, e), 1 − log(x) > 0, so f′(x) > 0: f(x) is increasing.
- For x ∈ (e, ∞), 1 − log(x) < 0, so f′(x) < 0: f(x) is decreasing.

Intervals: f(x) is strictly increasing on (0, e) and strictly decreasing on (e, ∞).

1. Find the derivative:
f′(x) = 2x − 2/x².

2. Critical points: Set f′(x) = 0:
2x − 2/x² = 0 ⇒ 2x = 2/x² ⇒ x² = 1 ⇒ x = 1.

(Only x = 2 is in [1, 2].)

3. Evaluate f(x) at endpoints and critical point:
At x = 1: f(1) = 1² + 2/1 = 2.5.
At x = 2: f(2) = 2² + 2/2 = 2 + 1 = 3.

4. Conclusion: The absolute maximum value is 2.5 at x = 1, and the absolute minimum value is 2 at x = 2.

Final Answer: Absolute maximum: 2.5 (at x = 1), Absolute minimum: 2 (at x = 2).

1. Find the derivative:
f′(x) = 4x³ − 124x + a.

2. Condition for a local maximum at x = 1: At x = 1, f′(1) = 0:
4(1)³ − 124(1) + a = 0 ⇒ 4 − 124 + a = 0 ⇒ a = −6.

3. Update the function with a = −6:
f(x) = x⁴ − 62x² − 6x + 9.

4. Find other critical points: Set f′(x) = 0:
4x³ − 124x − 6 = 0 ⇒ 2x(2x² − 62) − 6 = 0.
Factorize:
2x(2x² − 62) = 6 ⇒ x(2x² − 62) = 3.
Simplify further:
2x³ − 62x − 3 = 0.
This cubic equation has roots that can be solved numerically or through approximation.

5. Determine the nature of the critical points: Evaluate the second derivative:
f′′(x) = 12x² − 124.
- At x = 1: f′′(1) = 12(1)² − 124 = −112 < 0 (local maximum).

Conclusion:
- Local maximum at x = 1 with a = −6.
- Other critical points are approximations of the cubic equation 2x³ − 62x − 3 = 0.

Final Answer: The value of a is −6. The local maximum occurs at x = 1.

1. Define dimensions of the rectangle: Let the length be 2r and the width be h, where 2r is the circumference of the cylinder base and h is its height. The perimeter is:
2r + 2h = 300 ⇒ r + h = 150 ⇒ h = 150 − r.

2. Volume of the cylinder: The volume of the cylinder is:
V = πr²h = πr²(150 − r).

3. Maximize V: Differentiate V with respect to r:
dV/dr = π[2r(150 − r) − r²] = π(300r − 3r²).

Set dV/dr = 0:
300r − 3r² = 0 ⇒ 3r(100 − r) = 0.
Thus, r = 0 or r = 100. Discard r = 0 since it gives no volume.

4. Second derivative test:
d²V/dr² = π(300 − 6r).
At r = 100:
d²V/dr² = π(300 − 600) = −300π < 0.
Hence, V is maximum at r = 100.

5. Find h:
h = 150 − r = 150 − 100 = 50.

Final Answer: The dimensions of the rectangular sheet are 2r = 200 cm and h = 50 cm.

Step 1: Equation of the line in terms of y:
x − 2y = 4 ⇒ y = (x − 4) / 2.
The line intersects the x-axis at x = 4 (putting y = 0), and the integration is divided into two parts:
- From x = −1 to x = 4,
- From x = 4 to x = 6.

Step 2: Area computation using integration:
The area is given by:
Area = ∫[x=−1 to x=6] (x − 4) / 2 dx.
Split the integral into two parts based on the sign of (x − 4):

Area = ∫[x=−1 to x=4] (4 − x) / 2 dx + ∫[x=4 to x=6] (x − 4) / 2 dx.

Step 3: Evaluate the integrals:
1. Compute the first integral:
∫[x=−1 to x=4] (4 − x) / 2 dx = 1/2 ∫[x=−1 to x=4] (4 − x) dx = 1/2 [4x − x²/2] from −1 to 4.
= 1/2 [(16 − 8) − (−4 − 0.5)] = 1/2 [8 + 4.5] = 6.25.

2. Compute the second integral:
∫[x=4 to x=6] (x − 4) / 2 dx = 1/2 ∫[x=4 to x=6] (x − 4) dx = 1/2 [x²/2 − 4x] from 4 to 6.
= 1/2 [(36/2 − 24) − (16/2 − 16)] = 1/2 [−6 + 8] = 1.

Step 4: Total area:
Area = 6.25 + 1 = 7.25 = 29/4.

Final Answer: The area of the region is 29/4.

1. Point of intersection of the two lines: Let:
x − 1 / 1 = y − 2 / 2 = z − 2 / 3 = t1, and x − 1 / 0 = y − 3 / −3 = z − 7 / 2 = t2.
From the first line:
x = 1 + t1, y = 2 + 2t1, z = 2 + 3t1.
From the second line:
x = 1, y = 3 − 3t2, z = 7 + 2t2.
Equate x, y, z for consistency: - From x: 1 + t1 = 1 ⇒ t1 = 0. - From y: 2 + 2t1 = 3 − 3t2 ⇒ 2 = 3 − 3t2 ⇒ t2 = 1 / 3.
Substitute t1 = 0 into the first line:
x = 1, y = 2, z = 2.
Thus, the point of intersection is (1, 2, 2).

2. Direction vectors of the given lines: - First line: ⃗d1 = ⟨1, 2, 3⟩. - Second line: ⃗d2 = ⟨0, −3, 2⟩.

3. Direction of the required line: The required line is perpendicular to both ⃗d1 and ⃗d2.
Use the cross product:
⃗d = ⃗d1 × ⃗d2 = | i j k |
| 1 2 3 |
| 0 −3 2 |
Expand:
⃗d = i(4 + 9) − j(2 − 0) + k(−3 − 0) = ⟨13, −2, −3⟩.

4. Equation of the required line: The line passing through (1, 2, 2) with direction vector ⟨13, −2, −3⟩ is:
x − 1 / 13 = y − 2 / −2 = z − 2 / −3.

Final Answer: x − 1 / 13 = y − 2 / −2 = z − 2 / −3.

1. Vector representation of AB: The position vectors of A and B are:
A⃗ = ⟨−1, 2, 1⟩, B⃗ = ⟨1, −2, 5⟩.
The vector −→AB is:
−→AB = B⃗ − A⃗ = ⟨1 − (−1), −2 − 2, 5 − 1⟩ = ⟨2, −4, 4⟩.

2. Direction vector of CD: The direction vector of the line CD is:
⃗dCD = ⟨1, −2, 2⟩.

3. Find a point on CD: From the parametric equation of CD:
x = 4 + t, y = −7 − 2t, z = 8 + 2t.
At t = 0, a point on CD is:
C0 = (4, −7, 8).

4. Shortest distance between AB and CD: The formula for the shortest distance between skew lines is:
d = |(⃗r2 − ⃗r1) · (⃗d1 × ⃗d2)| / |⃗d1 × ⃗d2|.
Here: - ⃗r1 = ⟨−1, 2, 1⟩ (point on AB), - ⃗r2 = ⟨4, −7, 8⟩ (point on CD), - ⃗d1 = −→AB = ⟨2, −4, 4⟩, - ⃗d2 = ⟨1, −2, 2⟩.
Compute ⃗r2 − ⃗r1:
⃗r2 − ⃗r1 = ⟨4 − (−1), −7 − 2, 8 − 1⟩ = ⟨5, −9, 7⟩.
Compute ⃗d1 × ⃗d2:
⃗d1 × ⃗d2 = | i j k |
| 2 −4 4 |
| 1 −2 2 |
⃗d1 × ⃗d2 = i(8 − (−8)) − j(4 − 4) + k(−4 − (−4)) = ⟨16, 0, 0⟩.

Compute (⃗r2 − ⃗r1) · (⃗d1 × ⃗d2):
(⃗r2 − ⃗r1) · (⃗d1 × ⃗d2) = ⟨5, −9, 7⟩ · ⟨16, 0, 0⟩ = 5 · 16 + 0 + 0 = 80.

Compute |⃗d1 × ⃗d2|:
|⃗d1 × ⃗d2| = √(16² + 0² + 0²) = √256 = 16.

Shortest distance:
d = |80| / 16 = 5.

5. Area of the parallelogram: The area of the parallelogram is given by:
Area = |−→AB| · d,
where:
|−→AB| = √(2² + (−4)² + 4²) = √(4 + 16 + 16) = √36 = 6.
Substitute:
Area = 6 · 5 = 30.

Final Answer: The distance between AB and CD is 5, and the area of the parallelogram is 30.

Step 1: Reflexive property.
For any x ∈ A, x − x = 0, which is divisible by 5. Hence, (x, x) ∈ R, and R is reflexive.

Step 2: Symmetric property.
If (x, y) ∈ R, then x − y is divisible by 5. This implies y − x = −(x − y) is also divisible by 5. Hence, (y, x) ∈ R, and R is symmetric.

Step 3: Transitive property.
If (x, y) ∈ R and (y, z) ∈ R, then x − y and y − z are both divisible by 5. Adding these gives:
(x − y) + (y − z) = x − z, which is also divisible by 5. Hence, (x, z) ∈ R, and R is transitive.
Since R is reflexive, symmetric, and transitive, it is an equivalence relation.

Step 4: Equivalence class [5].
The equivalence class of 5 is the set of all elements y ∈ A such that (5, y) ∈ R. This means:
5 − y is divisible by 5.
Equivalently:
5 − y = 5k ⇒ y = 5 − 5k for integers k.
Within the range −10 ≤ y ≤ 10, the possible values of y are:
[5] = {−10, −5, 0, 5, 10}.

Final Answer: [5] = {−10, −5, 0, 5, 10}.

The given differential equation is:
dP/dt = kP.

Separate the variables P and t:
(1/P) dP = k dt.

Integrate both sides:

∫(1/P) dP = ∫k dt.

Solve the integrals:

ln |P| = kt + C,

where C is the constant of integration.

Rewrite in exponential form:
P = e^(kt + C) = e^C · e^(kt).

Let e^C = P₀, where P₀ is the initial population. Then:
P = P₀ e^(kt).

Final Answer (i): The general solution is:
P = P₀ e^(kt).

From the general solution:
P = P₀ e^(kt).

At t = 0, P = 1000:
1000 = P₀ e^(k(0)) ⇒ P₀ = 1000.

At t = 1, P = 2000:
2000 = 1000 e^(k(1)).

Simplify:
e^k = 2000 / 1000 = 2.

Take the natural logarithm:
k = ln(2).

Final Answer (ii): The value of k is:
k = ln(2).

• Step 1: Express the given information algebraically using matrices.
  Let: x = Number of girl students, y = Number of meritorious students.
  The given conditions are:
  x + y = 50 (Total students).
  3000x + 4000y = 180000 (Total expenditure).
  Write this system in matrix form:

      [  1   1  ] [ x ]   [   50   ]
      [3000 4000] [ y ] = [ 180000 ]
      

• Step 2: Check consistency of the system.
  Calculate the determinant of the coefficient matrix A:

      A = [  1   1  ]
          [3000 4000]
      

  det(A) = (1)(4000) - (1)(3000) = 4000 - 3000 = 1000.
  Since det(A) ≠ 0, the system is consistent and has a unique solution.
• Step 3: Find the number of scholarships of each kind.
  Solve the system using the inverse of A:

      [ x ]   = A⁻¹ [ 50  ]
      [ y ]         [180000]
      

  The inverse of A is:
  A⁻¹ = (1/det(A)) * [ 4000 -1 ] [ -3000 1 ]
  Substituting values:
  [ x ] = (1/1000) * [ 4000 -1 ] [ 50 ] [ -3000 1 ] [180000]
  Simplify:
  [ x ] = (1/1000) * [ 20000 ] [ 30000 ]
  Thus, x = 20, y = 30.
• Step 4: If the scholarship amounts are interchanged.
  New total expenditure calculation:
  4000x + 3000y = 4000(20) + 3000(30).
  Simplify:
  4000(20) + 3000(30) = 80000 + 90000 = 170000.
  Thus, the new expenditure is 1,70,000.

• Step 1: Express the probability distribution in table format:
• Step 2: Find the value of k.
  The total probability must sum to 1:
  k + 4k + 9k + 8k + 10k + 12k = 1
  44k = 1 ⇒ k = 1/44.
• Step 3: Find the mean number of hours spent.
  The mean is given by:
  μ = E(X) = ∑ x ⋅ P(X = x)
  Substituting values:
  E(X) = 1⋅k + 2⋅4k + 3⋅9k + 4⋅8k + 5⋅10k + 6⋅12k
  = k(1 + 8 + 27 + 32 + 50 + 72)
  = k(190).
  Substituting k = 1/44, we get:
  E(X) = 190/44 = 95/22.
• Step 4: Find P(1 < X < 6).
  The probability P(1 < X < 6) is the sum of probabilities for x = 2, 3, 4, and 5:
  P(1 < X < 6) = P(X = 2) + P(X = 3) + P(X = 4) + P(X = 5)
  = 4k + 9k + 8k + 10k = 31k.
  Substituting k = 1/44, we get:
  P(1 < X < 6) = 31 ⋅ (1/44) = 31/44.