CBSE Class 12 Maths Board Paper 2024 (65/1/1): Solutions

f : R₊ → R (R₊ = non-negative reals), f(x) = 4x + 3. Is f one-one, onto, both or neither?

1. To check if f is one-one, assume f(x₁) = f(x₂), which gives 4x₁ + 3 = 4x₂ + 3.
2. Subtracting 3 from both sides yields 4x₁ = 4x₂, and dividing by 4 gives x₁ = x₂, proving f is one-one.
3. To check if f is onto, note that for x ≥ 0, the minimum value of f(x) is 3, so values like 0 in the codomain R are never reached, meaning f is not onto.

Answer: 4x₁ + 3 = 4x₂ + 3 forces x₁ = x₂, so f is one-one; but values below 3 (e.g. 0) are never reached, so it is not onto — option (A).

A matrix has 36 elements. How many different orders m × n are possible?

1. The number of elements in a matrix of order m × n is given by the product mn.
2. Since the matrix has 36 elements, we must have mn = 36.
3. The possible orders correspond to the factor pairs of 36, which are determined by its divisors: 1, 2, 3, 4, 6, 9, 12, 18, and 36.
4. There are 9 such divisors, so there are 9 possible orders for the matrix.

Answer: Each order m × n needs mn = 36, i.e. one order per divisor of 36: 1, 2, 3, 4, 6, 9, 12, 18, 36 — so 9 orders — option (D).

f(x) = x² + 3 for x ≠ 0 and f(0) = 1. Where is f continuous and differentiable?

1. Calculate the limit of f(x) as x approaches 0, which is 0² + 3 = 3.
2. Compare this limit with the defined value f(0) = 1; since 3 ≠ 1, the function is discontinuous at x = 0.
3. Identify that for all x ≠ 0, f(x) is a polynomial, which is continuous and differentiable everywhere.
4. Conclude that f is continuous and differentiable on R − {0}, corresponding to option (C).

Answer: The limit at 0 is 3 but f(0) = 1, so f breaks only at x = 0; elsewhere it is a polynomial, so f is continuous and differentiable on R − {0} — option (C).

f is differentiable in the open interval (a, b) and continuous on its closure. Which condition makes f strictly increasing there?

1. By the Mean Value Theorem, for any two points x1 and x2 in the interval with x1 < x2, there exists a point c between them such that the difference f(x2) - f(x1) equals f'(c) multiplied by (x2 - x1).
2. If the derivative f'(x) is strictly positive for all x in the interval, then f'(c) is positive and (x2 - x1) is positive, making the product f(x2) - f(x1) positive.
3. Since f(x2) - f(x1) > 0 implies f(x2) > f(x1), the function is strictly increasing, which corresponds to option (B).

Answer: f′(x) > 0 for every x in (a, b) — option (B).

The 2 × 2 matrix [x + y, 2; 5, xy] equals [6, 2; 5, 8]. Find 24/x + 24/y.

1. Equate the corresponding elements of the matrices to get x + y = 6 and xy = 8.
2. Simplify the expression 24/x + 24/y to 24(x + y)/xy.
3. Substitute the values to get 24(6)/8 = 144/8 = 18.

Answer: x + y = 6 and xy = 8, so 24/x + 24/y = 24(x + y)/xy = 24 × 6/8 = 18 — option (D).

Which integral over [a, b] always equals the integral of f(x) from a to b?

1. Apply the substitution property of definite integrals by replacing x with a + b − x.
2. The limits of integration remain a and b, so the integral equals the integral of f(a + b − x) from a to b, which corresponds to option (B).

Answer: Replacing x by a + b − x keeps the limits: the integral equals the integral of f(a + b − x) from a to b — option (B).

Unit vectors â and b̂ make angle θ with sin θ = 3/5. Find â · b̂.

1. Use the dot product formula for unit vectors: â · b̂ = |â| |b̂| cos θ = cos θ.
2. Since sin θ = 3/5, compute cos θ = ±√(1 − (3/5)²) = ±√(16/25) = ±4/5.
3. Thus â · b̂ = ±4/5, which corresponds to option (C).

Answer: â · b̂ = cos θ = ±√(1 − 9/25) = ±4/5 — option (C).

Find the integrating factor of (1 − x²) dy/dx + xy = ax, where −1 < x < 1.

1. Divide the differential equation by (1 − x²) to write it in the standard linear form dy/dx + [x/(1 − x²)]y = [ax/(1 − x²)].
2. Identify P(x) = x/(1 − x²) and calculate the integrating factor as e^(∫P dx) = e^(−½ log(1 − x²)) = 1/√(1 − x²).

Answer: Dividing by 1 − x² gives P = x/(1 − x²); e^(∫P dx) = e^(−½ log(1 − x²)) = 1/√(1 − x²) — option (D).

A line has direction cosines √3k, √3k, √3k. Find k.

1. The sum of the squares of the direction cosines of a line is equal to 1.
2. Substitute the given direction cosines into the formula: (√3k)² + (√3k)² + (√3k)² = 1.
3. Simplify the equation: 3k² + 3k² + 3k² = 1, which becomes 9k² = 1.
4. Solve for k: k² = 1/9, so k = ±1/3.

Answer: 3k² + 3k² + 3k² = 1 gives k² = 1/9, so k = ±1/3 — option (D).

What kind of function does a linear programming problem optimise?

1. A linear programming problem involves optimizing a linear function, known as the objective function, subject to linear constraints.
2. Therefore, the correct option is (B), as it identifies the function as linear.

Answer: A linear (objective) function — option (B).

Given P(A | B) = P(A′ | B), which relation must hold?

1. Given P(A | B) = P(A′ | B), and knowing that P(A | B) + P(A′ | B) = 1, we have 2P(A | B) = 1, so P(A | B) = 1/2.
2. Using the definition P(A | B) = P(A ∩ B) / P(B), we get P(A ∩ B) / P(B) = 1/2, which implies P(A ∩ B) = ½ P(B).

Answer: P(A | B) + P(A′ | B) = 1, so each is 1/2; hence P(A ∩ B) = ½ P(B) — option (C).

Evaluate the determinant with first row x + 1, x − 1 and second row x² + x + 1, x² − x + 1.

1. Expand the determinant using the formula ad - bc: (x + 1)(x² − x + 1) − (x − 1)(x² + x + 1).
2. Simplify the terms using algebraic identities: (x³ + 1) − (x³ − 1) = 2.

Answer: (x + 1)(x² − x + 1) − (x − 1)(x² + x + 1) = (x³ + 1) − (x³ − 1) = 2 — option (B).

Find d/dx of sin(x²) at x = √π.

1. Differentiate sin(x²) using the chain rule to get d/dx [sin(x²)] = cos(x²) · 2x.
2. Substitute x = √π into the derivative: 2(√π) cos((√π)²) = 2√π cos(π).
3. Since cos(π) = −1, the value is 2√π(−1) = −2√π, which corresponds to option (C).

Answer: The derivative is 2x cos(x²); at x = √π it is 2√π cos π = −2√π — option (C).

State the order and degree of [1 + (dy/dx)²]³ = d²y/dx².

1. Identify the highest order derivative in the equation, which is d²y/dx², so the order is 2.
2. Observe that the term containing the highest order derivative is raised to the power of 1, so the degree is 1.
3. Thus, the order is 2 and the degree is 1, corresponding to option (C).

Answer: Highest derivative is d²y/dx² (order 2), appearing to power 1 (degree 1): 2, 1 — option (C).

Find the vector that starts at B(3, −4, 7) and ends at A(2, −3, 5).

1. The vector from B to A is calculated by subtracting the coordinates of the initial point B from the coordinates of the terminal point A.
2. BA = (2 − 3)î + (−3 − (−4))ĵ + (5 − 7)k̂ = −î + ĵ − 2k̂.

Answer: BA = (2 − 3)î + (−3 + 4)ĵ + (5 − 7)k̂ = −î + ĵ − 2k̂ — option (D).

How far is P(a, b, c) from the y-axis?

1. The foot of the perpendicular from P(a, b, c) to the y-axis is the point (0, b, 0).
2. Using the distance formula, the distance is √((a-0)² + (b-b)² + (c-0)²) = √(a² + c²).

Answer: The foot on the y-axis is (0, b, 0), so the distance is √(a² + c²) — option (C).

How many corner points does the region x ≥ 0, y ≥ 0, x + y ≥ 4 have?

1. The boundary lines x = 0 and y = 0 intersect the line x + y = 4 at the points (0, 4) and (4, 0) respectively.
2. Since the region is defined by x + y ≥ 4 in the first quadrant, it is unbounded and has no other intersection points, resulting in exactly 2 corner points.

Answer: The unbounded region has corners only at (4, 0) and (0, 4): 2 — option (C).

A, B are non-zero square matrices of equal order with (A + B)² = A² + B². What follows?

1. Expand the left-hand side of the given equation (A + B)² to obtain A² + AB + BA + B².
2. Equate this expansion to the right-hand side A² + B², which simplifies to AB + BA = O, implying AB = −BA.

Answer: (A + B)² = A² + AB + BA + B², so AB + BA = O, i.e. AB = −BA — option (B).

A: for A = [1, cos θ, 1; −cos θ, 1, cos θ; −1, −cos θ, 1], θ in [0, 2π], |A| lies in [2, 4]. R: cos θ always lies in [−1, 1].

1. Expand the determinant of A to get |A| = 2 + 2cos²θ.
2. Since cos²θ lies in [0, 1], the value of |A| lies in [2, 4], making Assertion A true.
3. Reason R is true because cos θ always lies in [−1, 1], which explains why cos²θ is bounded by 1.
4. Both A and R are true, and R is the correct explanation for A, so the answer is option (A).

Answer: Expanding, |A| = 2 + 2cos²θ, and since cos²θ is between 0 and 1, |A| is between 2 and 4. Both true and R explains A — option (A).

A: no line can be perpendicular to all three coordinate axes. R: for angles α, β, γ with the axes, cos²α + cos²β + cos²γ = 1.

1. If a line were perpendicular to all three axes, its direction cosines would be 0, 0, and 0.
2. This contradicts the identity cos²α + cos²β + cos²γ = 1, so Assertion A is true and Reason R explains it.

Answer: Perpendicular to all three would make every cosine 0, contradicting the sum of squares 1. Both true and R explains A — option (A).

Is f(x) = x²|x| differentiable at x = 0?

1. Write f(x) as x³ for x ≥ 0 and −x³ for x < 0.
2. Calculate the right-hand derivative: lim(h→0⁺) [f(0+h) − f(0)]/h = lim(h→0⁺) h³/h = 0.
3. Calculate the left-hand derivative: lim(h→0⁻) [f(0+h) − f(0)]/h = lim(h→0⁻) (−h³)/h = 0.
4. Since both one-sided limits are equal, f is differentiable at x = 0 with f′(0) = 0.

Answer: f(x) = x³ for x ≥ 0 and −x³ for x < 0. Both one-sided derivative limits at 0 are 0, so f is differentiable at 0 with f′(0) = 0.

Given y = √(tan √x), show √x · dy/dx = (1 + y⁴)/(4y).

1. Differentiate y = √(tan √x) with respect to x using the chain rule: dy/dx = (1/(2√(tan √x))) × sec²(√x) × (1/(2√x)).
2. Simplify the expression to dy/dx = sec²(√x) / (4√x √(tan √x)).
3. Multiply both sides by √x to get √x dy/dx = sec²(√x) / (4√(tan √x)).
4. Use the identity sec² θ = 1 + tan² θ to rewrite the numerator: √x dy/dx = (1 + tan²(√x)) / (4√(tan √x)).
5. Substitute y² = tan √x into the expression, noting that y⁴ = tan²(√x) and √(tan √x) = y, to obtain √x dy/dx = (1 + y⁴)/(4y).

Answer: dy/dx = sec²√x / (2√(tan√x) · 2√x). So √x dy/dx = (1 + tan²√x)/(4y) = (1 + y⁴)/(4y), since tan √x = y².

Show that f(x) = 4x³ − 18x² + 27x − 7 has no maximum or minimum.

1. Differentiate: f′(x) = 12x² − 36x + 27.
2. Factor: f′(x) = 3(4x² − 12x + 9) = 3(2x − 3)².
3. So f′(x) ≥ 0 for every real x, and f′(x) = 0 only at the single point x = 3/2.
4. f′ does not change sign at x = 3/2 (it is positive on both sides), so x = 3/2 is not a turning point.
5. Hence f is increasing on R and has no local maximum or minimum.

Answer: f′(x) = 12x² − 36x + 27 = 3(2x − 3)² ≥ 0 for all x, so f is increasing on R and has no local maxima or minima.

Integrate x√(1 + 2x) with respect to x.

1. Let 1 + 2x = t², so x = (t² − 1)/2 and dx = t dt.
2. Substitute to get ∫ [(t² − 1)/2] · t · t dt = ½ ∫ (t⁴ − t²) dt.
3. Integrate to get ½ [t⁵/5 − t³/3] + C = t⁵/10 − t³/6 + C.
4. Substitute t = √(1 + 2x) to obtain (1 + 2x)^(5/2)/10 − (1 + 2x)^(3/2)/6 + C.

Answer: Put 1 + 2x = t², dx = t dt: ½∫(t⁴ − t²) dt, giving (1 + 2x)^(5/2)/10 − (1 + 2x)^(3/2)/6 + C.

Evaluate the integral of sin√x / √x from 0 to π²/4.

1. Let √x = t, so x = t² and dx = 2t dt.
2. Change limits: when x = 0, t = 0; when x = π²/4, t = π/2.
3. Substitute to get ∫ (sin t / t) × 2t dt from 0 to π/2, which simplifies to 2∫ sin t dt from 0 to π/2.
4. Evaluate the integral: 2[−cos t] from 0 to π/2 = 2(−cos(π/2) + cos(0)) = 2(0 + 1) = 2.

Answer: Put √x = t, dx = 2t dt: 2∫ sin t dt from 0 to π/2 = 2[−cos t] = 2.

Non-zero vectors a, b satisfy (a + b) ⊥ a and (2a + b) ⊥ b. Prove |b| = √2 |a|.

1. Since (a + b) is perpendicular to a, their dot product is zero: (a + b) · a = 0.
2. Expanding the dot product gives a · a + b · a = 0, which simplifies to |a|² + a · b = 0.
3. Since (2a + b) is perpendicular to b, their dot product is zero: (2a + b) · b = 0.
4. Expanding this dot product gives 2(a · b) + b · b = 0, which simplifies to 2(a · b) + |b|² = 0.
5. From the first equation, a · b = −|a|²; substituting this into the second equation gives 2(−|a|²) + |b|² = 0.
6. Solving for |b|² yields |b|² = 2|a|², and taking the square root gives |b| = √2 |a|.

Answer: From the first, |a|² + a·b = 0; from the second, 2a·b + |b|² = 0. Substituting a·b = −|a|² gives |b|² = 2|a|², so |b| = √2|a|.

Parallelogram ABCD has AB = 2î − 4ĵ + 5k̂ and diagonal DB = 3î − 6ĵ + 2k̂. Find AD and the area of ABCD.

1. Going A → D → B gives AB = AD + DB, so AD = AB − DB.
2. AD = (2î − 4ĵ + 5k̂) − (3î − 6ĵ + 2k̂) = −î + 2ĵ + 3k̂.
3. The area of the parallelogram is |AD × AB| for adjacent sides AD and AB.
4. AD × AB = (2×5 − 3×(−4))î − ((−1)×5 − 3×2)ĵ + ((−1)×(−4) − 2×2)k̂ = 22î + 11ĵ + 0k̂.
5. Area = √(22² + 11²) = √(484 + 121) = √605 = 11√5 square units.

Answer: AD = AB − DB = −î + 2ĵ + 3k̂. AD × AB = 22î + 11ĵ, so the area is √605 = 11√5 square units.

On A = {1, 2, 3, 4, 5}, R = {(x, y) : |x² − y²| < 8}. Test R for reflexivity, symmetry and transitivity.

1. For reflexivity, check if (x, x) ∈ R for all x ∈ A. Since |x² − x²| = 0 and 0 < 8, R is reflexive.
2. For symmetry, if (x, y) ∈ R, then |x² − y²| < 8. Since |x² − y²| = |y² − x²|, it follows that (y, x) ∈ R, so R is symmetric.
3. For transitivity, consider (1, 2) and (2, 3). |1² − 2²| = 3 < 8 and |2² − 3²| = 5 < 8, so both pairs are in R.
4. Check if (1, 3) ∈ R. Calculate |1² − 3²| = |1 − 9| = 8. Since 8 is not less than 8, (1, 3) is not in R, so R is not transitive.

Answer: Reflexive: |x² − x²| = 0 < 8. Symmetric: |x² − y²| = |y² − x²|. Not transitive: (1, 2) and (2, 3) are in R (differences 3 and 5) but |1 − 9| = 8 is not < 8, so (1, 3) is not in R.

f : R → R, f(x) = ax + b; given f(2) = 3 and f(1) = 1, find f and test it for one-one and onto.

1. Given f(x) = ax + b, substitute x = 1 to get f(1) = a + b = 1.
2. Substitute x = 2 to get f(2) = 2a + b = 3.
3. Subtract the first equation from the second to find a = 2.
4. Substitute a = 2 into a + b = 1 to find b = -1, so f(x) = 2x - 1.
5. For one-one, assume f(x1) = f(x2), which gives 2x1 - 1 = 2x2 - 1, implying x1 = x2.
6. For onto, let y be any real number and solve y = 2x - 1 to get x = (y + 1)/2, which is a real number.
7. Since f is both one-one and onto, it is bijective.

Answer: a + b = 1, 2a + b = 3 give a = 2, b = −1, so f(x) = 2x − 1. 2x₁ − 1 = 2x₂ − 1 ⇒ x₁ = x₂ (one-one); any real y equals f((y + 1)/2) (onto). So f is both.

Given √(1 − x²) + √(1 − y²) = a(x − y), prove dy/dx = √((1 − y²)/(1 − x²)).

1. Put x = sin θ and y = sin φ, with θ, φ in [−π/2, π/2], so √(1 − x²) = cos θ and √(1 − y²) = cos φ.
2. The equation becomes cos θ + cos φ = a(sin θ − sin φ).
3. Using sum-to-product: 2cos((θ + φ)/2)cos((θ − φ)/2) = 2a cos((θ + φ)/2)sin((θ − φ)/2).
4. Cancel 2cos((θ + φ)/2): cot((θ − φ)/2) = a, so θ − φ = 2cot⁻¹a.
5. So sin⁻¹x − sin⁻¹y = 2cot⁻¹a, a constant.
6. Differentiate with respect to x: 1/√(1 − x²) − (1/√(1 − y²)) dy/dx = 0.
7. Hence dy/dx = √(1 − y²)/√(1 − x²) = √((1 − y²)/(1 − x²)).

Answer: Put x = sin θ, y = sin φ: cos θ + cos φ = a(sin θ − sin φ) gives cot((θ − φ)/2) = a, so sin⁻¹x − sin⁻¹y = 2cot⁻¹a. Differentiating, 1/√(1 − x²) − (1/√(1 − y²)) dy/dx = 0, which gives the result.

Find dy/dx when y = (tan x)^x.

1. Take the natural logarithm on both sides of the equation y = (tan x)^x to get log y = x log(tan x).
2. Differentiate both sides with respect to x, applying the product rule on the right side to obtain (1/y)(dy/dx) = x(sec²x / tan x) + log(tan x).
3. Multiply both sides by y to isolate dy/dx, resulting in dy/dx = (tan x)^x [x sec²x / tan x + log(tan x)].

Answer: log y = x log(tan x), so dy/dx = (tan x)^x [x sec²x / tan x + log(tan x)].

Integrate x² / ((x² + 4)(x² + 9)) with respect to x.

1. Let t = x², so the integrand becomes t / ((t + 4)(t + 9)).
2. Perform partial fraction decomposition: t / ((t + 4)(t + 9)) = −(4/5)/(t + 4) + (9/5)/(t + 9).
3. Substitute t = x² back into the expression to get −(4/5)/(x² + 4) + (9/5)/(x² + 9).
4. Integrate each term using the standard formula ∫ dx/(x² + a²) = (1/a) tan⁻¹(x/a).
5. The integral of −(4/5)/(x² + 4) is −(4/5) × (1/2) tan⁻¹(x/2) = −(2/5) tan⁻¹(x/2).
6. The integral of (9/5)/(x² + 9) is (9/5) × (1/3) tan⁻¹(x/3) = (3/5) tan⁻¹(x/3).
7. Combine the results and add the constant of integration C to obtain −(2/5) tan⁻¹(x/2) + (3/5) tan⁻¹(x/3) + C.

Answer: With t = x²: t/((t + 4)(t + 9)) = −(4/5)/(t + 4) + (9/5)/(t + 9). The integral is −(2/5) tan⁻¹(x/2) + (3/5) tan⁻¹(x/3) + C.

Evaluate the integral of |x − 1| + |x − 2| + |x − 3| from 1 to 3.

1. On [1, 3], |x − 1| = x − 1 and |x − 3| = 3 − x; |x − 2| = 2 − x on [1, 2] and x − 2 on [2, 3].
2. ∫ from 1 to 3 of (x − 1) dx = [(x − 1)²/2] = 4/2 = 2.
3. ∫ from 1 to 2 of (2 − x) dx + ∫ from 2 to 3 of (x − 2) dx = 1/2 + 1/2 = 1.
4. ∫ from 1 to 3 of (3 − x) dx = [−(3 − x)²/2] = 0 + 4/2 = 2.
5. Adding: 2 + 1 + 2 = 5.

Answer: Splitting at x = 2: the three parts give 2, 1/2 + 1/2 and 2, so the value is 5.

Solve x² dy/dx − xy = x² cos²(y/(2x)) given y = π/2 when x = 1.

1. Divide the equation by x² to get dy/dx − y/x = cos²(y/2x).
2. Substitute y = vx, so dy/dx = v + x(dv/dx), and simplify to x(dv/dx) = cos²(v/2).
3. Separate variables as sec²(v/2) dv = dx/x and integrate both sides.
4. The integration yields 2 tan(v/2) = log|x| + C.
5. Replace v with y/x to obtain the general solution 2 tan(y/2x) = log|x| + C.
6. Use the condition y = π/2 at x = 1 to find C: 2 tan(π/4) = log(1) + C, so C = 2.
7. The particular solution is 2 tan(y/2x) = log|x| + 2.

Answer: dy/dx = y/x + cos²(y/2x); with y = vx, sec²(v/2) dv = dx/x, so 2 tan(y/2x) = log|x| + C. At x = 1, y = π/2: C = 2. Particular solution: 2 tan(y/(2x)) = log|x| + 2.

By graph, maximise Z = 500x + 300y subject to x, y ≥ 0, 4x + 5y ≥ 20, 2x + y ≤ 12 and x + 2y ≤ 12.

1. Draw the lines 4x + 5y = 20, 2x + y = 12 and x + 2y = 12 in the first quadrant.
2. The feasible region lies above 4x + 5y = 20 and below both other lines; it is bounded.
3. Corner points: (5, 0) and (0, 4) on 4x + 5y = 20; (6, 0) on 2x + y = 12; (0, 6) on x + 2y = 12; and (4, 4) where 2x + y = 12 meets x + 2y = 12.
4. Z = 500x + 300y gives Z(5, 0) = 2500, Z(6, 0) = 3000, Z(4, 4) = 2000 + 1200 = 3200, Z(0, 6) = 1800, Z(0, 4) = 1200.
5. Maximum Z = 3200 at x = 4, y = 4.

Answer: Corners (0, 4), (0, 6), (4, 4), (6, 0), (5, 0) give Z = 1200, 1800, 3200, 3000, 2500. Maximum Z = 3200 at x = 4, y = 4.

E, F are independent with P(E ∪ F) = 0.6 and P(not E) = 0.6. Find P(F) and P(Ē ∪ F̄).

1. Given P(not E) = 0.6, we calculate P(E) = 1 - P(not E) = 1 - 0.6 = 0.4.
2. Using the formula for the union of independent events, P(E ∪ F) = P(E) + P(F) - P(E)P(F).
3. Substituting the known values into the equation: 0.6 = 0.4 + P(F) - 0.4 P(F).
4. Simplifying the equation gives 0.2 = 0.6 P(F), so P(F) = 0.2 / 0.6 = 1/3.
5. Using De Morgan's Law, P(Ē ∪ F̄) = 1 - P(E ∩ F).
6. Since E and F are independent, P(E ∩ F) = P(E)P(F) = 0.4 × (1/3) = 2/15.
7. Therefore, P(Ē ∪ F̄) = 1 - 2/15 = 13/15.

Answer: P(E) = 0.4; 0.6 = 0.4 + P(F) − 0.4 P(F) gives P(F) = 1/3. P(Ē ∪ F̄) = 1 − P(E)P(F) = 1 − 0.4/3, so P(F) = 1/3, P(Ē ∪ F̄) = 13/15.

For A = [1, −2, 0; 2, −1, −1; 0, −2, 1] find A⁻¹, then solve 2x − y − z = 8, x − 2y = 10, −2y + z = 7.

1. Reorder the equations as x − 2y = 10, 2x − y − z = 8, −2y + z = 7. Then the system is AX = B with the given A and B = (10, 8, 7).
2. |A| = 1((−1)(1) − (−1)(−2)) − (−2)(2×1 − 0) + 0 = −3 + 4 = 1, so A⁻¹ exists.
3. Cofactors: C11 = −3, C12 = −2, C13 = −4, C21 = 2, C22 = 1, C23 = 2, C31 = 2, C32 = 1, C33 = 3.
4. adj A is the transpose of the cofactor matrix: [−3, 2, 2; −2, 1, 1; −4, 2, 3]. Since |A| = 1, A⁻¹ = adj A.
5. X = A⁻¹B: x = (−3)(10) + 2(8) + 2(7) = 0, y = (−2)(10) + 8 + 7 = −5, z = (−4)(10) + 2(8) + 3(7) = −3.
6. So x = 0, y = −5, z = −3 (check: 2(0) + 5 + 3 = 8).

Answer: |A| = 1, A⁻¹ = adj A = [−3, 2, 2; −2, 1, 1; −4, 2, 3]. X = A⁻¹B with B = (10, 8, 7) gives (x, y, z) = (0, −5, −3).

A = [−1, a, 2; 1, 2, x; 3, 1, 1] has inverse [1, −1, 1; −8, 7, −5; b, y, 3]. Find (a + x) − (b + y).

1. Since A · A⁻¹ = I, equate the (1,3) element of the product to 0: (−1)(1) + a(−5) + 2(3) = 0, which simplifies to 5 − 5a = 0, giving a = 1.
2. Equate the (2,3) element of the product to 0: 1(1) + 2(−5) + x(3) = 0, which simplifies to 3x − 9 = 0, giving x = 3.
3. Equate the (3,1) element of the product to 0: 3(1) + 1(−8) + 1(b) = 0, which simplifies to −5 + b = 0, giving b = 5.
4. Equate the (3,2) element of the product to 0: 3(−1) + 1(7) + 1(y) = 0, which simplifies to 4 + y = 0, giving y = −4.
5. Calculate the required value: (a + x) − (b + y) = (1 + 3) − (5 − 4) = 4 − 1 = 3.

Answer: Comparing A·A⁻¹ with I: −5 + b = 0, 5 − 5a = 0, 4 + y = 0, 3x − 9 = 0, so b = 5, a = 1, y = −4, x = 3, and (a + x) − (b + y) = 4 − 1 = 3.

Evaluate the integral of (sin x + cos x)/(9 + 16 sin 2x) from 0 to π/4.

1. Put t = sin x − cos x. Then dt = (cos x + sin x) dx; x = 0 gives t = −1 and x = π/4 gives t = 0.
2. t² = 1 − 2 sin x cos x = 1 − sin 2x, so sin 2x = 1 − t² and 9 + 16 sin 2x = 25 − 16t².
3. The integral becomes ∫ from −1 to 0 of dt/(25 − 16t²) = ∫ dt/((5 − 4t)(5 + 4t)).
4. Partial fractions: 1/((5 − 4t)(5 + 4t)) = (1/10)[1/(5 − 4t) + 1/(5 + 4t)].
5. Integrating: (1/10) × (1/4)[log(5 + 4t) − log(5 − 4t)] = (1/40) log((5 + 4t)/(5 − 4t)).
6. From −1 to 0: (1/40)[log 1 − log(1/9)] = (1/40) log 9 = (1/20) log 3 ≈ 0.0549.

Answer: Put sin x − cos x = t, so sin 2x = 1 − t² and the integral is ∫ dt/(25 − 16t²) from −1 to 0 = (1/40) log 9 = (1/20) log 3 ≈ 0.0549.

Evaluate the integral of sin 2x · tan⁻¹(sin x) from 0 to π/2.

1. Let I = ∫ from 0 to π/2 of sin 2x tan⁻¹(sin x) dx and use sin 2x = 2 sin x cos x to write I = 2 ∫ from 0 to π/2 of sin x cos x tan⁻¹(sin x) dx.
2. Substitute t = sin x so dt = cos x dx, changing limits from x=0 to t=0 and x=π/2 to t=1, resulting in I = 2 ∫ from 0 to 1 of t tan⁻¹(t) dt.
3. Integrate by parts with u = tan⁻¹(t) and dv = t dt, giving du = dt/(1+t²) and v = t²/2, so ∫ t tan⁻¹(t) dt = (t²/2)tan⁻¹(t) - ∫ (t²/2)/(1+t²) dt.
4. Simplify the remaining integral as 1/2 ∫ (1 - 1/(1+t²)) dt, which evaluates to 1/2(t - tan⁻¹(t)).
5. Combine terms to get the antiderivative (t²/2)tan⁻¹(t) - t/2 + 1/2 tan⁻¹(t), and evaluate from 0 to 1 using tan⁻¹(1) = π/4 to obtain π/4 - 1/2.
6. Multiply by the constant factor 2 to find I = 2(π/4 - 1/2) = π/2 - 1.
7. The final answer is π/2 - 1, which is approximately 0.5708.

Answer: With sin x = t: 2∫ t tan⁻¹t dt from 0 to 1 = 2[(t²/2) tan⁻¹t − t/2 + ½ tan⁻¹t] = 2(π/4 − 1/2) = π/2 − 1 ≈ 0.5708.

By integration, find the area of the ellipse x²/16 + y²/4 = 1 lying between x = −2 and x = 2.

1. From x²/16 + y²/4 = 1, solve for y to get y = ±½√(16 − x²).
2. The required area is 2 ∫ from −2 to 2 of ½√(16 − x²) dx, which simplifies to ∫ from −2 to 2 of √(16 − x²) dx.
3. Using symmetry about the y-axis, rewrite the area as 2 ∫ from 0 to 2 of √(16 − x²) dx.
4. Apply the standard formula ∫√(a² − x²) dx = (x/2)√(a² − x²) + (a²/2)sin⁻¹(x/a) with a = 4.
5. Substitute the limits 0 and 2 into the antiderivative: 2 [ (2/2)√(16 − 4) + (16/2)sin⁻¹(2/4) − 0 ].
6. Evaluate the terms to get 2 [ √12 + 8sin⁻¹(1/2) ] = 2 [ 2√3 + 8(π/6) ].
7. Simplify the final expression to 4√3 + 8π/3, which is approximately 15.31 square units.

Answer: y = ½√(16 − x²); area = 4∫₀² ½√(16 − x²) dx = 2[(x/2)√(16 − x²) + 8 sin⁻¹(x/4)]₀² = 4√3 + 8π/3 ≈ 15.31 square units.

P′(1, 0, 7) is the image of P(x, y, z) in the line x/1 = (y − 1)/2 = (z − 2)/3. Find P.

1. Let the foot of the perpendicular from P to the line be M, which lies on the line x/1 = (y − 1)/2 = (z − 2)/3 = λ.
2. The coordinates of M are (λ, 2λ + 1, 3λ + 2).
3. Since P′(1, 0, 7) is the image of P, M is the midpoint of PP′, so the vector P′M is perpendicular to the direction vector of the line (1, 2, 3).
4. The vector P′M is (λ − 1, 2λ + 1, 3λ − 5).
5. Using the dot product for perpendicularity: (λ − 1)(1) + (2λ + 1)(2) + (3λ − 5)(3) = 0.
6. Simplifying the equation: λ − 1 + 4λ + 2 + 9λ − 15 = 0, which gives 14λ − 14 = 0, so λ = 1.
7. Substituting λ = 1, M is (1, 3, 5); using P = 2M − P′, we get P = 2(1, 3, 5) − (1, 0, 7) = (1, 6, 3).

Answer: Midpoint M(λ, 2λ + 1, 3λ + 2) with P′M ⊥ line gives λ = 1, M(1, 3, 5); then P = 2M − P′ = (1, 6, 3).

A camera sits 5 m up a pole; a car is x m from the pole's base and sees the camera at elevation θ. Write θ in terms of the height and x.

1. In the right-angled triangle formed by the pole, the ground, and the line of sight, the height of the camera is the opposite side to angle θ, which is 5 m.
2. The distance of the car from the base of the pole is the adjacent side to angle θ, which is x m.
3. Using the definition of tangent, tan θ = opposite / adjacent = 5 / x.
4. Therefore, θ = tan⁻¹(5 / x).

Answer: tan θ = 5/x, so θ = tan⁻¹(5/x).

With θ = tan⁻¹(5/x) as above, find dθ/dx.

1. Differentiate θ = tan⁻¹(5/x) with respect to x using the chain rule: dθ/dx = 1/(1 + (5/x)²) × d(5/x)/dx.
2. Simplify the derivative of the inner function: d(5/x)/dx = −5/x², so dθ/dx = 1/(1 + 25/x²) × (−5/x²).
3. Combine the terms to get the final result: dθ/dx = −5/(x² + 25).

Answer: dθ/dx = −5/(25 + x²).

The car moves away from the 5 m pole at 20 m/s. Find dθ/dt when it is 50 m from the pole.

1. Let x be the distance from the pole and h be the height of the pole, so tan θ = h/x = 5/x.
2. Differentiate both sides with respect to time t to get sec² θ (dθ/dt) = -5/x² (dx/dt).
3. When x = 50 m, calculate sec² θ = 1 + tan² θ = 1 + (5/50)² = 1 + 1/100 = 101/100.
4. Substitute dx/dt = 20 m/s, x = 50, and sec² θ = 101/100 into the equation: (101/100)(dθ/dt) = -5/2500 × 20.
5. Simplify the right side to -100/2500 = -1/25, then solve for dθ/dt: dθ/dt = (-1/25) × (100/101) = -4/101 rad/s.

Answer: dθ/dt = (−5/(25 + 2500)) × 20 = −100/2525 = −4/101 rad/s (the angle decreases at 4/101 rad/s).

Another car, 50 m from the 5 m pole, has its elevation angle changing at 3/101 rad/s. Find its speed.

1. From (ii), dθ/dt = (dθ/dx)(dx/dt) = −5/(25 + x²) × dx/dt.
2. At x = 50: 25 + x² = 2525, so dθ/dt = −(5/2525) dx/dt = −(1/505) dx/dt.
3. With dθ/dt = 3/101: dx/dt = −505 × 3/101 = −15 m/s (θ rising means the car is moving towards the pole).
4. The speed is the size of dx/dt, so the car's speed is 15 m/s.

Answer: 3/101 = (5/2525) × speed in size, so the speed is 15 m/s.

Note: The paper does not say whether the angle is increasing or decreasing; either way the speed is 15 m/s, and only the sign of dx/dt (direction) changes.

Severe, moderate and light turbulence are equally likely; a plane is late with chance 55%, 37%, 17% respectively. Find P(late).

1. Let S, M, and L denote the events of severe, moderate, and light turbulence respectively. Since they are equally likely, P(S) = P(M) = P(L) = 1/3.
2. Let E be the event that the plane is late. Using the law of total probability, P(E) = P(E|S)P(S) + P(E|M)P(M) + P(E|L)P(L).
3. Substitute the given probabilities: P(E) = (0.55)(1/3) + (0.37)(1/3) + (0.17)(1/3).
4. Factor out 1/3 and sum the terms: P(E) = (1/3)(0.55 + 0.37 + 0.17) = (1/3)(1.09) = 109/300.

Answer: P(late) = (1/3)(0.55 + 0.37 + 0.17) = 109/300.

In the same setting, given the plane was late, find the probability the turbulence was moderate.

1. Using Bayes' theorem, the probability of moderate turbulence given the plane was late is P(moderate | late) = P(late | moderate) × P(moderate) / P(late).
2. Substitute the given values into the formula: P(moderate | late) = (1/3 × 37/100) / (109/300) = 37/109.

Answer: P(moderate | late) = (1/3 × 37/100)/(109/300) = 37/109.

sin is restricted to a set A onto [−1, 1] so that sin⁻¹ exists. Give such an interval A other than the principal branch.

1. Identify that for sin⁻¹ to exist, the function sin x must be one-to-one and map the interval A onto the range [−1, 1].
2. Select an interval of length π where sine is monotonic and covers the full range, such as [π/2, 3π/2], noting that sin(π/2) = 1 and sin(3π/2) = −1.

Answer: For example [π/2, 3π/2] (any interval on which sine runs once through [−1, 1], such as [−3π/2, −π/2]).

Note: Any interval on which sine takes every value in [−1, 1] exactly once is correct, e.g. [π/2, 3π/2] or [−3π/2, −π/2].

Using principal values, find sin⁻¹(−1/2) − sin⁻¹(1).

1. The principal value of sin⁻¹(−1/2) is −π/6 since sin(−π/6) = −1/2.
2. The principal value of sin⁻¹(1) is π/2 since sin(π/2) = 1.
3. Substitute the values into the expression: −π/6 − π/2 = −2π/3 (≈ −2.0944).

Answer: −π/6 − π/2 = −2π/3 (≈ −2.0944).

Sketch y = sin⁻¹x on [−1, 1] with its principal value branch.

1. Identify the domain of the function as [−1, 1] and the principal range as [−π/2, π/2].
2. Plot the key points: (−1, −π/2), (0, 0), and (1, π/2).
3. Draw an increasing curve connecting these points, noting it is steepest at the ends and is the reflection of y = sin x on [−π/2, π/2] in the line y = x.

Answer: An increasing curve through the origin from (−1, −π/2) to (1, π/2), steepest at the ends; it is the reflection of sine on [−π/2, π/2] in the line y = x.

For f(x) = 2 sin⁻¹(1 − x), state its domain and its range.

1. Since the argument of sin⁻¹ must lie in [−1, 1], we set −1 ≤ 1 − x ≤ 1.
2. Solving the inequality gives 0 ≤ x ≤ 2, so the domain is [0, 2].
3. The range of sin⁻¹(t) is [−π/2, π/2], so −π/2 ≤ sin⁻¹(1 − x) ≤ π/2.
4. Multiplying by 2 gives −π ≤ 2 sin⁻¹(1 − x) ≤ π, so the range is [−π, π].

Answer: −1 ≤ 1 − x ≤ 1 gives domain [0, 2]; −π/2 ≤ sin⁻¹(1 − x) ≤ π/2 gives range [−π, π].