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

• Step 1: Differentiating y = cos⁻¹(e^x).
  We know that the derivative of cos⁻¹(u) is:
  d/dx [cos⁻¹(u)] = − 1 / √(1 − u²) · du/dx.
  Here, u = e^x. Therefore:
  dy/dx = − 1 / √(1 − (e^x)²) · d/dx (e^x).
• Step 2: Simplifying the expression.
  The derivative of e^x is e^x. Substituting:
  dy/dx = − e^x / √(1 − e^2x).
• Step 3: Simplify the denominator.
  In the denominator:
  1 − e^2x = e^−2x − 1.
  This can be rewritten as:
  √(1 − e^2x) = √(e^−2x − 1).
• Step 4: Final simplification.
  Combine the terms:
  dy/dx = − e^x / √(e^−2x − 1).
• Step 5: Adjust for the correct format.
  Using e^x · e^−x = 1, the final result becomes:
  dy/dx = − 1 / √(e^−2x − 1).

• Step 1: Understanding the definitions of order and degree.
  - The order of a differential equation is the highest derivative present in the equation.
  - The degree of a differential equation is defined only when all the derivatives are in polynomial form (i.e., no logarithms, trigonometric functions, etc.).
• Step 2: Identifying the order of the given equation.
  The given differential equation is:
  y''² + log(y') = x⁵.
  - The highest derivative present is y'', so the order is 2.
• Step 3: Checking the degree of the given equation.
  The equation contains log(y'), which is a non-polynomial function of the derivative y'. Therefore, the degree of the equation is not defined.

• Step 1: Cross product to find the perpendicular vector.
  The cross product of two vectors Ā = î + k̂ and B̄ = î - k̂ gives a vector perpendicular to both:
  | î   ĵ   k̂ |
  | 1   0   1 |
  | 1   0   -1 |
  
  Expanding the determinant:
  Ā × B̄ = î | 0 1 | - ĵ | 1 1 | + k̂ | 1 0 |
  | 0 -1 | | 1 -1 | | 1 0 |
• Step 2: Simplify the minors.
  - First minor:
  | 0 1 | = 0.
  | 0 -1 |
• Step 3: Conclusion.
  The resulting vector is along the ĵ direction. The unit vector is ĵ.

• Step 1: Parametrize the line.
  The given line equation is:
  (x−1)/2 = −y = (2z+1)/6.
  Let t be the parameter. From each equation:
  (x − 1)/2 = t ⇒ x − 1 = 2t ⇒ x = 2t + 1,
  −y = t ⇒ y = −t,
  (2z + 1)/6 = t ⇒ 2z + 1 = 6t ⇒ 2z = 6t − 1 ⇒ z = 3t − 1/2.
• Step 2: Extract direction ratios.
  The coefficients of t in x = 2t + 1, y = −t, and z = 3t − 1/2 are:
  2, −1, 3.
• Step 3: Conclusion.
  The direction ratios of the line are: 2, −1, 3.

• Step 1: Understanding the problem.
  The given matrix is:
  A = [tan x 1; −1 tan x].
  The transpose of A, denoted A', is:
  A' = [tan x −1; 1 tan x].
  The equation A + A' = 2√3I implies:
  [tan x 1; −1 tan x] + [tan x −1; 1 tan x] = 2√3[1 0; 0 1].
• Step 2: Simplifying A + A'.
  Adding A and A':
  [tan x + tan x 1−1; −1+1 tan x + tan x] = [2 tan x 0; 0 2 tan x].
• Step 3: Solve for x.
  The equation becomes:
  2 tan x = 2√3 ⇒ tan x = √3 ⇒ x = π/3.

• Step 1: Use the direction cosine property.
  The angles α, β, γ made by a line with the positive directions of the x-, y-, and z-axes satisfy the equation for direction cosines:
  cos²α + cos²β + cos²γ = 1.
  Here: α = 30°, β = 120°, γ = ?.
• Step 2: Compute cos α and cos β.
  cos α = cos 30° = √3 / 2,
  cos β = cos 120° = −1 / 2.
  Thus:
  cos²α = ( √3 / 2 )² = 3/4,
  cos²β = (−1/2)² = 1/4.
• Step 3: Solve for cos²γ.
  Using the equation for direction cosines:
  cos²α + cos²β + cos²γ = 1.
  3/4 + 1/4 + cos²γ = 1
  ⇒ cos²γ = 1 − (3/4 + 1/4).
• Step 4: Determine γ.
  Simplify:
  cos²γ = 1 − 1 = 0.
  If cos²γ = 0, then:
  cos γ = 0.
  The angle γ corresponding to cos γ = 0 is:
  γ = 90°.
• Step 5: Conclusion.
  The angle which the line makes with the positive direction of the z-axis is:
  90°.

• Step 1: Understanding the properties of scalar matrices.
  In a scalar matrix, all off-diagonal elements are zero, and diagonal elements are equal.
• Step 2: Calculate the diagonal elements.
  Let the diagonal elements be x, then the sum of all elements of the 3 × 3 scalar matrix is:
  3x = 9 ⇒ x = 3.
• Step 3: Compute the product.
  Since all off-diagonal elements are zero, the product of all elements is:
  0 (as multiplication by any zero results in zero).
• Step 4: Conclusion.
  The product of all elements is: 0.

• Step 1: Analyze the function for one-one property.
  The given function is:
  f(x) = 9x² + 6x − 5.
  Differentiating f(x) to check its monotonicity:
  f'(x) = 18x + 6.
  For x ≥ 0 (as x ∈ R₊), f'(x) > 0. This implies that f(x) is strictly increasing for x ≥ 0, and therefore, f(x) is one-one.
• Step 2: Analyze the function for onto property.
  The range of f(x) can be determined:
  f(x) = 9x² + 6x − 5,   x ≥ 0.
  At x = 0, f(0) = 9(0)² + 6(0) − 5 = −5.
  Since f(x) is strictly increasing for x ≥ 0, the function attains all values greater than or equal to −5. Therefore, the range of f(x) is [−5, ∞), which matches the codomain of f. Hence, f(x) is onto.
• Step 3: Conclude the nature of f.
  Since f(x) is both one-one and onto, f(x) is bijective.

• Step 1: Expand the determinant.
  The determinant of the given matrix is:
  |−a   b   c|
  |  a   −b   c|
  |  a   b   −c|.
  Expanding along the first row:
  −a |−b   c| − b |a   c| + c |a   −b|.
          |b   −c|       |a   −c|       |a   b|
• Step 2: Simplify the minors.
  - First minor:
  |−b   c|
  |  b   −c| = (−b)(−c) − (b)(c) = bc − bc = −2bc.
  - Second minor:
  |a   c|
  |a   −c| = (a)(−c) − (a)(c) = −ac − ac = −2ac.
  - Third minor:
  |a   −b|
  |a   b| = (a)(b) − (a)(−b) = ab + ab = 2ab.
• Step 3: Substitute back and simplify.
  −a(−2bc) − b(−2ac) + c(2ab) = 2abc + 2abc + 2abc = 4abc.
• Step 4: Conclusion.
  The value of k is: 4.

• Step 1: Identify the points of possible discontinuity.
  The given piecewise function has transitions at x = −3 and x = 3. These are the points where the function could be discontinuous.
• Step 2: Check continuity at x = −3.
  For x ≤ −3, f(x) = |x| + 3, and for −3 < x < 3, f(x) = −2x.
  Evaluate left-hand limit: f(−3) = |−3| + 3 = 6.
  Evaluate right-hand limit: f(−3) = −2(−3) = 6.
  Since both limits are equal, f(x) is continuous at x = −3.
• Step 3: Check continuity at x = 3.
  For −3 < x < 3, f(x) = −2x, and for x ≥ 3, f(x) = 6x + 2.
  Evaluate left-hand limit: f(3) = −2(3) = −6.
  Evaluate right-hand limit: f(3) = 6(3) + 2 = 20.
  Since the left-hand limit ≠ right-hand limit, f(x) is discontinuous at x = 3.
• Step 4: Conclusion.
  The number of points of discontinuity is 1.

• Step 1: Find the derivative of f(x).
  The derivative of f(x) is:
  f'(x) = 3x² − 6x + 12.
• Step 2: Analyze f'(x).
  Simplify f'(x):
  f'(x) = 3(x² − 2x + 4).
  The quadratic x² − 2x + 4 has a discriminant:
  Δ = (−2)² − 4(1)(4) = 4 − 16 = −12.
  Since the discriminant is negative, x² − 2x + 4 is always positive. Hence, f'(x) > 0 for all x ∈ R.
• Step 3: Conclusion about monotonicity.
  Since f'(x) > 0 for all x, the function f(x) is strictly increasing on R.
• Step 4: Final Answer.
  The function f(x) is: strictly increasing on R.

• Step 1: Simplify the given expression.
  We are tasked to find the anti-derivative of:
  √(1 + sin 2x).
  Using the trigonometric identity:
  sin 2x = 2 sin x cos x,
  the expression becomes:
  √(1 + 2 sin x cos x).
  This simplifies to:
  √((cos x + sin x)²).
• Step 2: Determine the anti-derivative.
  For x ∈ [0, π/4], cos x + sin x ≥ 0. Thus:
  √((cos x + sin x)²) = cos x + sin x.
  The anti-derivative of cos x + sin x is:
  ∫(cos x + sin x) dx = sin x − cos x + C.
• Step 3: Conclusion.
  The required anti-derivative is: − cos x + sin x.

• Step 1: Definition of a homogeneous function.
  A function F(x, y) is homogeneous of degree n if:
  F(tx, ty) = tⁿ F(x, y)    for all t > 0.
• Step 2: Analyze each option.
  - (A) cos x − sin (y/x): This term involves cos x, which depends only on x and not on the ratio y/x. Thus, F(x, y) is not homogeneous.
  - (B) y/x: This is a homogeneous function of degree 0, as it depends only on the ratio y/x.
  - (C) (x² + y²)/xy: Simplifying, this becomes:
  (x²/xy) + (y²/xy) = x/y + y/x.
  Each term depends only on x/y or y/x, so this is homogeneous.
  - (D) cos² (x/y): This depends only on the ratio x/y, so it is homogeneous.
• Step 3: Conclusion.
  Only (A) does not satisfy the condition for homogeneity.

• Step 1: Use the dot product definition.
  The dot product ā ⋅ b̄ is given by:
  ā ⋅ b̄ = |ā| |b̄| cos θ,
  where θ is the angle between ā and b̄.
• Step 2: Analyze the range of cos θ.
  Since −1 ≤ cos θ ≤ 1, it follows that:
  −|ā| |b̄| ≤ ā ⋅ b̄ ≤ |ā| |b̄|.
• Step 3: Conclusion.
  The correct answer is: ā ⋅ b̄ ≤ |ā| |b̄|.

• Step 1: Define the x-axis.
  The x-axis is defined by all points of the form (x,0,0).
• Step 2: Foot of the perpendicular.
  The foot of the perpendicular from (0,1,2) onto the x-axis is the closest point on the x-axis.
  Since x = 0, the coordinates are:
  (0,0,0).
• Step 3: Conclusion.
  The foot of the perpendicular is: (0,0,0).

• Step 1: Definition of feasible region.
  The feasible region is the common region determined by all constraints (inequalities) in a linear programming problem. It represents all possible solutions.
• Step 2: Analyze the options.
  - (A) Unbounded region: May or may not occur depending on constraints.
  - (B) Optimal region: Refers to the solution, not the region itself.
  - (C) Bounded region: A feasible region may or may not be bounded.
  - (D) Feasible region: Always the common region defined by constraints.
• Step 3: Conclusion.
  The correct answer is: a feasible region.

• Step 1: Understanding conditional probability.
  The conditional probability formula states:
  P(A|B) = P(A ∩ B) / P(B), provided P(B) > 0.
• Step 2: Apply the formula for P(S|E).
  Given that S is the sample space, we know that for any event E:
  S ∩ E = E, since the event E is a subset of the sample space S.
  Thus, P(S|E) = P(S ∩ E) / P(E) = P(E) / P(E) = 1.
• Step 3: Conclusion.
  The probability of the entire sample space given any event E with nonzero probability is always 1.

• Step 1: Compute the elements of the matrix A.
  The elements of the matrix A are defined as:
  a_ij = i − 3j.
  For a 3 × 3 matrix:

          A = | a11  a12  a13 |
              | a21  a22  a23 |
              | a31  a32  a33 |
          

  
  where:
  a₁₁ = 1 − 3(1) = −2,   a₁₂ = 1 − 3(2) = −5,   a₁₃ = 1 − 3(3) = −8,
  a₂₁ = 2 − 3(1) = −1,   a₂₂ = 2 − 3(2) = −4,   a₂₃ = 2 − 3(3) = −7,
  a₃₁ = 3 − 3(1) = 0,   a₃₂ = 3 − 3(2) = −3,   a₃₃ = 3 − 3(3) = −6.
• Step 2: Verify each option.
  - (A) a₁₁ < 0: Here, a₁₁ = −2, which is indeed less than 0. This statement is true.
  - (B) a₁₂ + a₂₁ = −6: Here, a₁₂ = −5 and a₂₁ = −1, so:
  a₁₂ + a₂₁ = −5 + (−1) = −6. This statement is true.
  - (C) a₁₃ > a₃₁: Here, a₁₃ = −8 and a₃₁ = 0. Clearly, −8 > 0 is false.
  - (D) a₃₁ = 0: From the calculations, a₃₁ = 0. This statement is true.
• Step 3: Conclusion.
  The only false statement is: (C) a₁₃ > a₃₁.

• Step 1: Analyze Assertion (A).
  If A is symmetric, A^T = A. However, in general, B^TAB is not skew-symmetric.
  Thus, the assertion is false.
• Step 2: Analyze Reason (R).
  A matrix P is skew-symmetric if P^T = −P. This definition is correct, making the reason true.
• Step 3: Conclusion.
  The correct answer is: (D) Assertion (A) is false, but Reason (R) is true.

• Step 1: Analyze the assertion.
  The expression (b̄ ⋅ c̄) ā involves the dot product b̄ ⋅ c̄, which is a scalar. However, this scalar is multiplied by the vector ā, resulting in a vector, not a scalar. Hence, the assertion that (b̄ ⋅ c̄) ā being a scalar is false.
• Step 2: Analyze the reason.
  The statement "Dot product of two vectors is a scalar quantity" is mathematically correct. Thus, the reason is true.
• Step 3: Conclusion.
  While the reason is true, it does not justify the incorrect assertion. Hence, the correct answer is:
  (D) Assertion (A) is false, but Reason (R) is true.

• Step 1: Find the derivative of f(x).
  The given function is:
  f(x) = x² − 6x + 3.
  The derivative of f(x) is:
  f'(x) = 2x − 6.
• Step 2: Analyze the sign of f'(x) on [4,6].
  Substitute values of x in f'(x):
  • At x = 4: f'(4) = 2(4) − 6 = 8 − 6 = 2 > 0.
  • At x = 6: f'(6) = 2(6) − 6 = 12 − 6 = 6 > 0.
  Since f'(x) > 0 for all x ∈ [4,6], the function is increasing in this interval.
• Step 3: Conclusion.
  From the above analysis, f(x) is increasing over [4,6].

• Step 1: Evaluate each term separately.
  1. sin⁻¹(−1/2):
  From the inverse sine definition:
  sin⁻¹(−1/2) = −π/6.
• 2. cos⁻¹(−√3/2):
  From the inverse cosine definition:
  cos⁻¹(−√3/2) = π − π/6 = 5π/6.
• 3. cot⁻¹(tan 4π/3):
  The cotangent and tangent functions are related as:
  cot⁻¹(tan(4π/3)) = cot⁻¹(tan(π + π/3)) = cot⁻¹(tan(π/3)).
  Since cot⁻¹(tan x) = x for principal values:
  cot⁻¹(tan 4π/3) = π/3.

Step 2: Add the values.

Combine the results:

sin⁻¹(−1/2) + cos⁻¹(−√3/2) + cot⁻¹(tan 4π/3) = −π/6 + 5π/6 + π/3.

Simplify:

−π/6 + 5π/6 + 2π/6 = 5π/6.

• Step 1: Find the domain of f(x).
  The function cos⁻¹(z) is defined when z ∈ [−1,1]. Here:
  1 − x² ∈ [−1,1].
  Simplify:
  −1 ≤ 1 − x² ≤ 1.
  Rearrange:
  0 ≤ x² ≤ 2.
  Since x² ≥ 0, the domain of f(x) is:
  x ∈ [−√2, √2].
• Step 2: Find the range of f(x).
  The range of cos⁻¹(z) is [0, π], so the range of f(x) = cos⁻¹(1 − x²) is also:
  [0, π].

• Step 1: Analyze the function f(x) = |tan 2x|.
  The absolute value function requires splitting into cases:
  f(x) =
  { tan 2x, if tan 2x ≥ 0
  { −tan 2x, if tan 2x < 0
• Step 2: Differentiate f(x) piecewise.
  For tan 2x ≥ 0:
  f'(x) = d/dx [tan 2x] = 2 sec² 2x.
  For tan 2x < 0:
  f'(x) = d/dx [−tan 2x] = −2 sec² 2x.
• Step 3: Evaluate tan 2x at x = π/3.
  At x = π/3:
  2x = 2π/3, tan 2π/3 = −tan π/3 = −√3.
  Since tan 2x < 0, we use:
  f'(x) = −2 sec² 2x.
• Step 4: Compute sec² 2x.
  sec² 2x = 1 + tan² 2x = 1 + (−√3)² = 1 + 3 = 4.
  Substitute:
  f'(x) = −2 ⋅ 4 = −8.

• Step 1: Express csc(cot⁻¹ x) in terms of x.
  Let θ = cot⁻¹ x, so:
  cot θ = x and csc θ = √(1 + cot² θ) = √(1 + x²).
  Thus:
  y = csc(cot⁻¹ x) = √(1 + x²).
• Step 2: Differentiate y with respect to x.
  dy/dx = d/dx (√(1 + x²)).
  Using the chain rule:
  dy/dx = (1 / (2√(1 + x²))) ⋅ 2x = x / √(1 + x²).
• Step 3: Verify the given expression.
  Substituting y = √(1 + x²) and dy/dx = x / √(1 + x²) into the equation:
  √(1 + x²) ⋅ (x / √(1 + x²)) − x = x − x = 0.

• Step 1: Differentiate f(x).
  The given function is:
  f(x) = x + 1/x.
  Differentiate f(x) to find critical points:
  f'(x) = 1 − 1/x².
• Step 2: Solve f'(x) = 0.
  Set f'(x) = 0:
  1 − 1/x² = 0 ⇒ 1/x² = 1 ⇒ x² = 1.
  Thus, x = ±1.
• Step 3: Determine the nature of critical points.
  Differentiate f'(x) to find f''(x):
  f''(x) = 2/x³.
  - At x = 1: f''(1) = 2/1³ = 2 (positive, so x = 1 is a local minimum).
  - At x = −1: f''(−1) = 2/(−1)³ = −2 (negative, so x = −1 is a local maximum).
• Step 4: Compute M and m.
  - At x = 1: f(1) = 1 + 1/1 = 2 (local minimum m).
  - At x = −1: f(−1) = −1 + 1/−1 = −2 (local maximum M).

• Step 1: Simplify the integrand.
  Let:
  I = ∫ (e^4x − 1) / (e^4x + 1) dx.
  Divide the numerator and denominator by e^2x:
  (e^4x − 1) / (e^4x + 1) = (e^2x − e^−2x) / (e^2x + e^−2x).
• Step 2: Use the hyperbolic identity.
  The numerator and denominator resemble hyperbolic sine and cosine:
  e^2x − e^−2x = 2 sinh(2x),
  e^2x + e^−2x = 2 cosh(2x).
  Thus:
  (e^4x − 1) / (e^4x + 1) = sinh(2x) / cosh(2x) = tanh(2x).
• Step 3: Solve the integral of tanh(2x).
  The derivative of log(cosh(2x)) is 2 tanh(2x). Thus:
  ∫ tanh(2x) dx = ½ log(cosh(2x)).
• Step 4: Express in terms of exponentials.
  Using the identity cosh(2x) = (e^2x + e^−2x) / 2, the integral becomes:
  I = ½ log ( (e^2x + e^−2x) / 2 ).
  Simplify by absorbing the constant ½ into C:
  I = ½ log |e^2x + e^−2x| + C.

• Step 1: Define the values of X.
  The possible absolute differences are:
  X = {0, 1, 2, 3, 4, 5}.
• Step 2: Count the outcomes for each X.
  For X = 0: Both dice show the same number. There are 6 outcomes:
  (1,1), (2,2), (3,3), (4,4), (5,5), (6,6).
  For X = 1: The numbers differ by 1. Outcomes are:
  (1,2), (2,1), (2,3), (3,2), …, (5,6), (6,5).
  This gives 2 × 5 = 10 outcomes.
  Repeat similar counting for X = 2, 3, 4, 5.
• Step 3: Calculate probabilities.
  The total number of outcomes is 6 × 6 = 36. Probabilities are:
  P(X = k) = (Number of favorable outcomes for X = k) / 36.
• Step 4: Write the probability distribution.
  P(X = 0) = 6/36,
  P(X = 1) = 10/36,
  P(X = 2) = 8/36, …

• Step 1: Use integration by parts.
  Let:
  u = log(x² − 1) and dv = x² dx.
  Then:
  du = (1 / (x² − 1)) ⋅ 2x dx and v = x³ / 3.
• Using the integration by parts formula:
  ∫ u dv = uv − ∫ v du,
  we have:
  ∫ x² log(x² − 1) dx = (x³ / 3) log(x² − 1) − ∫ (x³ / 3) ⋅ (2x / (x² − 1)) dx.
• Step 2: Simplify the remaining integral.
  Simplify the second term:
  ∫ (x³ / 3) ⋅ (2x / (x² − 1)) dx = (2/3) ∫ (x⁴ / (x² − 1)) dx.
  Rewrite x⁴ as x²(x² − 1) + x²:
  x⁴ / (x² − 1) = x² + x² / (x² − 1).
  Thus, the integral becomes:
  (2/3) ∫ (x² + x² / (x² − 1)) dx.
• Step 3: Evaluate each term.
  1. For (2/3) ∫ x² dx:
  ∫ x² dx = x³ / 3,
  (2/3) ∫ x² dx = (2/9) x³.
• 2. For (2/3) ∫ (x² / (x² − 1)) dx, split the fraction:
  x² / (x² − 1) = 1 + 1 / (x² − 1).
  Thus:
  ∫ (x² / (x² − 1)) dx = ∫ 1 dx + ∫ (1 / (x² − 1)) dx.
  The first term is:
  ∫ 1 dx = x.
  The second term is:
  ∫ (1 / (x² − 1)) dx = (1/2) log |x − 1| − (1/2) log |x + 1| = (1/2) log |(x − 1)/(x + 1)|.
  Combine:
  ∫ (x² / (x² − 1)) dx = x + (1/2) log |(x − 1)/(x + 1)|.
• Step 4: Combine all results.
  Substituting back into the original expression:
  ∫ x² log(x² − 1) dx = (x³ / 3) log(x² − 1) − (2/9) x³ − (2/3) (x + (1/2) log |(x − 1)/(x + 1)|).

• Step 1: Differentiate y = (log x)² once.
  Using the chain rule:
  y = (log x)² ⇒ y' = 2(log x) * (1/x) = (2 log x)/x.
• Step 2: Differentiate y' to find y''.
  Using the quotient rule:
  y' = (2 log x)/x ⇒ y'' = d/dx [(2 log x)/x].
  Let u = 2 log x and v = x. Then:
  y'' = (v*u' - u*v')/v² = (x * (2/x) - 2 log x * 1) / x² = (2 - 2 log x)/x².
• Step 3: Verify the given equation.
  Substitute y' = (2 log x)/x and y'' = (2 - 2 log x)/x² into x²y'' + xy':
  x²y'' = x² * (2 - 2 log x)/x² = 2 - 2 log x,
  xy' = x * (2 log x)/x = 2 log x.
  Add:
  x²y'' + xy' = (2 - 2 log x) + 2 log x = 2.

• Step 1: Differentiate y = sin(tan^-1 e^x).
  Using the chain rule:
  y = sin(tan^-1 e^x) ⇒ dy/dx = cos(tan^-1 e^x) * d/dx (tan^-1 e^x).
• Step 2: Differentiate tan^-1 e^x.
  The derivative of tan^-1u is 1/(1+u²), so: d/dx (tan^-1 e^x) = e^x / (1+(e^x)²) = e^x / (1+e^2x).
• Step 3: Simplify dy/dx.
  Substitute:
  dy/dx = cos(tan^-1 e^x) * e^x/(1+e^2x).
  Using the trigonometric identity for cos(tan^-1 u) = 1/√(1+u²):
  cos(tan^-1 e^x) = 1/√(1+(e^x)²) = 1/√(1+e^2x).
  Thus: dy/dx = (1/√(1+e^2x)) * (e^x/(1+e^2x)) = e^x / (1+e^2x)^3/2.
• Step 4: Evaluate at x = 0.
  At x = 0:
  e^x = e⁰ = 1, 1 + e^2x = 1 + e⁰ = 1+1 = 2.
  Substitute: dy/dx = 1/ (2)^3/2 = 1/(2√2)

• Step 1: Differentiate y = (log x)² once.
  Using the chain rule:
  y = (log x)² ⇒ y' = 2(log x) * (1/x) = (2 log x)/x.
• Step 2: Differentiate y' to find y''.
  Using the quotient rule:
  y' = (2 log x)/x ⇒ y'' = d/dx [(2 log x)/x].
  Let u = 2 log x and v = x. Then:
  y'' = (v*u' - u*v')/v² = (x * (2/x) - 2 log x * 1) / x² = (2 - 2 log x)/x².
• Step 3: Verify the given equation.
  Substitute y' = (2 log x)/x and y'' = (2 - 2 log x)/x² into x²y'' + xy':
  x²y'' = x² * (2 - 2 log x)/x² = 2 - 2 log x,
  xy' = x * (2 log x)/x = 2 log x.
  Add:
  x²y'' + xy' = (2 - 2 log x) + 2 log x = 2.

• Step 1: Differentiate y = sin(tan^-1 e^x).
  Using the chain rule:
  y = sin(tan^-1 e^x) ⇒ dy/dx = cos(tan^-1 e^x) * d/dx (tan^-1 e^x).
• Step 2: Differentiate tan^-1 e^x.
  The derivative of tan^-1u is 1/(1+u²), so: d/dx (tan^-1 e^x) = e^x / (1+(e^x)²) = e^x / (1+e^2x).
• Step 3: Simplify dy/dx.
  Substitute:
  dy/dx = cos(tan^-1 e^x) * e^x/(1+e^2x).
  Using the trigonometric identity for cos(tan^-1 u) = 1/√(1+u²):
  cos(tan^-1 e^x) = 1/√(1+(e^x)²) = 1/√(1+e^2x).
  Thus: dy/dx = (1/√(1+e^2x)) * (e^x/(1+e^2x)) = e^x / (1+e^2x)^3/2.
• Step 4: Evaluate at x = 0.
  At x = 0:
  e^x = e⁰ = 1, 1 + e^2x = 1 + e⁰ = 1+1 = 2.
  Substitute: dy/dx = 1/ (2)^3/2 = 1/(2√2)

• Step 1: Rearrange the given equation.
  The given equation is: 2xy + y² - 2x²(dy/dx) = 0. Rearrange to express dy/dx: dy/dx = (2xy + y²) / (2x²).
• Step 2: Separate variables.
  Divide through by y² (assuming y ≠ 0): (1/y²) dy = (2x + y) / (2x²) dx.
• Step 3: Integrate both sides.
  - For the left-hand side: ∫ (1/y²) dy = ∫ y^-2 dy = -1/y.
  - For the right-hand side: Split the fraction: ∫ (2x + y) / (2x²) dx = ∫ (2x / 2x²) dx + ∫ (y / 2x²) dx.
  - The first term: ∫ (2x / 2x²) dx = ∫ (1/x) dx = ln|x|.
  - The second term (not dependent on y): ∫ (y / 2x²) dx = y/2 ∫ x^-2 dx = y/2 * (-1/x) = -y / 2x.
• Step 4: Combine results.
  The general solution is: -1/y = ln|x| - y / 2x + C, where C is the constant of integration.
• Step 5: Apply the initial condition.
  When x = 1 and y = 2: -1/2 = ln|1| - 2 / (2 * 1) + C. Simplify: -1/2 = 0 - 1 + C => C = 1/2.
• Step 6: Write the particular solution.
  Substitute C = 1/2 into the general solution: -1/y = ln|x| - y / 2x + 1/2.
• Conclusion:
  The particular solution is: 1/y = -ln|x| + y / 2x - 1/2.

• Step 1: Rewrite the equation.
  The given equation is: y dx = (x + 2y²) dy. Rearrange to separate x and y: dx/dy = (x/y) + 2y. This can be written as dx/dy - x/y = 2y . It is a linear DE.
• Step 2: Find Integrating Factor.
  Integrating factor = e^(integral of -1/y dy) = e^(-lny) = e^(ln(y^-1)) = 1/y .
• Step 3: Multiply by Integrating Factor.
  Multiplying the DE by 1/y : 1/y dx/dy - x/y^2 = 2. This can be written as d(x/y) = 2dy
• Step 4: Integrate both sides.
  Integrating both sides with respect to y , gives x/y = 2y + C
  => x = 2y^2 + Cy => x = y(2y+C) where C is the constant of integration.

• Step 1: Find the vectors AB, BC, and AC.
  ⃗AB = ⃗B - ⃗A = (⃗i - 3⃗j - 5⃗k) - (2⃗i - ⃗j + ⃗k) = -⃗i - 2⃗j - 6⃗k.
  ⃗AC = ⃗C - ⃗A = (3⃗i - 4⃗j - 4⃗k) - (2⃗i - ⃗j + ⃗k) = ⃗i - 3⃗j - 5⃗k.
  ⃗BC = ⃗C - ⃗B = (3⃗i - 4⃗j - 4⃗k) - (⃗i - 3⃗j - 5⃗k) = 2⃗i - ⃗j + ⃗k.
• Step 2: Find the magnitudes of the vectors.
  - Magnitude of ⃗AB: ||⃗AB|| = √((-1)² + (-2)² + (-6)²) = √ (1+4+36) = √41.
  - Magnitude of ⃗AC: ||⃗AC|| = √((1)² + (-3)² + (-5)²) = √ (1+9+25) = √35.
  - Magnitude of ⃗BC: ||⃗BC|| = √((2)² + (-1)² + (1)²) = √(4+1+1) = √6.
• Step 3: Use the dot product to find cos C.
  To find ∠C, use the vectors ⃗AC and ⃗BC:
  ⃗AC . ⃗BC = (1)(2) + (-3)(-1) + (-5)(1) = 2 + 3 - 5 = 0.
  Thus:
  cos C = (⃗AC . ⃗BC) / (||⃗AC|| ||⃗BC||) = 0 / (√35 * √6) = 0.
• Step 4: Determine ∠C.
  Since cos C = 0: C = π/2.
• Step 5: Conclusion.
  The angle C of ΔABC is: C = π/2 (a right angle).

• Step 1: Equation of the parabola and area calculation.
  The given parabola is y² = 4x, which opens to the right. Its vertex is at the origin, and the curve is symmetric about the x-axis.
  The area of the region bounded by the curve y² = 4x, the line x = a, and the x-axis is given by: A = 2∫₀^a y dx, where y = √4x = 2√x (only the positive branch of the parabola is considered in the first quadrant).
• Step 2: Compute A₁.
  For A₁, the region is bounded by x = 1. Hence, A₁ = 2 ∫₀¹ 2√x dx = 4 ∫₀¹ x^1/2 dx.
• Step 3: Evaluate A₁.
  A₁ = 4 [x^3/2 / (3/2)]₀¹ = 4 * (2/3) [1^3/2 - 0^3/2] = 8/3.
• Step 4: Compute A₂.
  For A₂, the region is bounded by x = 4. Hence, A₂ = 2 ∫₀⁴ 2√x dx = 4 ∫₀⁴ x^1/2 dx.
• Step 5: Evaluate A₂.
  A₂ = 4 [x^3/2 / (3/2)]₀⁴ = 4 * (2/3) [4^3/2 - 0^3/2] = 8/3 * 8 = 64/3.
• Step 6: Find the ratio A₁ : A₂.
  A₁ : A₂ = (8/3) : (64/3) = 8 : 64 = 1 : 8.
• Conclusion:
  The ratio A₁ : A₂ = 1 : 8.

• Step 1: Check if f(x) is one-one.
  For f(x) to be one-one, f(a) = f(b) must imply a = b. Let: f(x) = x² + x + 1.
  Assume f(a) = f(b): a² + a + 1 = b² + b + 1.
  Rearranging: a² - b² + a - b = 0 => (a - b)(a + b + 1) = 0.
  This implies either a - b = 0 or a + b + 1 = 0. When a + b + 1 = 0, a ≠ b, so f(x) is not one-one.
• Step 2: Check if f(x) is onto.
  For f(x) to be onto, every real number y must have a corresponding x such that f(x) = y. Let:
  f(x) = y => x² + x + (1 - y) = 0.
  The discriminant of this quadratic equation is: Δ = 1² - 4(1)(1 - y) = 1 - 4 + 4y = 4y - 3.
  For f(x) to have a real solution, Δ ≥ 0: 4y - 3 ≥ 0 => y ≥ 3/4.
  Thus, f(x) is not onto because it cannot produce values less than 3/4.
• Step 3: Find all x such that f(x) = 3.
  Substitute f(x) = 3: x² + x + 1 = 3 => x² + x - 2 = 0.
  Factorize: (x + 2)(x - 1) = 0.
  Thus, x = -2 or x = 1.

• Step 1: Reflexivity.
  To prove R is an equivalence relation, we need to verify three properties: reflexivity, symmetry, and transitivity.
  For any (a, b) ∈ N × N, a/a = b/b =1. Thus, (a, b) R (a, b), so R is reflexive.
• Step 2: Symmetry.
  If (a, b) R (c, d), then: a/c = b/d.
  This implies: c/a = d/b, so (c, d) R (a, b). Thus, R is symmetric.
• Step 3: Transitivity.
  If (a, b) R (c, d) and (c, d) R (e, f), then: a/c = b/d and c/e = d/f.
  Multiply the equations: (a/c) * (c/e) = (b/d) * (d/f), which implies a/e = b/f, So, (a, b) R (e, f).
  Therefore, R is transitive.

• Step 1: Find the midpoints of the sides opposite to A and B.
  The midpoint of BC, denoted as M, is: M = ((1 + (-2)) / 2, (2 + 2) / 2, (1 + (-1)) / 2) = (-1/2, 2, 0).
  The midpoint of AC, denoted as N, is: N = ((1 + (-2)) / 2, (1 + 2) / 2, (0 + (-1)) / 2) = (-1/2, 3/2, -1/2).
• Step 2: Find the equations of the medians.
  The median through A joins A(1, 1, 0) to M(-1/2, 2, 0). The direction ratios of the line are:
  (-1/2 - 1, 2 - 1, 0 - 0) = (-3/2, 1, 0).
  The parametric equation of the median through A is: x = 1 - (3/2)t, y = 1 + t, z = 0.
  Eliminate t to get the symmetric form: (x - 1) / (-3/2) = (y - 1) / 1 = z / 0.
  The median through B joins B(1, 2, 1) to N(-1/2, 3/2, -1/2). The direction ratios of the line are:
  (-1/2 - 1, 3/2 - 2, -1/2 - 1) = (-3/2, -1/2, -3/2).
  The parametric equation of the median through B is: x = 1 - (3/2)t, y = 2 - (1/2)t, z = 1 - (3/2)t.
  Eliminate t to get the symmetric form: (x - 1) / (-3/2) = (y - 2) / (-1/2) = (z - 1) / (-3/2).
• Step 3: Find the coordinates of the centroid.
  The centroid of a triangle is given by: G = ((x1 + x2 + x3) / 3, (y1 + y2 + y3) / 3, (z1 + z2 + z3) / 3), where A(x1, y1, z1), B(x2, y2, z2), and C(x3, y3, z3).
  Substituting the coordinates: G = ((1 + 1 - 2) / 3, (1 + 2 + 2) / 3, (0 + 1 - 1) / 3) = (0, 5/3, 0).

• Step 1: Substitute and simplify.
  Let a = 1/x, b = 1/y, c = 1/z. Then, the system of equations becomes: 2a + 3b + 10c = 4, 4a - 6b + 5c = 1, 6a + 9b - 20c = 2.
• Step 2: Represent the system in matrix form.
  The system can be written as: [[2, 3, 10], [4, -6, 5], [6, 9, -20]] [[a], [b], [c]] = [[4], [1], [2]].
• Step 3: Compute the determinant of the coefficient matrix.
  The coefficient matrix is: A = [[2, 3, 10], [4, -6, 5], [6, 9, -20]].
  The determinant of A is: Det(A) = 2 * ((-6)(-20) - (5)(9)) - 3 * ((4)(-20) - (5)(6)) + 10 * ((4)(9) - (-6)(6)).
  Computing the minors: |[-6, 5], [9, -20]| = (-6)(-20) - (5)(9) = 120 - 45 = 75, |[4, 5], [6, -20]| = (4)(-20) - (5)(6) = -80 - 30 = -110, |[4, -6], [6, 9]| = (4)(9) - (-6)(6) = 36 + 36 = 72.
  Substituting into the determinant: Det(A) = 2(75) - 3(-110) + 10(72) = 150 + 330 + 720 = 1200.
• Step 4: Compute the inverse of A.
  The inverse of A is: A^-1 = 1/Det(A) * Adj(A), where Adj(A) is the adjugate matrix. After computation: A^-1 = 1/1200 * [[-270, -210, 120], [90, 60, 240], [-90, 30, 60]].
• Step 5: Solve X = A^-1B.
  The solution is: X = 1/1200 * [[-270, -210, 120], [90, 60, 240], [-90, 30, 60]] * [[4], [1], [2]].
  Performing the matrix multiplication: X = 1/1200 * [[-1080 - 210 + 240], [360 + 60 + 480], [-360 + 30 + 120]] = 1/1200 * [[-1050], [900], [-210]]. Simplifying: X = [[a], [b], [c]] = [[-7/8], [3/4], [-7/40]].
• Step 6: Back-substitute for x, y, z.
  From a = 1/x, b = 1/y, c = 1/z, we have: x = -8/7, y = 4/3, z = -40/7.

• Step 1: Compute the transpose of A.
  The matrix A is: A = [[1, cot x], [-cot x, 1]].
  The transpose of A is: A^T = [[1, -cot x], [cot x, 1]].
• Step 2: Compute the determinant of A.
  det(A) = (1)(1) - (-cot x)(cot x) = 1 + cot²x. Using the trigonometric identity 1 + cot²x = csc²x = 1/sin²x, we get: det(A) = 1/sin²x= -cos2x/sin²x.
• Step 3: Compute the inverse of A.
  The formula for the inverse of a 2x2 matrix is: A^-1 = 1/det(A) * [[d, -b], [-c, a]].
  For A = [[1, cot x], [-cot x, 1]], we have: A^-1 = 1/(1 + cot²x) * [[1, -cot x], [cot x, 1]].
  Using the identity 1 + cot²x = -cos2x/sin²x, the inverse becomes: A^-1 = 1/(-cos2x/sin²x) * [[1, -cot x], [cot x, 1]] = 1/-cos2x * [[1, -cot x], [cot x, 1]].
• Step 4: Compute A^TA^-1.
  Substitute A^T and A^-1:
  A^T = [[1, -cot x], [cot x, 1]], A^-1 = 1/-cos2x * [[1, -cot x], [cot x, 1]].
  Multiply A^T and A^-1: A^TA^-1 = [[1, -cot x], [cot x, 1]] * 1/(-cos2x) * [[1, -cot x], [cot x, 1]].
• Step 5: Simplify the product.
  Compute the product: A^TA^-1 = 1/(-cos2x) * [[1 - cot²x, -cotx - cotx], [cotx + cotx, -cot²x]].
  Using 1 - cot²x = -cos2x/sin²x and simplifying: A^TA^-1 = [[-cos 2x, -sin 2x], [sin 2x, -cos 2x]].
• Conclusion:
  We have shown that: A^TA^-1 = [[-cos 2x, -sin 2x], [sin 2x, -cos 2x]].

• Part (i): Probability that the airplane will not crash:
  The probability of a plane crash is:

  P(E₁) = 0.00001% = 0.00001 / 100 = 10⁻⁷.

  The probability that the airplane will not crash is:

  P(E₂) = 1 − P(E₁) = 1 − 10⁻⁷.

  Solution:

  P(E₂) = 1 − 10⁻⁷.

• Part (ii): Find P(A | E₁) + P(A | E₂):
  From the problem:

  P(A | E₁) = 0.95 (95% chance of survival after a crash),

  P(A | E₂) = 1 (all travelers survive if there is no crash).

  Thus:

  P(A | E₁) + P(A | E₂) = 0.95 + 1 = 1.95.

  Solution:

  P(A | E₁) + P(A | E₂) = 1.95.

• Part (iii)(a): Find P(A):
  Using the law of total probability:

  P(A) = P(A | E₁)P(E₁) + P(A | E₂)P(E₂).

  Substitute the values:

  P(A) = (0.95)(10⁻⁷) + (1)(1 − 10⁻⁷).

  Simplify:

  P(A) = 0.95 · 10⁻⁷ + 1 − 10⁻⁷ = 1 − 0.05 · 10⁻⁷.

  Solution:

  P(A) = 1 − 0.05 · 10⁻⁷.

• Part (iii)(b): Find P(E₂ | A):
  Using Bayes' theorem:

  P(E₂ | A) = P(A | E₂)P(E₂) / P(A).

  Substitute the values:

  P(E₂ | A) = (1)(1 − 10⁻⁷) / (1 − 0.05 · 10⁻⁷).

  Simplify:

  P(E₂ | A) = (1 − 10⁻⁷) / (1 − 0.05 · 10⁻⁷).

  Solution:

  P(E₂ | A) = (1 − 10⁻⁷) / (1 − 0.05 · 10⁻⁷).

• Part (i): Find F when V = 40 km/h:
  Substitute V = 40 into the equation:

  F = (40²/500) − (40/4) + 14.

  Simplify:

  F = (1600/500) − 10 + 14 = 3.2 − 10 + 14 = 7.2.

  Solution:

  F = 7.2 liters per 100 km.

• Part (ii): Find dF/dV:
  Differentiate F with respect to V:

  dF/dV = d/dV (V²/500) − d/dV (V/4) + d/dV (14).

  Simplify:

  dF/dV = (2V/500) − (1/4) + 0 = V/250 − 1/4.

  Solution:

  dF/dV = V/250 − 1/4.

• Part (iii)(a): Find the speed V for minimum F:
  For minimum fuel consumption, set:

  dF/dV = 0 ⇒ V/250 − 1/4 = 0.

  Solve for V:

  V/250 = 1/4 ⇒ V = 250/4 = 62.5.

  Solution:

  V = 62.5 km/h.

• Part (iii)(b): Find fuel required to travel 600 km:
  Given:

  dF/dV = −0.01 ⇒ V/250 − 1/4 = −0.01.

  Solve for V:

  V/250 = −0.01 + 1/4 = 0.24 ⇒ V = 250 * 0.24 = 60.

  At V = 60 km/h, substitute into the equation for F:

  F = (60²/500) − (60/4) + 14.

  Simplify:

  F = (3600/500) − 15 + 14 = 7.2 liters per 100 km.

  Total fuel required for 600 km:

  Fuel = (F/100) * 600 = (7.2/100) * 600 = 43.2 liters.

  Solution:

  Fuel required = 43.2 liters.

• Part (i): Constraints determining the feasible region:
  From Figure-2, the constraints are:

      3x + y ≤ 8,
      x + y ≥ 4,
      4x + 5y = 28,
      2x + y ≥ 10,
      x ≥ 0, y ≥ 0.
      

  Solution:

  The constraints are: 3x + y ≤ 8, x + y ≥ 4, 4x + 5y = 28, 2x + y ≥ 10, x ≥ 0, y ≥ 0.

• Part (ii): Minimize cost Z = 16x + 20y:
  Evaluate Z at the vertices of the feasible region (from Figure-2):

  Vertices: A(10, 0), B(2, 4), C(1, 5), D(0, 8).

  - At A(10, 0):

  Z = 16(10) + 20(0) = 160.

  - At B(2, 4):

  Z = 16(2) + 20(4) = 32 + 80 = 112.

  - At C(1, 5):

  Z = 16(1) + 20(5) = 16 + 100 = 116.

  - At D(0, 8):

  Z = 16(0) + 20(8) = 0 + 160 = 160.

  Minimum cost:

  Z = 112 at B(2, 4).

  Solution:

  The minimum cost is ₹112 at x = 2, y = 4.