10 questions · Form 5 Additional Mathematics Bab 2: Differentiation
10 questions · plays in fullscreen
Prefer reading to quizzing? All 10 questions are listed below with the answer and explanation under each one.
1. Find the equation of the normal to the curve y = 3x² - 5 at x = 1.
Answer: B
At x = 1, y = 3(1)² - 5 = -2. dy/dx = 6x => m_t = 6(1) = 6. Gradient of normal m_n = -16. Equation: y - (-2) = -16(x - 1) => 6y + 12 = -x + 1 => x + 6y + 11 = 0.
2. Given y = x³ - 6x² + 9x + 2, find the x-coordinates of the stationary points.
Answer: C
dy/dx = 3x² - 12x + 9 = 0 => x² - 4x + 3 = 0 => (x - 1)(x - 3) = 0 => x = 1, x = 3.
3. Differentiate y = x²(2x + 1)³ with respect to x.
Answer: B
Using Product Rule (u = x², v = (2x+1)³): dy/dx = x³ derivative... = x²[3(2x+1)²(2)] + (2x+1)³[2x] = 6x²(2x+1)² + 2x(2x+1)³.
4. Find the derivative of y = 2x + 1x - 3 with respect to x.
Answer: A
Using Quotient Rule (u = 2x+1, v = x-3): dy/dx = (x-3)(2) - (2x+1)(1)x-3² = 2x - 6 - 2x - 1x-3² = -7x-3².
5. Find the second derivative d²y/dx² for y = 2x⁵ - 4x³ + x.
Answer: A
dy/dx = 10x⁴ - 12x² + 1. d²y/dx² = d/dx(10x⁴ - 12x² + 1) = 40x³ - 24x.
6. Differentiate y = 4x³ - 5x² + 7x - 9 with respect to x.
Answer: B
dy/dx = d/dx(4x³) - d/dx(5x²) + d/dx(7x) - d/dx(9) = 12x² - 10x + 7.
7. Find the equation of the tangent to the curve y = x² + 3 at the point (1, 4).
Answer: A
dy/dx = 2x. At x = 1, m = 2(1) = 2. Equation: y - 4 = 2(x - 1) => y = 2x + 2.
8. The radius r of a circle increases at a rate of 0.2 cm s⁻¹. Find the rate of change of its area A when r = 5 cm.
Answer: D
A = π r² => dA/dr = 2π r. Using Chain Rule: dA/dt = (dA/dr)(dr/dt) = (2π × 5)(0.2) = 2.0π cm² s⁻¹.
9. Find the gradient of the normal to the curve y = x² - 4x + 5 at point (3, 2).
Answer: B
dy/dx = 2x - 4. At x = 3, m_t = 2(3) - 4 = 2. Gradient of normal m_n = -1m_t = -12.
10. Given y = 3x², evaluate dy/dx when x = 2.
Answer: A
y = 3x⁻² => dy/dx = -6x⁻³ = -6x³. At x = 2, dy/dx = -62³ = -68 = -34.