IBDP Maths SL 5.1 Concept of a limit AA HL Paper 2- Exam Style Questions- New Syllabus
▶️ Answer/Explanation
(a) [2 marks]
(i) Plant B:
\( h_B(t) = 8t + 32 \). At \( t = 0 \): \( h_B(0) = 8 \cdot 0 + 32 = 32 \) (A1).
Initial height: 32 cm.
(ii) Plant A:
\( h_A(t) = \sin(2t + 6) + 9t + 27 \). At \( t = 0 \): \( h_A(0) = \sin(6) + 27 \approx -0.2794 + 27 \approx 26.7206 \) (M1).
To three significant figures: 26.7 cm (A1).
(b) [3 marks]
Solve \( h_A(t) = h_B(t) \): \( \sin(2t + 6) + 9t + 27 = 8t + 32 \implies \sin(2t + 6) + t – 5 = 0 \) (M1).
Since \( \sin(2t + 6) = 5 – t \), and \( -1 \leq \sin(2t + 6) \leq 1 \), solve in \( 4 \leq t \leq 6 \). Using numerical methods, roots are approximately \( t = 4.0074, 4.7034, 5.8833 \).
Answer: \( t \approx 4.01, 4.70, 5.88 \) weeks (A2).
(c) [3 marks]
Prove \( h_A(t) > h_B(t) \) for \( t > 6 \).
Compute: \( h_A(t) – h_B(t) = \sin(2t + 6) + 9t + 27 – (8t + 32) = \sin(2t + 6) + t – 5 \) (A1).
Since \( \sin(2t + 6) \geq -1 \), \( h_A(t) – h_B(t) \geq -1 + t – 5 = t – 6 \). For \( t > 6 \), \( t – 6 > 0 \), so \( h_A(t) – h_B(t) > 0 \) (A1R1).
Alternatively: Minimum of \( \sin(2t + 6) = -1 \), so \( h_A(t) – h_B(t) \geq t – 6 > 0 \) for \( t > 6 \) (R1A1).
Thus, Plant A is always taller than Plant B for \( t > 6 \) (AG).
(d) [4 marks]
Find when \( h_B'(t) > h_A'(t) \).
Derivatives: \( h_A'(t) = 2\cos(2t + 6) + 9 \), \( h_B'(t) = 8 \) (M1).
Solve: \( 8 > 2\cos(2t + 6) + 9 \implies \cos(2t + 6) < -0.5 \) (M1).
General solution: \( \frac{2\pi}{3} + 2k\pi < 2t + 6 < \frac{4\pi}{3} + 2k\pi \implies \frac{\pi}{3} – 3 + k\pi < t < \frac{2\pi}{3} – 3 + k\pi \).
For \( k = 1 \): \( t \in (1.18879, 2.23598) \).
For \( k = 2 \): \( t \in (4.33038, 5.37758) \).
For \( k = 3 \): \( t \in (7.47197, 8.51917) \) (A1).
Each interval length: \( \frac{2\pi}{3} – \frac{\pi}{3} = \frac{\pi}{3} \approx 1.0472 \). Total: \( 3 \times \frac{\pi}{3} = \pi \approx 3.14 \) weeks (A1M1).
Markscheme Answers:
(a) (i) 32 cm (A1)
(ii) 26.7 cm (M1A1)
(b) \( t \approx 4.01, 4.70, 5.88 \) weeks (M1A2)
(c) Plant A always taller than Plant B for \( t > 6 \) (A1R1A1)
(d) \( \pi \approx 3.14 \) weeks (M1M1A1A1)
Total [12 marks]
▶️ Answer/Explanation
(a) [7 marks]
Differentiate \( x^2 y = 5 – y^4 \) implicitly (M1):
Left: \( \frac{d}{dx}(x^2 y) = 2xy + x^2 \frac{dy}{dx} \) (A1).
Right: \( \frac{d}{dx}(5 – y^4) = -4y^3 \frac{dy}{dx} \) (A1).
Equation: \( 2xy + x^2 \frac{dy}{dx} = -4y^3 \frac{dy}{dx} \).
Set gradient to zero: \( \frac{dy}{dx} = 0 \implies 2xy = 0 \implies x = 0 \) or \( y = 0 \) (M1A1).
If \( x = 0 \): \( 0 = 5 – y^4 \implies y^4 = 5 \implies y = \pm \sqrt[4]{5} \) (R1).
If \( y = 0 \): \( x^2 \cdot 0 = 5 \implies 0 = 5 \), which is impossible (R1).
Thus, only \( x = 0 \), \( y = \pm \sqrt[4]{5} \) gives two points: \( (0, \sqrt[4]{5}) \), \( (0, -\sqrt[4]{5}) \) (AG).
(b) [5 marks]
At \( P(2, 1) \), substitute into derivative: \( 2 \cdot 2 \cdot 1 + 2^2 \frac{dy}{dx} = -4 \cdot 1^3 \frac{dy}{dx} \implies 4 + 4 \frac{dy}{dx} = -4 \frac{dy}{dx} \) (M1).
Solve: \( 4 + 4 \frac{dy}{dx} + 4 \frac{dy}{dx} = 0 \implies 8 \frac{dy}{dx} = -4 \implies \frac{dy}{dx} = -\frac{1}{2} \) (A1).
Normal gradient: \( -\frac{1}{-\frac{1}{2}} = 2 \) (M1).
Equation of normal: \( y – 1 = 2 (x – 2) \implies y = 2x – 4 + 1 \implies y = 2x – 3 \) (M1A1).
(c) [3 marks]
Normal: \( y = 2x – 3 \). Substitute into curve: \( x^2 (2x – 3) = 5 – (2x – 3)^4 \) (M1).
Simplify: \( 2x^3 – 3x^2 = 5 – (16x^4 – 96x^3 + 216x^2 – 216x + 81) \).
\( 2x^3 – 3x^2 – 5 + 16x^4 – 96x^3 + 216x^2 – 216x + 81 = 0 \).
\( 16x^4 – 94x^3 + 213x^2 – 216x + 76 = 0 \) (A1).
Solve numerically, excluding \( x = 2 \): \( x \approx 0.724 \) (A1).
(d) [7 marks]
Volume by rotation about \( y \)-axis requires two regions (M1).
From curve: \( x^2 = \frac{5 – y^4}{y} \).
Volume 1: For \( y \) from 1 to \( \sqrt[4]{5} \):
\( V_1 = \pi \int_1^{\sqrt[4]{5}} \frac{5 – y^4}{y} \, dy \) (M1A1 for integrand, A1 for limits).
Compute: \( \int \frac{5 – y^4}{y} \, dy = \int (5y^{-1} – y^3) \, dy = 5 \ln y – \frac{y^4}{4} \).
Evaluate: \( \left[ 5 \ln y – \frac{y^4}{4} \right]_1^{\sqrt[4]{5}} \approx 10.1\pi \approx 31.78 \) (A1).
Volume 2: From normal \( y = 2x – 3 \), so \( x = \frac{y + 3}{2} \), for \( y \) from -3 to 1:
EITHER: Disk from \( x = 0 \) to \( x = 2 \) at \( y = 1 \): \( V_2 = \frac{1}{3} \pi \cdot 2^2 \cdot 4 = \frac{16\pi}{3} \approx 16.75 \) (M1A1).
OR: \( V_2 = \pi \int_{-3}^1 \left( \frac{y + 3}{2} \right)^2 \, dy = \pi \int_{-3}^1 \frac{y^2 + 6y + 9}{4} \, dy = \frac{\pi}{4} \left[ \frac{y^3}{3} + 3y^2 + 9y \right]_{-3}^1 = \frac{16\pi}{3} \approx 16.75 \) (M1A1).
Total volume: \( 10.1\pi + \frac{16\pi}{3} \approx 19.9 \) cubic units (A1).
Markscheme Answers:
(a) Exactly two points at \( (0, \sqrt[4]{5}) \), \( (0, -\sqrt[4]{5}) \) (M1A1A1M1A1R1R1)
(b) \( y = 2x – 3 \) (M1A1M1A1)
(c) \( x \approx 0.724 \) (M1A1A1)
(d) 19.9 cubic units (M1M1A1A1M1A1A1)
Total [22 marks]