Practice for: HSC Mathematics Advanced (NESA)
HSC Mathematics Advanced practice paper
18 minutes · 13 marks · 5 questions
Not an official NESA paper — for practice only.
- 1
Differentiate y = x²eˣ.
2 marks - 2
Evaluate ∫₁ᵉ (1/x) dx.
2 marks - 3
The population P of a town grows so that dP/dt = 0.02P. If P = 5000 when t = 0 years, find the population after 10 years, to the nearest whole number.
3 marks - 4
Find the area of the region enclosed between the curve y = x² and the line y = 2x.
3 marks - 5
Sketch the graph of y = x² − 4x + 3, showing the intercepts and the coordinates of the turning point.
3 marks
Mark scheme
1. dy/dx = 2xeˣ + x²eˣ = xeˣ(x + 2). 1 mark for applying the product rule; 1 mark for the simplified result.
2. [ln x]₁ᵉ = ln e − ln 1 = 1. 1 mark for ln x as the primitive; 1 mark for the value 1.
3. P = 5000e^0.02t; P(10) = 5000e^0.2 ≈ 6107. 1 mark for the exponential form; 1 mark for substituting t = 10; 1 mark for 6107.
4. Intersections at x = 0 and x = 2; ∫₀² (2x − x²) dx = 4/3 square units. 1 mark for the intersections; 1 mark for the correct integral; 1 mark for 4/3.
5. x-intercepts (1, 0) and (3, 0); y-intercept (0, 3); turning point (2, −1). 1 mark for the intercepts; 1 mark for the correct parabola shape; 1 mark for the turning point.
Worked solution · Question 1
Use the product rule with u = x² and v = eˣ, so u′ = 2x and v′ = eˣ.
dy/dx = u′v + uv′ = 2xeˣ + x²eˣ.
Factorise: dy/dx = xeˣ(x + 2).
Worked solution · Question 2
A primitive of 1/x is ln x (for x > 0).
∫₁ᵉ (1/x) dx = [ln x]₁ᵉ = ln e − ln 1.
= 1 − 0 = 1.
Worked solution · Question 3
dP/dt = kP has the solution P = P₀eᵏᵗ. Here P₀ = 5000 and k = 0.02, so P = 5000e^0.02t.
At t = 10: P = 5000e^(0.02 × 10) = 5000e^0.2.
e^0.2 ≈ 1.2214, so P ≈ 6107.0.
The population after 10 years is about 6107.
Worked solution · Question 4
Find where the curve and line meet: x² = 2x ⇒ x² − 2x = 0 ⇒ x(x − 2) = 0, so x = 0 or x = 2.
On 0 ≤ x ≤ 2 the line lies above the curve (test x = 1: 2 > 1), so the area is ∫₀² (2x − x²) dx.
= [x² − x³/3]₀² = (4 − 8/3) − 0 = 4/3.
The enclosed area is 4/3 square units.
Worked solution · Question 5
y = x² − 4x + 3 = (x − 1)(x − 3), so the x-intercepts are x = 1 and x = 3, and the y-intercept is 3.
The turning point lies midway between the intercepts at x = 2. Substituting gives y = 2² − 4(2) + 3 = −1, so the vertex is (2, −1).
The parabola is concave up because the coefficient of x² is positive.