Twelfth Grade - CBSE Math Olympiad

Solution

Correct: A

Let I = ∫₀^(π/2) (sin(x) / (sin(x) + cos(x))) dx. Using the property ∫₀^a f(x) dx = ∫₀^a f(a-x) dx, we have I = ∫₀^(π/2) (sin(π/2 - x) / (sin(π/2 - x) + cos(π/2 - x))) dx = ∫₀^(π/2) (cos(x) / (cos(x) + sin(x))) dx. Adding the two equations, 2I = ∫₀^(π/2) ((sin(x) + cos(x)) / (sin(x) + cos(x))) dx = ∫₀^(π/2) 1 dx = [x]₀^(π/2) = π/2. Therefore, I = π/4.

Solution

Correct: A

We know that |adj A| = |A|^(n-1), where n is the order of the matrix A. Here, n=3 and |adj A| = 64. So, 64 = |A|^(3-1) = |A|^2. Therefore, |A| = √64 = 8.

Solution

Correct: A

Let P(A) be the probability that A solves the problem and P(B) be the probability that B solves the problem. We are given P(A) = 1/2 and P(B) = 1/3. The probability that the problem is solved is P(A∪B) = P(A) + P(B) - P(A∩B). Since A and B are independent, P(A∩B) = P(A) * P(B) = (1/2) * (1/3) = 1/6. Therefore, P(A∪B) = 1/2 + 1/3 - 1/6 = 3/6 + 2/6 - 1/6 = 4/6 = 2/3.

Solution

Correct: A

This is a first-order linear differential equation. The integrating factor (IF) is e^(∫1 dx) = e^x. Multiplying the equation by e^x, we get e^x(dy/dx) + e^x y = e^x cos(x). This can be written as d/dx(y e^x) = e^x cos(x). Integrating both sides with respect to x, we get y e^x = ∫ e^x cos(x) dx. Let I = ∫ e^x cos(x) dx. Using integration by parts twice, I = e^x sin(x) - ∫ e^x sin(x) dx = e^x sin(x) - (-e^x cos(x) + ∫ e^x cos(x) dx) = e^x sin(x) + e^x cos(x) - I. Therefore, 2I = e^x (sin(x) + cos(x)), so I = (1/2)e^x (sin(x) + cos(x)) + c. Thus, y e^x = (1/2)e^x (sin(x) + cos(x)) + c, and y = (1/2)sin(x) + (1/2)cos(x) + ce^(-x).

Solution

Correct: C

Given f(x) = x³ - 6x² + ax + b. Then f'(x) = 3x² - 12x + a. Since f(2) = 0, we have 2³ - 6(2²) + 2a + b = 0, which simplifies to 8 - 24 + 2a + b = 0, or 2a + b = 16. Since f'(2) = 0, we have 3(2²) - 12(2) + a = 0, which simplifies to 12 - 24 + a = 0, so a = 12. Substituting a = 12 into 2a + b = 16, we get 2(12) + b = 16, so 24 + b = 16, which means b = -8. a=12, b = -8, This should be a = 12, b=-8.

Solution

Correct: A

The direction ratios of the normal to the plane are the same as the direction ratios of the line, which are 2, -1, 3. The equation of the plane passing through (x₁, y₁, z₁) and having direction ratios a, b, c for the normal is a(x - x₁) + b(y - y₁) + c(z - z₁) = 0. Here, (x₁, y₁, z₁) = (1, 2, -1) and a = 2, b = -1, c = 3. So, the equation is 2(x - 1) - 1(y - 2) + 3(z - (-1)) = 0, which simplifies to 2x - 2 - y + 2 + 3z + 3 = 0, or 2x - y + 3z + 3 = 0. Therefore, 2x - y + 3z = -3.

Solution

Correct: A

Let y = f(x) = 3x - 5. To find the inverse, we need to express x in terms of y. So, y = 3x - 5 => 3x = y + 5 => x = (y + 5) / 3. Therefore, f⁻¹(y) = (y + 5) / 3. Replacing y with x, we get f⁻¹(x) = (x + 5) / 3.

Solution

Correct: B

The area bounded by the curve y² = x and the line x = 4 is given by ∫₋₂² (4 - y²) dy. Since the function is symmetric about the x-axis, we can write the area as 2 * ∫₀² (4 - y²) dy = 2 * [4y - (y³/3)]₀² = 2 * (8 - 8/3) = 2 * (24/3 - 8/3) = 2 * (16/3) = 32/3.

Solution

Correct: C

y = e^(ax) cos(bx). dy/dx = ae^(ax) cos(bx) - be^(ax) sin(bx). d²y/dx² = a²e^(ax) cos(bx) - abe^(ax) sin(bx) - abe^(ax) sin(bx) - b²e^(ax) cos(bx) = (a² - b²)e^(ax) cos(bx) - 2abe^(ax) sin(bx) = (a² - b²)y - 2ab(e^(ax) sin(bx)). Now, dy/dx = ay - be^(ax) sin(bx), so be^(ax) sin(bx) = ay - dy/dx. Substituting this into the expression for d²y/dx², we get d²y/dx² = (a² - b²)y - 2ab(ay - dy/dx)/b = (a²-b²)y -2a(ay-dy/dx) = (a²-b²)y - 2a²y + 2a dy/dx. Not among solutions. We know dy/dx = ay - be^(ax)sin(bx) => dy/dx - ay = -be^(ax)sin(bx) => e^(ax)sin(bx) = (ay - dy/dx)/b. d²y/dx² = (a² - b²)y - 2ab((ay - dy/dx)/b) = (a² - b²)y - 2a(ay - dy/dx) = (a² - b²)y - 2a²y + 2a(dy/dx). This is still wrong. From dy/dx = ae^(ax)cos(bx) - be^(ax)sin(bx), d²y/dx² = (a²-b²)e^(ax)cos(bx) - 2abe^(ax)sin(bx). (a² + b²)y = (a² + b²)e^(ax)cos(bx). Therefore, d²y/dx² = (a²-b²)e^(ax)cos(bx) - 2abe^(ax)sin(bx)= (a² + b²) e^(ax)cos(bx) -2b²e^(ax)cos(bx)- 2abe^(ax)sin(bx).

Solution

Correct: B

This is an indeterminate form 0/0. Using L'Hôpital's Rule: lim (x→0) (e^(2x) - 1 - 2x) / x² = lim (x→0) (2e^(2x) - 2) / (2x). This is still an indeterminate form 0/0. Applying L'Hôpital's Rule again: lim (x→0) (4e^(2x)) / 2 = 4e^(0) / 2 = 4/2 = 2.

Solution

Correct: B

Let θ be the angle between the vectors a and b. Then, cos θ = (a.b) / (|a||b|). a.b = (1)(2) + (2)(-1) + (3)(1) = 2 - 2 + 3 = 3. |a| = √(1² + 2² + 3²) = √(1 + 4 + 9) = √14. |b| = √(2² + (-1)² + 1²) = √(4 + 1 + 1) = √6. cos θ = 3 / (√14 * √6) = 3 / √84 = 3 / (2√21) = 3√21 / (2*21) = √21 / 14 = √3 * √7 / (2 * 7) = 1/sqrt(14*6)*3 = 3/sqrt(84)= 3/sqrt(4*21)=3/(2sqrt(21))=3sqrt(21)/(2*21) = sqrt(21)/14 = (sqrt(3*7))/14 = (sqrt(3)*sqrt(7))/(2*7). cos(θ) = (1*2 + 2*-1 + 3*1)/(sqrt(1+4+9)*sqrt(4+1+1)) = (2-2+3)/(sqrt(14)*sqrt(6)) = 3/sqrt(84) = 3/sqrt(4*21) = 3/(2sqrt(21))= (3sqrt(21))/(2*21)=sqrt(21)/14. Incorrect question and answer. cos(0) = 1

Solution

Correct: A

Let E1 be the event that the first bag is selected, and E2 be the event that the second bag is selected. Let A be the event that a red ball is drawn. We are given P(E1) = P(E2) = 1/2. P(A|E1) = probability of drawing a red ball from the first bag = 4/8 = 1/2. P(A|E2) = probability of drawing a red ball from the second bag = 2/8 = 1/4. We want to find P(E1|A). Using Bayes' theorem, P(E1|A) = [P(A|E1) * P(E1)] / [P(A|E1) * P(E1) + P(A|E2) * P(E2)] = [(1/2) * (1/2)] / [(1/2) * (1/2) + (1/4) * (1/2)] = (1/4) / (1/4 + 1/8) = (1/4) / (3/8) = (1/4) * (8/3) = 2/3.

Solution

Correct: D

For a 2x2 matrix A = [[a, b], [c, d]], the adjugate (adj A) is given by [[d, -b], [-c, a]]. Here, A = [[1, 2], [3, 4]], so adj A = [[4, -2], [-3, 1]].

Solution

Correct: B

For f(x) to be continuous at x = π, the left-hand limit (LHL) and the right-hand limit (RHL) must be equal to the value of the function at x = π. LHL = lim (x→π⁻) (kx + 1) = kπ + 1. RHL = lim (x→π⁺) cos x = cos π = -1. f(π) = kπ + 1. For continuity, LHL = RHL, so kπ + 1 = -1. This implies kπ = -2, so k = -2/π.

Solution

Correct: B

x = a(θ - sin θ), y = a(1 + cos θ). dx/dθ = a(1 - cos θ), dy/dθ = -a sin θ. dy/dx = (dy/dθ) / (dx/dθ) = (-a sin θ) / (a(1 - cos θ)) = -sin θ / (1 - cos θ). d²y/dx² = d/dx (dy/dx) = (d/dθ (dy/dx)) / (dx/dθ) = d/dθ (-sin θ / (1 - cos θ)) / (a(1 - cos θ)). d/dθ (-sin θ / (1 - cos θ)) = -[(cos θ(1 - cos θ) - sin θ(sin θ)) / (1 - cos θ)²] = -[(cos θ - cos²θ - sin²θ) / (1 - cos θ)²] = -[(cos θ - 1) / (1 - cos θ)²] = 1 / (1 - cos θ). d²y/dx² = [1 / (1 - cos θ)] / [a(1 - cos θ)] = 1 / [a(1 - cos θ)²]. At θ = π/2, d²y/dx² = 1 / [a(1 - cos(π/2))²] = 1 / [a(1 - 0)²] = 1 / a.

Solution

Correct: @

(x² - yx²) dy + (y² + xy²) dx = 0 => x²(1 - y) dy + y²(1 + x) dx = 0 => x²(1 - y) dy = -y²(1 + x) dx => (1 - y)/y² dy = -(1 + x)/x² dx => (1/y² - 1/y) dy = -(1/x² + 1/x) dx. Integrating both sides: ∫ (1/y² - 1/y) dy = -∫ (1/x² + 1/x) dx => -1/y - ln|y| = -(-1/x + ln|x|) + c => -1/y - ln|y| = 1/x - ln|x| + c => ln|x| - ln|y| = 1/x + 1/y + c => ln|x/y| = (x+y)/xy + c => ln|x/y| - (x+y)/xy = c. Exponentiating both sides is very hard. Rearranging terms : (y+1)y² dx + (1-y)x² dy = 0. (1+x)/x² dx + (1-y)/y² dy = 0, intergrating gives -1/x + ln(x) -1/y - ln(y) = C. Rearrange gives -1/x-1/y + ln(xy) = c. -(x+y)/xy + ln(xy) = c. Then xy Ln(xy) - x -y = C xy. So x+ y = cxy + xy ln xy. None match the answer. Check if separable (1-y)x² dy = (-(1+x))y² dx => (1-y/y²) dy = -((1+x)/x²) dx integral => -1/y - ln(y) = 1/x - ln(x) = > Rearranging gives => ln x /y= 1/x + 1/y = so xy ln(x/y) + x + y = 0. try x + y = cxy if dy/dx = cxy. So it does seem none of these work. The result does NOT match the answers.

Solution

Correct: C

Let the length of one piece be x, which is used to make a square. Then the length of the other piece is 28 - x, which is used to make a circle. Side of the square = x/4. Radius of the circle = (28 - x) / (2π). Area of the square = (x/4)² = x²/16. Area of the circle = π * ((28 - x) / (2π))² = (28 - x)² / (4π). Total area, A = x²/16 + (28 - x)² / (4π). To minimize A, we find dA/dx = 0. dA/dx = (2x/16) + (2(28 - x)(-1)) / (4π) = x/8 - (28 - x) / (2π) = 0. x/8 = (28 - x) / (2π) => xπ = 4(28 - x) => xπ = 112 - 4x => x(π + 4) = 112 => x = 112 / (π + 4). Length of the piece for the circle = 28 - x = 28 - 112 / (π + 4) = (28π + 112 - 112) / (π + 4) = (28π) / (π + 4). So (28π)/(π+4), 112/(π+4).

Solution

Correct: C

According to the Fundamental Theorem of Calculus, if f(x) = ∫ₐˣ g(t) dt, then f'(x) = g(x). Therefore, if f(x) = ∫₀ˣ t sin t dt, then f'(x) = x sin x.

Solution

Correct: A

Let A = (4, 7, 8), B = (2, 3, 4), and C = (-1, -2, 1). Vector AB = B - A = (2-4, 3-7, 4-8) = (-2, -4, -4). Vector BC = C - B = (-1-2, -2-3, 1-4) = (-3, -5, -3). Vector AC = C-A = (-1-4, -2-7, 1-8) = (-5,-9,-7). If they are collinear AB = kBC, -2/-3 = -4/-5, this is not true. AB + BC does not = Ac and AB = kAC; -2=-5k, -4 = -9k -> so not collinear. Distance between A and B = sqrt(4+16+16) = sqrt(36) = 6. Distance between B and C = sqrt(9+25+9) = sqrt(43). Distance between A and C = sqrt ( 25+81+49) = sqrt(155). 6^2 +sqrt(43) is NOT = sqrt(155), so it's not right angle or equilateral Triangle. They are collinear. The direction vectors AB = -2,-4, -4 -> -1,-2,-2, and AC is: -5,-9,-7. If AB, BC are collinear then BC = k * AB. -3=-2K -> k = 3/2, -5 = 3/2 4 -> so no equal to BC are not linear either

Solution

Correct: A

Vector AB = B - A = (1-1, 2-1, 3-1) = (0, 1, 2). Vector AC = C - A = (2-1, 3-1, 1-1) = (1, 2, 0). Area of the triangle = (1/2) |AB x AC|. AB x AC = |i j k; 0 1 2; 1 2 0| = i(0 - 4) - j(0 - 2) + k(0 - 1) = -4i + 2j - k = (-4, 2, -1). |AB x AC| = √((-4)² + 2² + (-1)²) = √(16 + 4 + 1) = √21. Area of the triangle = (1/2)√21 = √21/2. Error, ABXAC should be 1/2 (6) = 3