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Quiz Chapter 2: Differentiation

10 questions · Form 5 Additional Mathematics Bab 2: Differentiation

10 questions · plays in fullscreen

Full Question List & Answer Key

Prefer reading to quizzing? All 10 questions are listed below with the answer and explanation under each one.

1. 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.

2. Find the gradient of the tangent to the curve y = 2x² - 3x + 1 at the point x = 2.

  1. 8
  2. 3
  3. 5
  4. 1
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Answer: C

dy/dx = 4x - 3. At x = 2, dy/dx = 4(2) - 3 = 5.

3. Evaluate lim (x → 3) [x² - 9x - 3].

  1. 3
  2. 0
  3. 6
  4. Undefined
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Answer: C

lim (x → 3) [(x - 3)x + 3x - 3] = lim (x → 3) (x + 3) = 3 + 3 = 6.

4. If y = (3x - 2)⁴, find dy/dx.

  1. 12(3x - 2)³
  2. 4(3x - 2)³
  3. 3(3x - 2)⁴
  4. 12(3x - 2)⁴
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Answer: A

Using Chain Rule: dy/dx = 4(3x - 2)³ × d/dx(3x - 2) = 4(3x - 2)³(3) = 12(3x - 2)³.

5. 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.

6. Given y = x³, find the small change in y, δy, when x increases from 2 to 2.01.

  1. 0.12
  2. 0.06
  3. 0.01
  4. 0.24
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Answer: A

dy/dx = 3x². At x = 2, dy/dx = 3(2²) = 12. δx = 2.01 - 2 = 0.01. δy ≈ (dy/dx) · δx = 12 × 0.01 = 0.12.

7. 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.

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.

  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.

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. The side length x of a cube increases from 5 cm to 5.02 cm. Find the approximate percentage increase in its volume V.

  1. 1.2%
  2. 0.4%
  3. 0.8%
  4. 1.6%
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Answer: A

V = x³ => dV/dx = 3x². δV ≈ (3x²)δx. Percentage change in V = (δVV) × 100% = [3x² δxx³] × 100% = [3(δxx)] × 100% = 3(0.025) × 100% = 1.2%.

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