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A Level H1 Mathematics Calculus Quiz

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Questions

A-Level Maths H1 Quiz - Calculus

Name: _________________________ Class: _________________________ Date: _________________________ Score: ______ / 50

Duration: 1 hour 15 minutes Total Marks: 50

Instructions:

Answer ALL questions.
Show all working clearly. Marks are awarded for method.
You may use an approved graphing calculator (without CAS) unless stated otherwise.
Where unsupported answers from a GC are obtained, show appropriate mathematical notation.
Give non-exact answers to 3 significant figures unless otherwise stated.

Section A: Differentiation Techniques (Questions 1–5, 12 marks)

1. Differentiate ( y = 5x^4 - 3x^2 + 2x - 7 ) with respect to ( x ).

[2 marks]

2. Find ( \frac{dy}{dx} ) for ( y = \frac{2x + 1}{x^2 - 3} ). Simplify your answer.

[3 marks]

3. Differentiate ( y = e^{3x} \ln x ) with respect to ( x ). Give your answer in its simplest factorised form.

[3 marks]

4. Find the derivative of ( y = \sqrt{2x + 5} ) with respect to ( x ).

[2 marks]

5. A curve has equation ( y = x^3 - 6x^2 + 9x + 4 ). Find the ( x )-coordinates of the stationary points on the curve.

[2 marks]

Section B: Applications of Differentiation (Questions 6–10, 13 marks)

6. Find the equation of the tangent to the curve ( y = 2x^3 - x^2 + 4 ) at the point where ( x = 1 ). Give your answer in the form ( y = mx + c ).

[3 marks]

7. A curve has equation ( y = x^4 - 4x^3 + 4x^2 ). Determine the nature of each stationary point on the curve.

[3 marks]

8. The profit ( P ) (in thousands of dollars) from selling ( x ) hundred units of a product is modelled by [ P(x) = 80x - 2x^2 - 200, \quad 0 \leq x \leq 30. ] Find the number of units that should be sold to maximise profit, and state the maximum profit.

[3 marks]

9. A rectangular enclosure is to be built against an existing wall. The other three sides require fencing. Given that 120 metres of fencing is available, find the dimensions of the enclosure that maximise the enclosed area.

[2 marks]

10. The cost ( C ) (in dollars) of producing ( q ) items is given by ( C(q) = 500 + 20q + 0.5q^2 ). Find the production level that minimises the average cost per item, ( \frac{C(q)}{q} ).

[2 marks]

Section C: Integration (Questions 11–15, 12 marks)

11. Find ( \displaystyle \int (6x^2 - 4x + 3) , dx ).

[2 marks]

12. Evaluate ( \displaystyle \int_1^4 \left( 2x + \frac{3}{x^2} \right) dx ).

[3 marks]

13. Find ( \displaystyle \int e^{2x+1} , dx ).

[2 marks]

14. Find ( \displaystyle \int (3x - 2)^5 , dx ).

[2 marks]

15. Evaluate ( \displaystyle \int_0^2 3e^{0.5x} , dx ). Give your answer correct to 3 significant figures.

[3 marks]

Section D: Applications of Integration (Questions 16–20, 13 marks)

16. The diagram shows the curve ( y = x^2 + 1 ) and the line ( y = 5 ).

Find the area of the region bounded by the curve, the line ( y = 5 ), and the ( y )-axis.

[3 marks]

17. A curve has equation ( y = 4x - x^2 ). Find the area of the region bounded by the curve and the ( x )-axis.

[3 marks]

18. Find the area enclosed between the curve ( y = e^x ), the ( x )-axis, and the lines ( x = 0 ) and ( x = \ln 3 ).

[2 marks]

19. The marginal revenue function for a product is given by ( R'(q) = 50 - 0.4q ), where ( q ) is the quantity sold. Find the total revenue function ( R(q) ), given that ( R(0) = 0 ).

[2 marks]

20. The rate of change of a population ( P ) (in thousands) with respect to time ( t ) (in years) is modelled by [ \frac{dP}{dt} = 0.5e^{0.1t}. ] Given that the initial population is 20 thousand, find the population after 10 years. Give your answer correct to the nearest thousand.

[3 marks]

END OF QUIZ

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Answers

A-Level Maths H1 Quiz - Calculus — Answers and Marking Scheme

Total Marks: 50

Section A: Differentiation Techniques

1. ( y = 5x^4 - 3x^2 + 2x - 7 )

( \frac{dy}{dx} = 20x^3 - 6x + 2 )

[M2] Award M1 for correct differentiation of at least two terms; A1 for fully correct simplified answer.

2. ( y = \frac{2x + 1}{x^2 - 3} )

Using quotient rule: ( u = 2x + 1, \quad u' = 2; \quad v = x^2 - 3, \quad v' = 2x )

( \frac{dy}{dx} = \frac{(x^2 - 3)(2) - (2x + 1)(2x)}{(x^2 - 3)^2} )

( = \frac{2x^2 - 6 - 4x^2 - 2x}{(x^2 - 3)^2} )

( = \frac{-2x^2 - 2x - 6}{(x^2 - 3)^2} = \frac{-2(x^2 + x + 3)}{(x^2 - 3)^2} )

[M3] M1 for correct application of quotient rule; M1 for correct expansion of numerator; A1 for fully simplified answer.

3. ( y = e^{3x} \ln x )

Using product rule: ( u = e^{3x}, \quad u' = 3e^{3x}; \quad v = \ln x, \quad v' = \frac{1}{x} )

( \frac{dy}{dx} = 3e^{3x} \ln x + e^{3x} \cdot \frac{1}{x} )

( = e^{3x}\left(3\ln x + \frac{1}{x}\right) )

[M3] M1 for correct product rule setup; M1 for correct derivatives of ( e^{3x} ) and ( \ln x ); A1 for fully factorised answer.

4. ( y = \sqrt{2x + 5} = (2x + 5)^{1/2} )

Using chain rule: ( \frac{dy}{dx} = \frac{1}{2}(2x + 5)^{-1/2} \cdot 2 = \frac{1}{\sqrt{2x + 5}} )

[M2] M1 for correct application of chain rule; A1 for correct simplified answer.

5. ( y = x^3 - 6x^2 + 9x + 4 )

( \frac{dy}{dx} = 3x^2 - 12x + 9 )

Set ( \frac{dy}{dx} = 0 ): ( 3x^2 - 12x + 9 = 0 )

( x^2 - 4x + 3 = 0 )

( (x - 1)(x - 3) = 0 )

( x = 1 ) or ( x = 3 )

[M2] M1 for correct derivative and setting to zero; A1 for both correct ( x )-coordinates.

Section B: Applications of Differentiation

6. ( y = 2x^3 - x^2 + 4 )

( \frac{dy}{dx} = 6x^2 - 2x )

At ( x = 1 ): ( \frac{dy}{dx} = 6(1)^2 - 2(1) = 4 )

( y )-coordinate: ( y = 2(1)^3 - (1)^2 + 4 = 5 )

Point: ( (1, 5) ), gradient ( m = 4 )

Tangent: ( y - 5 = 4(x - 1) )

( y = 4x + 1 )

[M3] M1 for correct derivative; M1 for evaluating gradient and ( y )-coordinate at ( x = 1 ); A1 for correct tangent equation.

7. ( y = x^4 - 4x^3 + 4x^2 )

( \frac{dy}{dx} = 4x^3 - 12x^2 + 8x = 4x(x^2 - 3x + 2) = 4x(x - 1)(x - 2) )

Stationary points at ( x = 0, 1, 2 ).

( \frac{d^2y}{dx^2} = 12x^2 - 24x + 8 )

At ( x = 0 ): ( \frac{d^2y}{dx^2} = 8 > 0 ) → minimum.

At ( x = 1 ): ( \frac{d^2y}{dx^2} = 12 - 24 + 8 = -4 < 0 ) → maximum.

At ( x = 2 ): ( \frac{d^2y}{dx^2} = 48 - 48 + 8 = 8 > 0 ) → minimum.

[M3] M1 for correct first derivative and stationary points; M1 for second derivative and evaluation; A1 for correct nature of all three points.

8. ( P(x) = 80x - 2x^2 - 200 )

( P'(x) = 80 - 4x )

Set ( P'(x) = 0 ): ( 80 - 4x = 0 \Rightarrow x = 20 )

( P''(x) = -4 < 0 ) → maximum.

Maximum profit: ( P(20) = 80(20) - 2(20)^2 - 200 = 1600 - 800 - 200 = 600 )

Sell ( 20 \times 100 = 2000 ) units. Maximum profit = $600,000.

[M3] M1 for correct derivative and critical point; M1 for verifying maximum; A1 for correct quantity and profit.

9. Let width (perpendicular to wall) = ( x ) m, length (parallel to wall) = ( y ) m.

Constraint: ( 2x + y = 120 \Rightarrow y = 120 - 2x )

Area: ( A = xy = x(120 - 2x) = 120x - 2x^2 )

( \frac{dA}{dx} = 120 - 4x ); set to zero: ( x = 30 )

( \frac{d^2A}{dx^2} = -4 < 0 ) → maximum.

Dimensions: width = 30 m, length = ( 120 - 60 = 60 ) m.

[M2] M1 for correct area expression and differentiation; A1 for correct dimensions.

10. Average cost: ( A(q) = \frac{C(q)}{q} = \frac{500}{q} + 20 + 0.5q )

( A'(q) = -\frac{500}{q^2} + 0.5 )

Set ( A'(q) = 0 ): ( \frac{500}{q^2} = 0.5 \Rightarrow q^2 = 1000 \Rightarrow q = \sqrt{1000} \approx 31.6 )

( A''(q) = \frac{1000}{q^3} > 0 ) for ( q > 0 ) → minimum.

Production level ≈ 31.6 items (accept 32 items or ( 10\sqrt{10} )).

[M2] M1 for correct average cost function and derivative; A1 for correct production level.

Section C: Integration

11. ( \displaystyle \int (6x^2 - 4x + 3) , dx = 2x^3 - 2x^2 + 3x + C )

[M2] M1 for correct integration of at least two terms; A1 for fully correct answer including constant.

12. ( \displaystyle \int_1^4 \left( 2x + \frac{3}{x^2} \right) dx = \int_1^4 (2x + 3x^{-2}) , dx )

( = \left[ x^2 - 3x^{-1} \right]_1^4 = \left[ x^2 - \frac{3}{x} \right]_1^4 )

( = \left(16 - \frac{3}{4}\right) - \left(1 - 3\right) = 15.25 - (-2) = 17.25 ) or ( \frac{69}{4} )

[M3] M1 for correct integration; M1 for correct substitution of limits; A1 for correct value.

13. ( \displaystyle \int e^{2x+1} , dx = \frac{1}{2}e^{2x+1} + C )

[M2] M1 for recognising chain rule in reverse; A1 for correct answer including constant.

14. ( \displaystyle \int (3x - 2)^5 , dx = \frac{1}{3} \cdot \frac{(3x - 2)^6}{6} + C = \frac{(3x - 2)^6}{18} + C )

[M2] M1 for correct application of reverse chain rule; A1 for correct simplified answer.

15. ( \displaystyle \int_0^2 3e^{0.5x} , dx = \left[ 3 \cdot \frac{1}{0.5} e^{0.5x} \right]_0^2 = \left[ 6e^{0.5x} \right]_0^2 )

( = 6e^1 - 6e^0 = 6e - 6 = 6(e - 1) )

( = 6(2.71828... - 1) = 6(1.71828...) = 10.3097... \approx 10.3 ) (3 s.f.)

[M3] M1 for correct integration; M1 for correct evaluation; A1 for correct answer to 3 s.f.

Section D: Applications of Integration

16. Curve: ( y = x^2 + 1 ); line: ( y = 5 )

Intersection: ( x^2 + 1 = 5 \Rightarrow x^2 = 4 \Rightarrow x = 2 ) (positive, since bounded by ( y )-axis).

Area = ( \displaystyle \int_0^2 \left[ 5 - (x^2 + 1) \right] dx = \int_0^2 (4 - x^2) , dx )

( = \left[ 4x - \frac{x^3}{3} \right]_0^2 = \left(8 - \frac{8}{3}\right) - 0 = \frac{16}{3} ) square units.

[M3] M1 for correct setup of integral; M1 for correct integration; A1 for correct area.

17. ( y = 4x - x^2 )

Intersection with ( x )-axis: ( 4x - x^2 = 0 \Rightarrow x(4 - x) = 0 \Rightarrow x = 0, 4 )

Area = ( \displaystyle \int_0^4 (4x - x^2) , dx = \left[ 2x^2 - \frac{x^3}{3} \right]_0^4 )

( = \left(32 - \frac{64}{3}\right) - 0 = \frac{96 - 64}{3} = \frac{32}{3} ) square units.

[M3] M1 for correct limits; M1 for correct integration; A1 for correct area.

18. Area = ( \displaystyle \int_0^{\ln 3} e^x , dx = \left[ e^x \right]_0^{\ln 3} = e^{\ln 3} - e^0 = 3 - 1 = 2 ) square units.

[M2] M1 for correct integration and limits; A1 for correct area.

19. ( R'(q) = 50 - 0.4q )

( R(q) = \displaystyle \int (50 - 0.4q) , dq = 50q - 0.2q^2 + C )

Given ( R(0) = 0 ): ( C = 0 )

( R(q) = 50q - 0.2q^2 )

[M2] M1 for correct integration; A1 for correct revenue function with constant evaluated.

20. ( \frac{dP}{dt} = 0.5e^{0.1t} )

( P(t) = \displaystyle \int 0.5e^{0.1t} , dt = 0.5 \cdot \frac{1}{0.1} e^{0.1t} + C = 5e^{0.1t} + C )

Given ( P(0) = 20 ): ( 5e^0 + C = 20 \Rightarrow 5 + C = 20 \Rightarrow C = 15 )

( P(t) = 5e^{0.1t} + 15 )

At ( t = 10 ): ( P(10) = 5e^{1} + 15 = 5(2.71828...) + 15 = 13.5914... + 15 = 28.5914... )

≈ 29 thousand (to nearest thousand).

[M3] M1 for correct integration; M1 for evaluating constant and substituting ( t = 10 ); A1 for correct answer to nearest thousand.

END OF ANSWER KEY