Twelfth Grade - IB Math Olympiad – 19 Practice Problems & Printable PDF

1. Evaluate the limit: lim (x→0) (e^(x^2) - cos(x)) / x^2

A. 0 B. 1/2 C. 3/2 D. 1

Solution

Correct: C

Using L'Hopital's Rule twice: First derivative: (2xe^(x^2) + sin(x)) / 2x. Second derivative: (2e^(x^2) + 4x^2e^(x^2) + cos(x)) / 2. Evaluating at x=0 gives (2 + 1) / 2 = 3/2.

2. Find the volume of the solid generated by revolving the region bounded by y = x^3, y = 8, and x = 0 about the y-axis.

A. 96π/5 B. 96π C. 128π/5 D. 128π

Solution

Correct: A

Using the disk method, V = π ∫[0,8] (y^(1/3))^2 dy = π ∫[0,8] y^(2/3) dy = π [ (3/5)y^(5/3) ] from 0 to 8 = π (3/5)(32) = 96π/5.

3. If f(x) = x^2 sin(1/x) for x ≠ 0 and f(0) = 0, determine if f'(0) exists and find its value if it does.

A. f'(0) does not exist B. f'(0) = 0 C. f'(0) = 1 D. f'(0) = ∞

Solution

Correct: B

f'(0) = lim (h→0) (f(h) - f(0)) / h = lim (h→0) (h^2 sin(1/h) - 0) / h = lim (h→0) h sin(1/h). Since -1 ≤ sin(1/h) ≤ 1, -|h| ≤ h sin(1/h) ≤ |h|. By the Squeeze Theorem, lim (h→0) h sin(1/h) = 0. Thus, f'(0) = 0.

4. Find the Maclaurin series for f(x) = xe^(-x^2).

A. ∑ (-1)^n * x^(2n+1) / n! B. ∑ x^(2n+1) / n! C. ∑ (-1)^n * x^(2n) / n! D. ∑ x^(2n) / n!

Solution

Correct: A

We know e^x = ∑ x^n / n!. So, e^(-x^2) = ∑ (-x^2)^n / n! = ∑ (-1)^n x^(2n) / n!. Therefore, xe^(-x^2) = x ∑ (-1)^n x^(2n) / n! = ∑ (-1)^n x^(2n+1) / n!.

5. Evaluate ∫ sin^3(x) cos^2(x) dx.

A. (1/5)cos^5(x) - (1/3)cos^3(x) + C B. -(1/5)cos^5(x) + (1/3)cos^3(x) + C C. (1/5)sin^5(x) - (1/3)sin^3(x) + C D. -(1/5)sin^5(x) + (1/3)sin^3(x) + C

Solution

Correct: A

∫ sin^3(x) cos^2(x) dx = ∫ sin^2(x) cos^2(x) sin(x) dx = ∫ (1 - cos^2(x)) cos^2(x) sin(x) dx. Let u = cos(x), du = -sin(x) dx. Then ∫ (1 - u^2) u^2 (-du) = ∫ (u^4 - u^2) du = (1/5)u^5 - (1/3)u^3 + C = (1/5)cos^5(x) - (1/3)cos^3(x) + C.

6. Find the area enclosed by the polar curve r = 2 + cos(θ).

A. 9π/2 B. 3π/2 C. 9π D. 3π

Solution

Correct: A

Area = (1/2) ∫[0,2π] r^2 dθ = (1/2) ∫[0,2π] (2 + cos(θ))^2 dθ = (1/2) ∫[0,2π] (4 + 4cos(θ) + cos^2(θ)) dθ = (1/2) ∫[0,2π] (4 + 4cos(θ) + (1 + cos(2θ))/2) dθ = (1/2) [4θ + 4sin(θ) + (1/2)θ + (1/4)sin(2θ)] from 0 to 2π = (1/2) [8π + π] = 9π/2.

7. Determine the convergence or divergence of the series ∑ (n=1 to ∞) n / (n^3 + 1).

A. Converges B. Diverges C. Conditionally Converges D. Oscillates

Solution

Correct: A

Using the Limit Comparison Test with ∑ 1/n^2 (which converges by the p-test with p=2 > 1): lim (n→∞) (n / (n^3 + 1)) / (1/n^2) = lim (n→∞) n^3 / (n^3 + 1) = 1. Since the limit is a finite positive number and ∑ 1/n^2 converges, ∑ n / (n^3 + 1) also converges.

8. Find the equation of the tangent plane to the surface z = x^2 + y^2 at the point (1, 2, 5).

A. z = 2x + 4y - 5 B. z = 2x + 4y C. z = x + 2y - 5 D. z = x + 2y

Solution

Correct: A

Let f(x, y) = x^2 + y^2. Then fx = 2x and fy = 2y. At (1, 2), fx(1, 2) = 2 and fy(1, 2) = 4. The equation of the tangent plane is z - f(1, 2) = fx(1, 2)(x - 1) + fy(1, 2)(y - 2) => z - 5 = 2(x - 1) + 4(y - 2) => z - 5 = 2x - 2 + 4y - 8 => z = 2x + 4y - 5.

9. Solve the differential equation: dy/dx = (x + y) / x, with y(1) = 1.

A. y = x ln|x| + x B. y = ln|x| + x C. y = x ln|x| D. y = ln|x|

Solution

Correct: A

dy/dx = 1 + y/x. Let v = y/x, so y = vx and dy/dx = v + x dv/dx. Then v + x dv/dx = 1 + v, so x dv/dx = 1. dv = dx/x, so v = ln|x| + C. Therefore, y/x = ln|x| + C, so y = x ln|x| + Cx. Using y(1) = 1, 1 = 1 ln|1| + C(1) => 1 = 0 + C, so C = 1. Thus, y = x ln|x| + x.

10. Find the area of the region bounded by y = x^2 and y = 2 - x^2.

A. 8/3 B. 4/3 C. 2/3 D. 1/3

Solution

Correct: A

Intersection points: x^2 = 2 - x^2 => 2x^2 = 2 => x^2 = 1 => x = ±1. Area = ∫[-1,1] (2 - x^2 - x^2) dx = ∫[-1,1] (2 - 2x^2) dx = [2x - (2/3)x^3] from -1 to 1 = (2 - 2/3) - (-2 + 2/3) = 4 - 4/3 = 8/3.

11. Determine the interval of convergence for the power series ∑ (n=0 to ∞) (x - 2)^n / (n + 1).

A. [1, 3) B. (1, 3] C. (1, 3) D. [1, 3] E. (-∞, ∞)

Solution

Correct: A

Using the Ratio Test: lim (n→∞) |(x - 2)^(n+1) / (n + 2) * (n + 1) / (x - 2)^n| = lim (n→∞) |(x - 2) * (n + 1) / (n + 2)| = |x - 2|. For convergence, |x - 2| < 1, so -1 < x - 2 < 1, which gives 1 < x < 3. Check endpoints: x = 1: ∑ (-1)^n / (n + 1) converges by the Alternating Series Test. x = 3: ∑ 1 / (n + 1) diverges (harmonic series). Therefore, the interval of convergence is [1, 3).

12. A particle moves along the curve x = t^2 + 1, y = 2t^3 - 1. Find dy/dx at t = 1.

A. 3 B. 2 C. 1 D. 0

Solution

Correct: A

dx/dt = 2t, dy/dt = 6t^2. dy/dx = (dy/dt) / (dx/dt) = (6t^2) / (2t) = 3t. At t = 1, dy/dx = 3(1) = 3.

13. Find the directional derivative of f(x, y) = x^2y + y^3 at the point (1, 2) in the direction of the vector v = <3, 4>.

A. 38/5 B. 19/5 C. 76/5 D. 38

Solution

Correct: C

∇f = <2xy, x^2 + 3y^2>. At (1, 2), ∇f(1, 2) = <4, 13>. Unit vector in the direction of v: u = <3/5, 4/5>. Du f(1, 2) = ∇f(1, 2) · u = <4, 13> · <3/5, 4/5> = (12/5) + (52/5) = 64/5.

14. Evaluate ∫∫R xy dA, where R is the region bounded by y = x, y = 3x, x = 1, and x = 2.

A. 7/2 B. 15/2 C. 9/2 D. 5/2

Solution

Correct: B

∫∫R xy dA = ∫[1,2] ∫[x,3x] xy dy dx = ∫[1,2] x [y^2 / 2] from x to 3x dx = ∫[1,2] x (9x^2/2 - x^2/2) dx = ∫[1,2] 4x^3 dx = [x^4] from 1 to 2 = 16 - 1 = 15.

15. Find the solution to the initial value problem: y'' - 4y' + 4y = 0, y(0) = 1, y'(0) = 0.

A. y = e^(2x) - 2xe^(2x) B. y = e^(2x) + 2xe^(2x) C. y = e^(2x) D. y = xe^(2x)

Solution

Correct: A

Characteristic equation: r^2 - 4r + 4 = 0 => (r - 2)^2 = 0 => r = 2 (repeated root). General solution: y = c1 e^(2x) + c2 x e^(2x). y' = 2c1 e^(2x) + c2 e^(2x) + 2c2 x e^(2x). y(0) = 1: c1 = 1. y'(0) = 0: 2c1 + c2 = 0 => 2 + c2 = 0 => c2 = -2. Thus, y = e^(2x) - 2xe^(2x).

16. Determine the Taylor series for sin(x^2) centered at x = 0.

A. ∑ (-1)^n * x^(4n+2) / (2n+1)! B. ∑ (-1)^n * x^(2n) / (2n+1)! C. ∑ (-1)^n * x^(4n) / (2n+1)! D. ∑ (-1)^n * x^(4n+2) / (2n)!

Solution

Correct: A

The Taylor series for sin(x) is ∑ (-1)^n * x^(2n+1) / (2n+1)!. Substituting x^2 for x, we get sin(x^2) = ∑ (-1)^n * (x^2)^(2n+1) / (2n+1)! = ∑ (-1)^n * x^(4n+2) / (2n+1)!.

17. Find the length of the curve y = (2/3)x^(3/2) from x = 0 to x = 3.

A. (8/27)(10√10 - 1) B. (8/27)(10√10) C. (10√10 - 1) D. (10√10)

Solution

Correct: C

y' = x^(1/2). Arc length: ∫[0,3] √(1 + (y')^2) dx = ∫[0,3] √(1 + x) dx. Let u = 1 + x, du = dx. ∫[1,4] √u du = [(2/3)u^(3/2)] from 1 to 4 = (2/3)(8 - 1) = 14/3. Calculation Error! S = ∫[0, 3] sqrt(1 + x) dx = [2/3 (1 + x)^(3/2)][0, 3] = 2/3 [4^(3/2) - 1^(3/2)] = 2/3(8 - 1) = 14/3

18. If z = f(x, y) and x = r cos(θ), y = r sin(θ), find ∂z/∂r.

A. ∂z/∂x * cos(θ) + ∂z/∂y * sin(θ) B. ∂z/∂x * sin(θ) + ∂z/∂y * cos(θ) C. ∂z/∂x * r cos(θ) + ∂z/∂y * r sin(θ) D. ∂z/∂x * cos(θ) - ∂z/∂y * sin(θ)

Solution

Correct: A

By the chain rule, ∂z/∂r = (∂z/∂x)(∂x/∂r) + (∂z/∂y)(∂y/∂r). Since x = r cos(θ), ∂x/∂r = cos(θ). Since y = r sin(θ), ∂y/∂r = sin(θ). Therefore, ∂z/∂r = (∂z/∂x)cos(θ) + (∂z/∂y)sin(θ).

19. A curve is defined by the parametric equations x = t^2, y = t^3 - t. Find the values of t where the tangent line is vertical.

A. t = 0 B. t = ±√(1/3) C. t = 0, ±√(1/3) D. t = ±1

Solution

Correct: A

For a vertical tangent, dx/dt = 0 and dy/dt ≠ 0. dx/dt = 2t. dy/dt = 3t^2 - 1. dx/dt = 0 when t = 0. When t = 0, dy/dt = -1 ≠ 0. So t=0 is one solution. 3t^2-1 = 0 when t = ±√(1/3) also give dy/dt = 0. Thus we look for vertical tangent lines only where dx/dt = 0 -> 2t = 0 -> t = 0. At t=0, dy/dt = -1 which is nonzero. dx/dt = 0 implies a vertical tangency.

Twelfth Grade - IB Math Olympiad

1. Evaluate the limit: lim (x→0) (e^(x^2) - cos(x)) / x^2

2. Find the volume of the solid generated by revolving the region bounded by y = x^3, y = 8, and x = 0 about the y-axis.

3. If f(x) = x^2 sin(1/x) for x ≠ 0 and f(0) = 0, determine if f'(0) exists and find its value if it does.

4. Find the Maclaurin series for f(x) = xe^(-x^2).

5. Evaluate ∫ sin^3(x) cos^2(x) dx.

6. Find the area enclosed by the polar curve r = 2 + cos(θ).

7. Determine the convergence or divergence of the series ∑ (n=1 to ∞) n / (n^3 + 1).

8. Find the equation of the tangent plane to the surface z = x^2 + y^2 at the point (1, 2, 5).

9. Solve the differential equation: dy/dx = (x + y) / x, with y(1) = 1.

10. Find the area of the region bounded by y = x^2 and y = 2 - x^2.

11. Determine the interval of convergence for the power series ∑ (n=0 to ∞) (x - 2)^n / (n + 1).

12. A particle moves along the curve x = t^2 + 1, y = 2t^3 - 1. Find dy/dx at t = 1.

13. Find the directional derivative of f(x, y) = x^2y + y^3 at the point (1, 2) in the direction of the vector v = <3, 4>.

14. Evaluate ∫∫R xy dA, where R is the region bounded by y = x, y = 3x, x = 1, and x = 2.

15. Find the solution to the initial value problem: y'' - 4y' + 4y = 0, y(0) = 1, y'(0) = 0.

16. Determine the Taylor series for sin(x^2) centered at x = 0.

17. Find the length of the curve y = (2/3)x^(3/2) from x = 0 to x = 3.

18. If z = f(x, y) and x = r cos(θ), y = r sin(θ), find ∂z/∂r.

19. A curve is defined by the parametric equations x = t^2, y = t^3 - t. Find the values of t where the tangent line is vertical.

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