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Secondary 3 Additional Mathematics Calculus Quiz

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Questions

Secondary 3 Additional Mathematics Quiz - Calculus

Name: _______________________________
Class: _______________________________
Date: _______________________________
Score: ________ / 60

Duration: 1 hour 15 minutes
Total Marks: 60

Instructions:

Answer ALL questions.
Show all working clearly. Marks are awarded for method.
Non-exact answers should be given to 3 significant figures unless otherwise stated.
Calculators are allowed.
The number of marks for each question or part is shown in brackets [ ].

Section A: Differentiation Techniques (15 marks)

Answer all questions in this section.

1. Differentiate with respect to (x):

(a) (y = 3x^4 - 2x^3 + 5x - 7)
[2]

(b) (y = \frac{2}{x^3} + \sqrt{x})
[2]

2. Find (\frac{dy}{dx}) for each of the following:

(a) (y = (2x + 1)(x^2 - 3))
[3]

(b) (y = \frac{x^2 + 1}{x - 2})
[3]

3. Use the Chain Rule to differentiate:

(a) (y = (3x^2 - 5)^4)
[2]

(b) (y = \sqrt{2x + 7})
[3]

4. Differentiate (y = \frac{1}{x^2} + 3x) with respect to (x).
[2]

5. Find the derivative of (y = x(2x - 1)^3).
[3]

Section B: Applications of Differentiation (15 marks)

Answer all questions in this section.

6. Find the equation of the tangent to the curve (y = x^3 - 6x^2 + 9x + 1) at the point where (x = 2).
[4]

7. Find the equation of the normal to the curve (y = \frac{4}{x}) at the point where (x = 2).
[4]

8. A curve has equation (y = 2x^3 - 3x^2 - 12x + 5).

(a) Find the coordinates of the stationary points of the curve.
[4]

(b) Determine the nature of each stationary point using the second derivative test.
[3]

9. The curve (y = x^3 + ax^2 + bx + 2) has a stationary point at ((1, 4)). Find the values of (a) and (b).
[5]

10. The volume, (V) cm³, of a spherical balloon is increasing at a constant rate of (4\pi) cm³/s. Find the rate of increase of the radius when the radius is 5 cm.
[4]

Section C: Integration (15 marks)

Answer all questions in this section.

11. Find each of the following indefinite integrals:

(a) (\displaystyle \int (4x^3 - 6x^2 + 2x - 1) , dx)
[2]

(b) (\displaystyle \int \left( \frac{3}{x^2} + \frac{1}{\sqrt{x}} \right) dx)
[3]

12. Evaluate:

(a) (\displaystyle \int_1^3 (2x^2 - 4x + 1) , dx)
[3]

(b) (\displaystyle \int_0^2 (x^3 - 3x^2 + 2) , dx)
[3]

13. The gradient of a curve is given by (\frac{dy}{dx} = 6x^2 - 2x + 3). The curve passes through the point ((1, 8)). Find the equation of the curve.
[4]

14. Find (\displaystyle \int (x + 1)(x - 2) , dx).
[3]

15. Evaluate (\displaystyle \int_0^1 (3x^2 - 2x + 1) , dx).
[2]

Section D: Applications of Integration (15 marks)

Answer all questions in this section.

16. Find the area of the region bounded by the curve (y = x^2 - 4x + 3), the (x)-axis, and the lines (x = 1) and (x = 4).
[5]

17. The diagram shows part of the curve (y = 4x - x^2).

(a) Find the coordinates of the points where the curve meets the (x)-axis.
[2]

(b) Calculate the area of the region enclosed by the curve and the (x)-axis.
[3]

18. Find the area bounded by the curve (y = x^2) and the line (y = 2x + 3).
[5]

19. A particle moves along a straight line. Its velocity, (v) m/s, is given by (v = 3t^2 - 4t + 2), where (t) is the time in seconds. Find the displacement of the particle between (t = 1) and (t = 3).
[3]

20. The curve (y = x^3 - 3x^2 + 2x) is shown. Find the total area enclosed between the curve and the (x)-axis from (x = 0) to (x = 2).
[4]

END OF QUIZ

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Answers

Secondary 3 Additional Mathematics Quiz - Calculus - ANSWERS

Section A: Differentiation Techniques (15 marks)

1. (a) (y = 3x^4 - 2x^3 + 5x - 7)
(\frac{dy}{dx} = 12x^3 - 6x^2 + 5) [2]

(b) (y = \frac{2}{x^3} + \sqrt{x} = 2x^{-3} + x^{1/2})
(\frac{dy}{dx} = -6x^{-4} + \frac{1}{2}x^{-1/2} = -\frac{6}{x^4} + \frac{1}{2\sqrt{x}}) [2]

2. (a) (y = (2x + 1)(x^2 - 3))
Using product rule: (u = 2x + 1, v = x^2 - 3)
(u' = 2, v' = 2x)
(\frac{dy}{dx} = 2(x^2 - 3) + (2x + 1)(2x) = 2x^2 - 6 + 4x^2 + 2x = 6x^2 + 2x - 6) [3]

(b) (y = \frac{x^2 + 1}{x - 2})
Using quotient rule: (u = x^2 + 1, v = x - 2)
(u' = 2x, v' = 1)
(\frac{dy}{dx} = \frac{2x(x - 2) - (x^2 + 1)(1)}{(x - 2)^2} = \frac{2x^2 - 4x - x^2 - 1}{(x - 2)^2} = \frac{x^2 - 4x - 1}{(x - 2)^2}) [3]

3. (a) (y = (3x^2 - 5)^4)
Let (u = 3x^2 - 5), then (y = u^4)
(\frac{dy}{du} = 4u^3, \frac{du}{dx} = 6x)
(\frac{dy}{dx} = 4(3x^2 - 5)^3 \cdot 6x = 24x(3x^2 - 5)^3) [2]

(b) (y = \sqrt{2x + 7} = (2x + 7)^{1/2})
Let (u = 2x + 7), then (y = u^{1/2})
(\frac{dy}{du} = \frac{1}{2}u^{-1/2}, \frac{du}{dx} = 2)
(\frac{dy}{dx} = \frac{1}{2}(2x + 7)^{-1/2} \cdot 2 = \frac{1}{\sqrt{2x + 7}}) [3]

4. (y = \frac{1}{x^2} + 3x = x^{-2} + 3x)
(\frac{dy}{dx} = -2x^{-3} + 3 = -\frac{2}{x^3} + 3) [2]

5. (y = x(2x - 1)^3)
Using product rule: (u = x, v = (2x - 1)^3)
(u' = 1, v' = 3(2x - 1)^2 \cdot 2 = 6(2x - 1)^2)
(\frac{dy}{dx} = 1 \cdot (2x - 1)^3 + x \cdot 6(2x - 1)^2 = (2x - 1)^2[(2x - 1) + 6x] = (2x - 1)^2(8x - 1)) [3]

Section B: Applications of Differentiation (15 marks)

6. (y = x^3 - 6x^2 + 9x + 1)
(\frac{dy}{dx} = 3x^2 - 12x + 9)
At (x = 2): (\frac{dy}{dx} = 3(4) - 12(2) + 9 = 12 - 24 + 9 = -3)
(y = 2^3 - 6(2^2) + 9(2) + 1 = 8 - 24 + 18 + 1 = 3)
Point: ((2, 3)), gradient = -3
Equation: (y - 3 = -3(x - 2))
(y = -3x + 6 + 3 = -3x + 9) [4]

7. (y = \frac{4}{x} = 4x^{-1})
(\frac{dy}{dx} = -4x^{-2} = -\frac{4}{x^2})
At (x = 2): (\frac{dy}{dx} = -\frac{4}{4} = -1)
(y = \frac{4}{2} = 2)
Point: ((2, 2)), gradient of tangent = -1
Gradient of normal = 1
Equation: (y - 2 = 1(x - 2))
(y = x) [4]

8. (y = 2x^3 - 3x^2 - 12x + 5)
(a) (\frac{dy}{dx} = 6x^2 - 6x - 12 = 6(x^2 - x - 2) = 6(x - 2)(x + 1))
Stationary points when (\frac{dy}{dx} = 0): (x = 2) or (x = -1)
At (x = 2): (y = 2(8) - 3(4) - 12(2) + 5 = 16 - 12 - 24 + 5 = -15)
At (x = -1): (y = 2(-1) - 3(1) - 12(-1) + 5 = -2 - 3 + 12 + 5 = 12)
Coordinates: ((2, -15)) and ((-1, 12)) [4]

(b) (\frac{d^2y}{dx^2} = 12x - 6)
At (x = 2): (\frac{d^2y}{dx^2} = 24 - 6 = 18 > 0) → minimum point ((2, -15))
At (x = -1): (\frac{d^2y}{dx^2} = -12 - 6 = -18 < 0) → maximum point ((-1, 12)) [3]

9. (y = x^3 + ax^2 + bx + 2)
(\frac{dy}{dx} = 3x^2 + 2ax + b)
At stationary point ((1, 4)):
(\frac{dy}{dx} = 0): (3(1)^2 + 2a(1) + b = 0) → (3 + 2a + b = 0) ... (1)
Point lies on curve: (4 = 1^3 + a(1)^2 + b(1) + 2) → (4 = 1 + a + b + 2) → (a + b = 1) ... (2)
From (2): (b = 1 - a)
Sub into (1): (3 + 2a + (1 - a) = 0) → (4 + a = 0) → (a = -4)
Then (b = 1 - (-4) = 5)
(a = -4, b = 5) [5]

10. Volume of sphere: (V = \frac{4}{3}\pi r^3)
(\frac{dV}{dt} = 4\pi r^2 \frac{dr}{dt})
Given (\frac{dV}{dt} = 4\pi) and (r = 5):
(4\pi = 4\pi (5)^2 \frac{dr}{dt})
(4\pi = 100\pi \frac{dr}{dt})
(\frac{dr}{dt} = \frac{4\pi}{100\pi} = 0.04) cm/s [4]

Section C: Integration (15 marks)

11. (a) (\int (4x^3 - 6x^2 + 2x - 1) dx = x^4 - 2x^3 + x^2 - x + C) [2]

(b) (\int \left( \frac{3}{x^2} + \frac{1}{\sqrt{x}} \right) dx = \int (3x^{-2} + x^{-1/2}) dx)
(= -3x^{-1} + 2x^{1/2} + C = -\frac{3}{x} + 2\sqrt{x} + C) [3]

12. (a) (\int_1^3 (2x^2 - 4x + 1) dx = \left[ \frac{2}{3}x^3 - 2x^2 + x \right]_1^3)
(= \left( \frac{2}{3}(27) - 2(9) + 3 \right) - \left( \frac{2}{3}(1) - 2(1) + 1 \right))
(= (18 - 18 + 3) - \left( \frac{2}{3} - 2 + 1 \right) = 3 - \left( -\frac{1}{3} \right) = 3 + \frac{1}{3} = \frac{10}{3}) [3]

(b) (\int_0^2 (x^3 - 3x^2 + 2) dx = \left[ \frac{1}{4}x^4 - x^3 + 2x \right]_0^2)
(= \left( \frac{1}{4}(16) - 8 + 4 \right) - 0 = 4 - 8 + 4 = 0) [3]

13. (\frac{dy}{dx} = 6x^2 - 2x + 3)
(y = \int (6x^2 - 2x + 3) dx = 2x^3 - x^2 + 3x + C)
Passes through ((1, 8)): (8 = 2(1)^3 - (1)^2 + 3(1) + C)
(8 = 2 - 1 + 3 + C = 4 + C) → (C = 4)
Equation: (y = 2x^3 - x^2 + 3x + 4) [4]

14. ((x + 1)(x - 2) = x^2 - x - 2)
(\int (x^2 - x - 2) dx = \frac{1}{3}x^3 - \frac{1}{2}x^2 - 2x + C) [3]

15. (\int_0^1 (3x^2 - 2x + 1) dx = \left[ x^3 - x^2 + x \right]_0^1)
(= (1 - 1 + 1) - 0 = 1) [2]

Section D: Applications of Integration (15 marks)

16. Area bounded by (y = x^2 - 4x + 3), (x)-axis, (x = 1) and (x = 4).
Check for crossing (x)-axis: (x^2 - 4x + 3 = 0) → ((x - 1)(x - 3) = 0) → (x = 1, 3)
Between (x = 1) and (x = 3), curve is below (x)-axis (negative).
Between (x = 3) and (x = 4), curve is above (x)-axis (positive).
Area = (\int_1^3 -(x^2 - 4x + 3) dx + \int_3^4 (x^2 - 4x + 3) dx)
(= \int_1^3 (-x^2 + 4x - 3) dx + \int_3^4 (x^2 - 4x + 3) dx)
(= \left[ -\frac{1}{3}x^3 + 2x^2 - 3x \right]_1^3 + \left[ \frac{1}{3}x^3 - 2x^2 + 3x \right]_3^4)
First part: (\left( -\frac{27}{3} + 18 - 9 \right) - \left( -\frac{1}{3} + 2 - 3 \right) = (-9 + 18 - 9) - \left( -\frac{4}{3} \right) = 0 + \frac{4}{3} = \frac{4}{3})
Second part: (\left( \frac{64}{3} - 32 + 12 \right) - \left( \frac{27}{3} - 18 + 9 \right) = \left( \frac{64}{3} - 20 \right) - (9 - 18 + 9) = \left( \frac{64}{3} - \frac{60}{3} \right) - 0 = \frac{4}{3})
Total area = (\frac{4}{3} + \frac{4}{3} = \frac{8}{3}) square units [5]

17. (y = 4x - x^2)
(a) Meets (x)-axis when (y = 0): (4x - x^2 = 0) → (x(4 - x) = 0) → (x = 0, 4)
Coordinates: ((0, 0)) and ((4, 0)) [2]

(b) Area = (\int_0^4 (4x - x^2) dx = \left[ 2x^2 - \frac{1}{3}x^3 \right]_0^4)
(= \left( 2(16) - \frac{64}{3} \right) - 0 = 32 - \frac{64}{3} = \frac{96 - 64}{3} = \frac{32}{3}) square units [3]

18. Intersection of (y = x^2) and (y = 2x + 3):
(x^2 = 2x + 3) → (x^2 - 2x - 3 = 0) → ((x - 3)(x + 1) = 0) → (x = -1, 3)
Area = (\int_{-1}^3 [(2x + 3) - x^2] dx = \int_{-1}^3 (-x^2 + 2x + 3) dx)
(= \left[ -\frac{1}{3}x^3 + x^2 + 3x \right]_{-1}^3)
(= \left( -\frac{27}{3} + 9 + 9 \right) - \left( \frac{1}{3} + 1 - 3 \right))
(= (-9 + 18) - \left( \frac{1}{3} - 2 \right) = 9 - \left( -\frac{5}{3} \right) = 9 + \frac{5}{3} = \frac{32}{3}) square units [5]

19. Displacement = (\int_1^3 v , dt = \int_1^3 (3t^2 - 4t + 2) dt)
(= \left[ t^3 - 2t^2 + 2t \right]_1^3)
(= (27 - 18 + 6) - (1 - 2 + 2) = 15 - 1 = 14) m [3]

20. (y = x^3 - 3x^2 + 2x = x(x^2 - 3x + 2) = x(x - 1)(x - 2))
Curve crosses (x)-axis at (x = 0, 1, 2).
From (x = 0) to (x = 1), curve is above (x)-axis.
From (x = 1) to (x = 2), curve is below (x)-axis.
Total area = (\int_0^1 (x^3 - 3x^2 + 2x) dx + \int_1^2 -(x^3 - 3x^2 + 2x) dx)
(= \left[ \frac{1}{4}x^4 - x^3 + x^2 \right]_0^1 + \left[ -\frac{1}{4}x^4 + x^3 - x^2 \right]_1^2)
First part: (\left( \frac{1}{4} - 1 + 1 \right) - 0 = \frac{1}{4})
Second part: (\left( -\frac{16}{4} + 8 - 4 \right) - \left( -\frac{1}{4} + 1 - 1 \right) = (-4 + 8 - 4) - \left( -\frac{1}{4} \right) = 0 + \frac{1}{4} = \frac{1}{4})
Total area = (\frac{1}{4} + \frac{1}{4} = \frac{1}{2}) square units [4]

END OF ANSWERS