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

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:

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

Substitute these values into the equation:

Simplify:

This implies:

Final Answer: π/2

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

1. Line 1: x + 2y = 76 The region above this line is shaded, indicating the constraint: x + 2y ≥ 76.
2. Line 2: 2x + y = 104 The region below this line is shaded, indicating the constraint: 2x + y ≤ 104.
3. 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.

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

For , the diagonal elements are 2, 3, and 5. Thus:

This matches option (A).

Step 1: Definition of an identity matrix.

An identity matrix A = [a_ij] is a square matrix in which all the diagonal elements are 1, and all off-diagonal elements are 0. Mathematically:

Step 2: Analyze each option.

• (A) : This is incorrect because it contradicts the definition of an identity matrix.
• (B) a_ij = 1, ∀ i, j: This is incorrect because an identity matrix has 0 for all off-diagonal elements.
• (C) a_ij = 0, ∀ i, j: This is incorrect because it implies all elements are 0, which is not an identity matrix.
• (D) : This is correct, as it matches the definition of an identity matrix.

Step 1: Check if the function is one-one (injective). A function f(x) is one-one if f(a) = f(b) implies a = b.

For f(x) = x³ − 1, assume f(a) = f(b):

Since a and b are integers, a³ = b³ implies a = b. Thus, f(x) is one-one.

Step 2: Check if the function is onto (surjective). A function f(x) is onto if for every y ∈ Z, there exists an x ∈ Z such that f(x) = y.

Suppose y ∈ Z. Then:

For x³ = y + 1, y + 1 must be a perfect cube. However, not all integers y result in y + 1 being a perfect cube. For example, if y = 2, y + 1 = 3, which is not a perfect cube of any integer. Thus, f(x) is not onto.

Step 3: Conclusion. The function f(x) = x³ − 1 is one-one but not onto.

For a 2 × 2 matrix , the adjugate matrix is:

If adj A = A, then:

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:

The sum of the elements is: a + b + c + d = a + 0 + 0 + a = 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. Since f(0+) = f(0-) = f(0) = 1, the function is continuous at x = 0.

The expression |1-2x| represents a "V" shaped graph with the vertex at (1/2, 0), and it is continuous.

Thus, f(x) is continuous everywhere.

Step 1: Surface area of a sphere. The surface area S of a sphere is given by:

Step 2: Rate of change of surface area with respect to radius. Differentiate S with respect to r:

Step 3: Calculate at r = 4 cm. Substitute r = 4 into the derivative:

Step 1: Symmetry property of definite integrals. For the integral ∫^a_-a f(x) dx to equal zero, the function f(x) must satisfy:

f(−x) = −f(x) (odd function).

This is because, for odd functions:

Step 2: Check the condition. The given condition f(−x) = −f(x) matches the requirement for odd functions.

Rewriting the equation:

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

where P(x) = 1/(x log x) and Q(x) = 2/x

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

Since the dot product is not zero the vectors are not perpendicular.

To check if the vectors are parallel, express one as a scalar multiple of the other. If →a = k→b where k is a scalar, then the vectors are parallel.

Let's check: 2î − ĵ + k̂ = k(î − 2ĵ + k̂)

From the i components: 2 = k.

Now substitute k = 2: 2(î − 2ĵ + k̂) = 2î - 4ĵ + 2k̂. This doesn't equal 2î − ĵ + k̂, so they are not parallel.

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

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. Also, using the trigonometric identity sin²θ + cos²θ = 1, we can write:

and the identity cos 2θ= 2cos²θ -1. So cos 2α + cos 2β + cos 2γ = 2(cos²α + cos²β + cos²γ)-3 = 2(1)-3 = -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.

The probability of the union is given by:

Substitute the values:

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

The conditional probability is:

Step 1: Definition of skew-symmetric matrices. A matrix A is skew-symmetric if: A^T = −A. Similarly, B^T = −B for skew-symmetric matrix B.

Step 2: Transpose property of AB + BA. Consider the transpose of AB + BA:

(AB + BA)^T = (AB)^T + (BA)^T

Using the property (XY)^T = Y^TX^T:

(AB)^T = B^TA^T and (BA)^T = A^TB^T

Substituting A^T = −A and B^T = −B:

(AB)^T = (−B)(−A) = BA, (BA)^T = (−A)(−B) = AB.

Thus: (AB + BA)^T = BA + AB = AB + BA.

Step 3: Symmetry condition. Since (AB + BA)^T = AB + BA, the matrix AB + BA is symmetric.

Step 1: Analyze the assertion (A). The dot product of two vectors →u and →v is given by: →u · →v = |→u||→v| cos θ, where θ is the angle between the vectors.

For →a · (−→a):

→a · (−→a) = |→a||-→a| cos π = 1 · 1 · (−1) = −1. Hence, the assertion →a · (−→a) = −1 is true.

Step 2: Analyze the reason (R). The angle θ between →a and −→a satisfies cos θ = −1. This occurs when θ = π, not π/2. Thus, the reason "angle between →a and −→a is π/2" is false.

Step 3: Conclusion. The assertion is true, but the reason is false.

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

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.

Step 1: Dot product formula. The cosine of the angle between two vectors →u and →v is given by:

Step 2: Identify the vectors. Here, →u = →a and →v = 2→a + 3→b + 6→c.

Step 3: Compute the dot product. Using the linearity of the dot product and the fact that →a, →b, →c are mutually perpendicular unit vectors (→a · →b = →a · →c = →b · →c = 0 and →a · →a = 1):

Step 4: Compute the magnitude of →v. The magnitude of →v = 2→a + 3→b + 6→c is:

Step 5: Compute cos θ. Substitute into the formula:

Step 1: Simplify cot²(csc⁻¹3). Let θ = csc⁻¹3. Then csc θ = 3 ⇒ sin θ = 1/3.

Using the identity cot²θ = cos²θ / sin²θ and cos²θ = 1 − sin²θ:

Step 2: Simplify sin²(cos⁻¹(1/3)). Let φ = cos⁻¹(1/3). Then cos φ = 1/3.

Using the Pythagorean identity sin²φ = 1 − cos²φ:

Step 3: Combine the results. Add the two results:

1. Start with the given equation: x = e^y/2. Take the natural logarithm on both sides: log x = y/2.
2. Differentiate both sides with respect to x: (1/x) = (1/2)(dy/dx).
3. Rearrange for dy/dx: dy/dx = 2/x.
4. Since log x = y/2, then y = 2log x. Substitute this into the equation from step 3: dy/dx = 1/(x log x)
5. Since x = e^y/2, substitute x = e^{log x} into the above equation:
6. Differentiating y = 2 log x with respect to x gives: . Also from the first step, we have y=2 log x. Substituting this value of y in the result of Step 4 gives
7. The given solution simplifies to the above solution. The given solution appears to have conflated multiple different differentiation paths.

1. Continuity at x = 1:

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:

Find the right-hand derivative:

Since f′(1−) ≠ f′(1+), the function is not differentiable at x = 1.

Using the trigonometric identity: , we rewrite the integral as:

Simplify: sin(−x) = −sin(x), so the integral becomes:

1. Evaluate ∫ sin(5x) dx:

At the limits:

2. Evaluate ∫ sin(x) dx:

At the limits:

Substitute back:

We are given and F(1) = 0. To find F(x), we integrate:

Complete the square in the denominator:

So the integral becomes:

This is a standard integral of the form where u = x - 1. Therefore:

Now, apply the initial condition F(1) = 0:

Thus, C = 0, and the final answer is:

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:

Here, →r_A = î + 2ĵ − k̂, →r_B = −î + ĵ + k̂, m = 4, and n = 1. Substitute:

Simplify:

Thus: →r_C = −(5/3)î + (2/3)ĵ + (5/3)k̂.

2. Find |→AB| : |→BC|: First, calculate →AB = →r_B − →r_A:

→AB = (−î + ĵ + k̂) − (î + 2ĵ − k̂) = −2î − ĵ + 2k̂.

Magnitude: |→AB| = √((−2)² + (−1)² + 2²) = √(4 + 1 + 4) = √9 = 3.

Now calculate →BC = →r_C − →r_B:

Simplify:

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.

Step 1: Write the constraints as equations. 2x + 5y = 100 and 8x + 5y = 200.

Step 2: Plot the constraints. Find the intercepts of each line:

For 2x + 5y = 100:

• Let x = 0, then y = 100/5 = 20.
• Let y = 0, then x = 100/2 = 50. Thus, the line passes through points (0, 20) and (50, 0).

For 8x + 5y = 200:

• Let x = 0, then y = 200/5 = 40.
• Let y = 0, then x = 200/8 = 25. Thus, the line passes through points (0, 40) and (25, 0).

Step 3: Determine the feasible region. The inequalities 2x + 5y ≤ 100 and 8x + 5y ≤ 200 represent the regions below the respective lines. The non-negativity constraints x ≥ 0 and y ≥ 0 restrict the region to the first quadrant.

Step 4: Find the corner points of the feasible region. The corner points are (0, 20), (0, 40), (25,0), and the intersection of 2x+5y=100 and 8x+5y=200

Solve simultaneously:

2x + 5y = 100

8x + 5y = 200

Subtracting the first equation from the second: 6x = 100 ⇒ x = 50/3. Substitute x = 50/3 into 2x + 5y = 100: 2(50/3) + 5y = 100 ⇒ 100/3 + 5y = 100 ⇒ 5y = 200/3 ⇒ y = 40/3. The intersection point is (50/3, 40/3).

Step 5: Evaluate z = x + y at the corner points.

• At (0, 20): z = 0 + 20 = 20.
• At (0, 40): z = 0 + 40 = 40.
• At (50/3, 40/3): z = 50/3 + 40/3 = 90/3 = 30.
• At (25, 0): z = 25 + 0 = 25.

Step 6: Conclusion. The maximum value of z is 40, which occurs at (0, 40).

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:
3. Calculate the prior probabilities: From the given ratio 4 : 1 : 2:
4. Calculate the total probability P(E):

   Substitute the given probabilities:

5. Calculate P(P₃ | E):

   Substitute:

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.

Since x cos(p + y) + cos p sin(p + y) = 0, we can solve for x: x = -cos p tan(p + y). Substitute this value of x into the above derivative:

1. Rearrange the equation: x cos(p + y) = - cos p sin(p + y)
2. Differentiate both sides with respect to x, using the product and chain rules:
3. Equating the LHS and RHS and rearranging:

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: As x → 2⁻: LHL = −1 + a.

2. Right-hand limit (RHL): For x > 2: 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.

1. Find the derivative:
2. Critical points: The critical points are obtained by solving f′(x) = 0: x − log x = 0 ⇒ x = e (approximately 2.718).
3. Sign of f′(x):
   • For x ∈ (0, e), x − log x > 0, so f′(x) > 0: f(x) is increasing.
   • For x ∈ (e, ∞), x − 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:
2. Critical points: Set f′(x) = 0: (1/2) − (2/x²) = 0 ⇒ x² = 4 ⇒ x = ±2. Since the interval is [1,2], we consider x=2.
3. Evaluate f(x) at endpoints and critical point:
   • At x = 1: f(1) = (1/2) + (2/1) = 2.5.
   • At x = 2: f(2) = (2/2) + (2/2) = 2.
4. Conclusion: The absolute maximum value is 2.5 at x = 1, and the absolute minimum value is 2 at x = 2.

Step 1: Simplify the given integral. Let √x = t, so x = t² and dx = 2t dt. Substitute x = t² into the integral:

Step 2 & 3: Rewrite and perform Partial Fraction Decomposition:

Step 4 & 5: Simplify and Integrate:

Step 6: Back-substitute t = √x:

1. Simplify the trigonometric expression: Using the identity 1 + cos 2x = 2 cos²x and sin 2x = 2 sin x cos x, rewrite the expression:
2. Rewrite the integral:
3. Solve the integral: We can use integration by parts, or recognize that the integrand is of the form e^x(f(x) + f′(x)), where f(x) = tan x. The integral of this form is simply e^xf(x) + C. Therefore:

1. Simplify the denominator: Use the identity sin x + cos x = √2 sin(x + π/4).
2. Rewrite the integral:
3. Substitute: Let u = x + π/4, so du = dx. Change the limits: x = 0 → u = π/4, x = π/4 → u = π/2.
4. The integral becomes:
5. Integrate csc u:
6. Evaluate at the limits:
7. Substitute values: csc(π/2) = 1, cot(π/2) = 0, csc(π/4) = √2, cot(π/4) = 1.
8. Simplify: ln|1 - 0| = ln(1) = 0, ln|√2 - 1| = ln(√2 - 1).
9. Final Result:

1. Point of intersection: Let (x − 1)/1 = (y − 2)/2 = (z − 2)/3 = t₁ and (x − 1)/0 = (y − 3)/(−3) = (z − 7)/2 = t₂.
2. From the first line: x = 1 + t₁, y = 2 + 2t₁, z = 2 + 3t₁.
3. From the second line: x = 1, y = 3 − 3t₂, z = 7 + 2t₂.
4. Equate x, y, z:
   • From x: 1 + t₁ = 1 ⇒ t₁ = 0.
   • Substituting t₁ = 0 into the equations for the first line gives x=1, y=2, z=2. Therefore, the point of intersection is (1, 2, 2).
5. Direction vectors: →d₁ = ⟨1, 2, 3⟩ and →d₂ = ⟨0, −3, 2⟩.
6. Direction of the required line: The required line's direction vector →d is perpendicular to both →d₁ and →d₂, so it's given by their cross product:
7. Equation of the required line: The line passes through (1, 2, 2) with direction vector ⟨13, −2, −3⟩:

1. Vector representation of AB: The position vectors of A and B are →A = ⟨−1, 2, 1⟩ and →B = ⟨1, −2, 5⟩. The vector →AB is →AB = →B − →A = ⟨2, −4, 4⟩.
2. Direction vector of CD: The direction vector of the line CD is →d_CD = ⟨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: Here, →r₁ = ⟨−1, 2, 1⟩, →r₂ = ⟨4, −7, 8⟩, →d₁ = →AB = ⟨2, −4, 4⟩, and →d₂ = ⟨1, −2, 2⟩.

   →r₂ − →r₁ = ⟨5, −9, 7⟩.

   Since →d₁ = 2→d₂ the lines are parallel, so the cross product →d₁ × →d₂=⟨0,0,0⟩. The distance formula is not applicable when the lines are parallel. Instead, find a vector perpendicular to both lines. One such vector is →n = ⟨2, 1, 0⟩.

   Let the point C be on the line CD: C(4+s, -7-2s, 8+2s). Since ABCD is a parallelogram, the midpoint of AC must be the same as the midpoint of BD.

   Midpoint of AC:

   Midpoint of BD:

   From the line equation, we can parameterize D as (4 + t, -7 - 2t, 8 + 2t). The midpoint of BD is then

   Equating the midpoints of AC and BD, we get s = 2 and t = -2. This means C=(6, -11, 12) and D = (2, -3, 4). →CD = ⟨-4, 8, -8⟩.

   The area of the parallelogram is the magnitude of the cross product of →AB and →AC.

   →AB = ⟨2, -4, 4⟩ and →AC = ⟨7, -13, 11⟩

   →AB × →AC = ⟨12, -6, 2⟩.

   Area of parallelogram = |→AB × →AC| = = =

   The distance between the parallel lines AB and CD is:

   →r₂ - →r₁ = (4, -7, 8)-(-1, 2, 1) = (5, -9, 7)

   →n = ⟨2, -4, 4⟩ × ⟨1, -2, 2⟩ = ⟨0, 0, 0⟩. As the cross product is 0, the vectors are parallel. Since the lines are parallel, choose →n = (2,1,0) which is perpendicular to the direction vector ⟨1,-2,2⟩.

Step 1: To find a such that f is onto. For f(x) to be onto, every y ∈ B must have a pre-image x ∈ A. Given , rearrange to express x in terms of y:

For x to be valid in A = R − {3}, we must have y − 1 ≠ 0, so y ≠ 1. Thus, the range of f(x) excludes a = 1, and B = R − {1}.

Step 2: Check if the function is one-one. A function f(x) is one-one if f(x₁) = f(x₂) implies x₁ = x₂.

Suppose f(x₁) = f(x₂):

Cross-multiply and simplify:

This simplifies to x₁ = x₂. Thus, f(x) is one-one.

Step 3: Conclusion. The function f(x) = (x-2)/(x-3) is one-one, and for f(x) to be onto, a=1.

1. Find the derivative:
2. Condition for local maximum: At x = 1, f′(1) = 0, so 4(1)³ − 124(1) + a = 0 ⇒ a = 120.
3. Update the function: f(x) = x⁴ − 62x² + 120x + 9.
4. Find other critical points: Set f′(x) = 0: 4x³ − 124x + 120 = 0 ⇒ x³ − 31x + 30 = 0. Factoring gives (x-1)(x+6)(x-5) = 0. So, x = 1, x = -6, and x=5 are the critical points.
5. Second derivative test: f″(x) = 12x² − 124.
   • At x = 1: f″(1) = 12 − 124 = −112 < 0 (local maximum).
   • At x = −6: f″(−6) = 12(36) - 124 = 308 > 0 (local minimum).
   • At x = 5: f″(5) = 12(25) - 124 = 176 > 0 (local minimum).
6. Conclusion: Local maximum at x = 1 with a = 120. Local minimums at x = -6 and x = 5.

1. Define dimensions: Let the length be 2r (circumference of the cylinder base) and the width be h (height). Perimeter: 2(2r + h) = 300 ⇒ h = 150 − 2r.
2. Volume of the cylinder:
3. Maximize V: Differentiate V with respect to r: Set dV/dr = 0: 300r − 6r² = 0 ⇒ 6r(50 − r) = 0. Thus, r = 0 or r = 50. Discard r = 0.
4. Second derivative test: At r = 50: d²V/dr² = π(300 − 600) = −300π < 0. Hence, V is maximum at r = 50.
5. Find h: h = 150 − 2r = 150 − 2(50) = 50.

Final Answer: The dimensions are 2r = 100 cm (rolled side) and h = 50 cm.

Step 1: The curve y = √(4 − x²) is the upper half of a circle with radius 2 centered at the origin.

Step 2: Setting up the integral:

Step 3 & 4: Substitution and Simplification: Let x = 2 sin θ, so dx = 2 cos θ dθ. Then √(4 − x²) = 2 cos θ. The limits of integration change: x = -1 ⇒ θ = -π/6, x = 1 ⇒ θ = π/6.

The integral becomes:

Step 5: Solve the integral:

(i) 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:

(ii) Check whether the system of matrix equations is consistent:

To check consistency, calculate the determinant of the coefficient matrix:

The determinant of A is: det(A) = (1)(4000) − (1)(3000) = 4000 − 3000 = 1000.

Since det(A) ≠ 0, the system is consistent and has a unique solution.

(iii)(a) Find the number of scholarships of each kind:

Solve the system using the inverse of A:

The inverse of A is:

Substitute A⁻¹ and B:

Multiply and simplify:

Thus: x = 20, y = 30.

(iii)(b) If the scholarship amounts are interchanged:

If the amounts are interchanged, the equation becomes 4000x + 3000y = Total Expenditure.

Substitute x = 20 and y = 30:

Total Expenditure = 4000(20) + 3000(30) = 80000 + 90000 = ₹170,000.

(i) Probability Distribution Table:

(ii) Find the value of k:

The total probability must equal 1: Σ⁶_x=1 P(X = x) = 1.

Substitute the probabilities: k + 4k + 9k + 8k + 10k + 12k = 1 ⇒ 44k = 1 ⇒ k = 1/44.

(iii)(a) Mean number of hours spent:

The mean is given by: μ = E(X) = Σ⁶_x=1 x · P(X = x).

Substitute P(X = x) and k=1/44: E(X) = (1/44)(1 + 8 + 27 + 32 + 50 + 72) = (1/44)(190) = 95/22.

(iii)(b) Find P(1 < X < 6):

P(1 < X < 6) = P(X = 2) + P(X = 3) + P(X = 4) + P(X = 5) = 4k + 9k + 8k + 10k = 31k = 31/44.

(i) General solution:

The given differential equation is:

Separate variables and integrate:

Rewrite in exponential form:

Let e^C = P₀ (initial population):

(ii) Find the value of k:

At t = 0, P = 1000: 1000 = P₀e⁰ ⇒ P₀ = 1000.

At t = 1, P = 2000: 2000 = 1000e^k ⇒ e^k = 2 ⇒ k = ln(2).