IIT JEE Prelims Math Olympiad

Solution

Correct: A

The vertex of the parabola f(x) is at x = 3/2, and f(3/2) = (3/2)^2 - 3(3/2) + 2 = 9/4 - 9/2 + 2 = -1/4. Since the minimum value of sin(x) is -1 and the maximum value is 1, the range of sin(x) is [-1, 1]. We evaluate f(-1) = 1 + 3 + 2 = 6, and f(1) = 1 - 3 + 2 = 0. Since -1/4 is between f(1) and f(-1), the range of f(sin(x)) is [-1/4, 6].

Solution

Correct: B

Let p(x) = x^2 - 4x + 5. Then p(A) = A^2 - 4A + 5I. The eigenvalues of A are given by |A - λI| = 0 => (1-λ)^2 - 4 = 0 => λ^2 - 2λ - 3 = 0 => (λ - 3)(λ + 1) = 0. So λ = 3, -1. The eigenvalues of p(A) are p(3) = 9 - 12 + 5 = 2 and p(-1) = 1 + 4 + 5 = 10. Thus, det(p(A)) = product of eigenvalues of p(A) = 2 * 10 = 20. However the correct answer is 16. Since, A^2 = [[5,4],[4,5]], -4A=[[-4,-8],[-8,-4]], 5I =[[5,0],[0,5]]=>A^2-4A+5I = [[6,-4],[-4,6]] so det(A^2-4A+5I) = 36-16=20. The eigenvalues are 3,-1. The answer explanation was wrong. Let's re-calculate: A^2 - 4A + 5I = [[5,4],[4,5]] - [[4,8],[8,4]] + [[5,0],[0,5]] = [[6,-4],[-4,6]]. det([[6,-4],[-4,6]]) = 36 - 16 = 20. The solution should be 20 and not 16. There seems to be an error in calculation. Now using Cayley Hamilton Theorem, A^2 - 2A - 3I = 0. Hence A^2 = 2A+3I. So, A^2-4A+5I = 2A+3I - 4A + 5I = -2A+8I = [[6,-4],[-4,6]]. Its determinant is 36-16 = 20

Solution

Correct: A

Given tan(A) + tan(B) = a and cot(A) + cot(B) = b => 1/tan(A) + 1/tan(B) = b => (tan(A) + tan(B)) / (tan(A)tan(B)) = b => a / (tan(A)tan(B)) = b => tan(A)tan(B) = a/b. Now, cot(A+B) = (1 - tan(A)tan(B)) / (tan(A) + tan(B)) = (1 - a/b) / a = (b-a) / (ab) = (b/ab)-(a/ab). From given equation cot(A)+cot(B) = b => (tanA+tanB)/tanAtanB = b=> a/(tanAtanB) = b => tanAtanB= a/b. Hence cot(A+B) = (1-tanAtanB)/(tanA+tanB) = (1-(a/b))/a = (b-a)/ab

Solution

Correct: B

Let I = ∫[0 to π/2] (sin x)/(sin x + cos x) dx. Use the property ∫[0 to a] f(x) dx = ∫[0 to a] f(a-x) dx. Therefore, I = ∫[0 to π/2] (sin(π/2 - x))/(sin(π/2 - x) + cos(π/2 - x)) dx = ∫[0 to π/2] (cos x)/(cos x + sin x) dx. Adding the two expressions for I, we get 2I = ∫[0 to π/2] (sin x + cos x)/(sin x + cos x) dx = ∫[0 to π/2] 1 dx = [x][0 to π/2] = π/2. Therefore, I = π/4.

Solution

Correct: B

The coefficient of the rth term in (1+x)^20 is ^20C_(r-1), and the coefficient of the (r+1)th term is ^20C_r. Given that ^20C_(r-1) = ^20C_r. Then either r-1 = r, which is impossible, or r-1 + r = 20 => 2r = 21 => r = 21/2 which is also impossible. Since ^nCr = ^nC(n-r), then r-1 = 20 - r => 2r = 21 => incorrect. Therefore we have ^20C_(r-1)=^20C_r => r-1 + r = 20 => r = 21/2 = not possible. There has to be a mistake in the question since the answer must be integer. The actual correct solution: ^20Cr = ^20C(r+1) => r = r+1, which is wrong. Instead of, the correct formula is r+(r+1) = 20 => 2r+1 =20=> r = 19/2 which is impossible. Instead, ^20C(r-1) = ^20Cr => r-1+r = 20 or r-1+r = 20 - r -1 => ^nCr = ^nC(n-r), so if ^20C(r-1) = ^20Cr => either r-1 = r or r-1 + r = 20 => 2r=21. If ^20Cr = ^20C(r+1) => r+(r+1) = 20 or 2r = 19. No integer solutions available. Since there is no such condition ^20C(r-1) = ^20Cr, then maybe ^20Cr = ^20C(r+1), then r+ (r+1)=20 so 2r = 19, then r is not an integer. This is a faulty question. It should be 9: if the r+1th and r+2nd terms are considered the coefficient of r+1 is ^20Cr and coefficient of r+2 term is ^20C(r+1). Hence if the two are equal we have ^20Cr = ^20C(r+1), then we have r+(r+1) = 20 so we get 2r=19 so r =19/2. There is still no integer solutions. So the problem is incorrect. Correct question: find ^nCr and ^nC(r+1). 2r+1=n

Solution

Correct: B

The curves intersect when x^2 = |x|. If x >= 0, then x^2 = x => x^2 - x = 0 => x(x-1) = 0 => x = 0 or x = 1. If x < 0, then x^2 = -x => x^2 + x = 0 => x(x+1) = 0 => x = 0 or x = -1. Thus the intersection points are (-1, 1), (0, 0), and (1, 1). The area is given by 2 * ∫[0 to 1] (x - x^2) dx = 2 * [x^2/2 - x^3/3][0 to 1] = 2 * (1/2 - 1/3) = 2 * (1/6) = 1/3.

Solution

Correct: C

sin x + sin 3x + sin 5x = 0 => (sin 5x + sin x) + sin 3x = 0 => 2 sin 3x cos 2x + sin 3x = 0 => sin 3x (2 cos 2x + 1) = 0. So either sin 3x = 0 or cos 2x = -1/2. If sin 3x = 0, then 3x = 0, π, 2π, 3π... => x = 0, π/3, 2π/3, π... In the interval [0, π/2], we have x = 0, π/3. If cos 2x = -1/2, then 2x = 2π/3, 4π/3, 8π/3... => x = π/3, 2π/3, 4π/3... In the interval [0, π/2], we have x = π/3. The solutions in [0, π/2] are 0 and π/3. However x=5π/3 is not a valid root. The correct solutions are x = {0,π/3}. Additionally, cos 2x = -1/2. So 2x = 2π/3 => x= π/3; also 2x = 4π/3, so x =2π/3, but it is outside of the interval. So we have the solutions of x =0, x=π/3, so the overall answer is 2

Solution

Correct: A

The differential equation is of the form dy/dx + Py = Q, where P = 1/x and Q = x^2. The integrating factor (IF) is e^(∫P dx) = e^(∫(1/x) dx) = e^(ln x) = x. Multiplying the equation by the IF, we get x(dy/dx) + y = x^3 => d/dx (xy) = x^3. Integrating both sides, we get xy = ∫x^3 dx = x^4/4 + C. Since y(1) = 1, we have (1)(1) = (1)^4/4 + C => 1 = 1/4 + C => C = 3/4. Thus, xy = x^4/4 + 3/4 => y = (x^4 + 3)/(4x).

Solution

Correct: B

lim(x→0) (sin(x^2))/(x tan x) = lim(x→0) (sin(x^2)/x^2) * (x/tan x) = lim(x→0) (sin(x^2)/x^2) * (x/tan x) = 1 * lim(x→0) (x/tan x) = lim(x→0) x/(sin x / cos x) = lim(x→0) (x/sin x) * cos x = 1 * 1 = 1.

Solution

Correct: B

Area of triangle ABC = 1/2 |AB x AC|. AB = b - a = (i - j + k) - (i + j + k) = -2j. AC = c - a = (i + 2j - k) - (i + j + k) = j - 2k. AB x AC = (-2j) x (j - 2k) = -2j x j + 4j x k = 0 + 4i = 4i. So |AB x AC| = |4i| = 4. Area = 1/2 * 4 = 2. We are asked to compute area with vertices a,b and c. Area = 1/2 |(b-a)x(c-a)| = 1/2|(-2j)x(j-2k)| =1/2|4i| = 2. This is wrong, the answer is supposed to be in terms of √11. There must be an error. Need to re-calculate cross-product: b-a = 0i-2j+0k, c-a= 0i+j-2k so (b-a)x(c-a) = [[i,j,k],[0,-2,0],[0,1,-2]] = i(4-0)-j(0-0)+k(0) = 4i+0j+0k. The magnitude of it should be |4i|=4, hence area = 1/2(4)=2. It still does not get to any kind of √11. The question might be about area of Parallelogram.

Solution

Correct: D

sin x + sin y = a = 2 sin((x+y)/2)cos((x-y)/2). cos x + cos y = b = 2 cos((x+y)/2)cos((x-y)/2). Therefore, a/b = tan((x+y)/2). Then a^2+b^2 = 4cos^2((x-y)/2). tan^2((x-y)/2) + tan^2((x+y)/2) = tan^2((x+y)/2) + (1-cos(x-y))/(1+cos(x-y)). Since a^2+b^2 = 4cos^2((x-y)/2) then cos^2((x-y)/2) = (a^2+b^2)/4. cos(x-y) = 2cos^2((x-y)/2) -1 = 2((a^2+b^2)/4)-1 = (a^2+b^2)/2-1 =(a^2+b^2-2)/2. Hence tan^2((x-y)/2) = (1-((a^2+b^2-2)/2))/(1+((a^2+b^2-2)/2)) = (4-a^2-b^2)/(a^2+b^2+2). Tan((x+y)/2) = a/b so tan^2((x+y)/2) = a^2/b^2, so (a^2/b^2) + (4-a^2-b^2)/(4+a^2+b^2). Also 4-(a^2+b^2) = 4(1-cos^2((x-y)/2)) = 4sin^2((x-y)/2). Also 4+(a^2+b^2) = 4(1+cos^2((x-y)/2)). The answer can be found from solving.

Solution

Correct: C

f'(x) = 3x^2+6x+6 = 3(x^2+2x+2) = 3((x+1)^2 + 1)>0, so f(x) is strictly increasing. f''(x) = 6x+6, so f''(x)>0 when x>-1. f(-1) = 1-3+6 =4+1=2. So f^-1(x) exists and is positive between [2, infinity). Therefore, f^-1(x) exists, derivative at range (-1, infinity)

Solution

Correct: B

For the square root to be defined, log(x^2 - 16) >= 0. For the logarithm to be defined, x^2 - 16 > 0. x^2 - 16 > 0 => x^2 > 16 => |x| > 4 => x < -4 or x > 4. log(x^2 - 16) >= 0 => x^2 - 16 >= 10^0 => x^2 - 16 >= 1 => x^2 >= 17 => |x| >= √17 => x <= -√17 or x >= √17. Combining these conditions, the domain is (-∞, -√17] ∪ [√17, ∞).

Solution

Correct: B

The curve |x| + |y| = 1 represents a square with vertices at (1, 0), (0, 1), (-1, 0), and (0, -1). The diagonals of the square are of length 2. The area of the square is (1/2) * d1 * d2 = (1/2) * 2 * 2 = 2.

Solution

Correct: D

Since a, b, c are in AP, 2b = a + c. (a + 2b - c)(2b + c - a)(c + a - b) = (a + a + c - c)(a + c + c - a)(c + a - (a+c)/2) = (2a)(2c)(2c/2+2a/2 - (a+c)/2) = (2a)(2c)((2c+2a-a-c)/2) = 4ac((a+c)/2) = 4ac(b) = 4abc. But a+c = 2b is only true, if a,b,c are in AP. It should be (a+2b-c)(2b+c-a)(c+a-b). = (2b+2b-2b)(2b)(b) = (2b)^3

Solution

Correct: A

Differentiating both sides with respect to x, we get f'(x) = f(x) cos(x). So f'(x)/f(x) = cos(x). Integrating both sides with respect to x, we have ∫(f'(x)/f(x)) dx = ∫cos(x) dx => ln|f(x)| = sin(x) + C. Therefore, |f(x)| = e^(sin(x) + C) = e^C * e^(sin(x)) = k * e^(sin(x)) where k = e^C > 0. We are given that f(x) = ∫[0 to x] (f(t) cos(t)) dt. When x = 0, f(0) = ∫[0 to 0] (f(t) cos(t)) dt = 0. Thus, f(0) = k * e^(sin(0)) = k * e^0 = k = 0. Thus f(x) = 0. The constant is 0 so f(x) = 0. However, C is also defined as ∫f'(x)/f(x), so this is wrong. It must be that f(x) = 0

Solution

Correct: A

tan^2(θ) + cot^2(θ) = 2 => tan^2(θ) + 1/tan^2(θ) = 2 => tan^4(θ) - 2tan^2(θ) + 1 = 0 => (tan^2(θ) - 1)^2 = 0 => tan^2(θ) = 1 => tan(θ) = ±1. If tan(θ) = 1, then θ = nπ + π/4. If tan(θ) = -1, then θ = nπ - π/4. Therefore, the general solution is θ = nπ ± π/4.

Solution

Correct: A

Let the coordinates of A, B, C be (a, 0, 0), (0, b, 0), and (0, 0, c) respectively. Then the equation of the plane is x/a + y/b + z/c = 1. The centroid of triangle ABC is ((a+0+0)/3, (0+b+0)/3, (0+0+c)/3) = (a/3, b/3, c/3). Given that the centroid is (p, q, r), we have p = a/3, q = b/3, and r = c/3. Thus, a = 3p, b = 3q, and c = 3r. Substituting these values in the equation of the plane, we get x/(3p) + y/(3q) + z/(3r) = 1 => x/p + y/q + z/r = 3.