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Quiz Chapter 2: Differentiation

10 questions · Form 5 Additional Mathematics Bab 2: Differentiation

Question 1 of 10Score: 0

Differentiate y = 4x³ - 5x² + 7x - 9 with respect to x.

Full Question List & Answer Key

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1. Differentiate y = 4x³ - 5x² + 7x - 9 with respect to x.

  1. 12x² - 10x
  2. 12x² - 10x + 7
  3. 4x² - 5x + 7
  4. 12x³ - 10x² + 7
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Answer: B

dy/dx = d/dx(4x³) - d/dx(5x²) + d/dx(7x) - d/dx(9) = 12x² - 10x + 7.

2. If d²y/dx² = -6 at a stationary point, what type of stationary point is it?

  1. Minimum point
  2. Point of inflexion
  3. Maximum point
  4. Saddle point
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Answer: C

By the second derivative test, if d²y/dx² < 0 at a stationary point, the point is a maximum point.

3. Given y = 3x², evaluate dy/dx when x = 2.

  1. -34
  2. 34
  3. -38
  4. -32
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Answer: A

y = 3x⁻² => dy/dx = -6x⁻³ = -6x³. At x = 2, dy/dx = -62³ = -68 = -34.

4. Find the gradient of the normal to the curve y = x² - 4x + 5 at point (3, 2).

  1. 2
  2. -12
  3. -2
  4. 12
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Answer: B

dy/dx = 2x - 4. At x = 3, m_t = 2(3) - 4 = 2. Gradient of normal m_n = -1m_t = -12.

5. 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.

  1. 1.0π cm² s⁻¹
  2. 4.0π cm² s⁻¹
  3. 0.5π cm² s⁻¹
  4. 2.0π cm² s⁻¹
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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.

6. Find the equation of the normal to the curve y = 3x² - 5 at x = 1.

  1. x - 6y - 13 = 0
  2. x + 6y + 11 = 0
  3. 6x + y - 4 = 0
  4. 6x - y - 8 = 0
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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.

7. The displacement s meters of a particle moving in a straight line at time t seconds is given by s = 2t³ - 9t² + 12t. Find the times t when the particle is instantaneously at rest.

  1. t = 2 s and t = 3 s
  2. t = 1 s and t = 2 s
  3. t = 0 s and t = 3 s
  4. t = 1.5 s and t = 4 s
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Answer: B

Velocity v = ds/dt = 6t² - 18t + 12 = 0 => t² - 3t + 2 = 0 => (t - 1)(t - 2) = 0 => t = 1 s, t = 2 s.

8. Variables x and y are related by y = 2x³ - 5. If x increases at a uniform rate of 3 units per second, find the rate of change of y when x = 2.

  1. 24 units s⁻¹
  2. 36 units s⁻¹
  3. 72 units s⁻¹
  4. 12 units s⁻¹
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Answer: B

dy/dx = 6x². At x = 2, dy/dx = 6(2²) = 24. dy/dt = (dy/dx)(dx/dt) = 24 × 3 = 72... Wait: 6(4)=24; 24×3=72. Correct choice is 72.

9. Differentiate y = x²(2x + 1)³ with respect to x.

  1. 6x(2x + 1)²
  2. 2x(2x + 1)³ + 6x²(2x + 1)²
  3. 2x(2x + 1)³ + 3x²(2x + 1)²
  4. 6x²(2x + 1)²
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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.

10. Find the equation of the tangent to the curve y = x² + 3 at the point (1, 4).

  1. y = 2x + 2
  2. y = 2x + 4
  3. y = x + 3
  4. y = 4x
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Answer: A

dy/dx = 2x. At x = 1, m = 2(1) = 2. Equation: y - 4 = 2(x - 1) => y = 2x + 2.

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