IBDP Maths SL 5.7 The second derivative AA HL Paper 2- Exam Style Questions- New Syllabus
▶️ Answer/Explanation
[21 marks]
(a) \( f'(x) = 30e^{-\frac{x^2}{400}} \cdot -\frac{2x}{400} = -\frac{3x}{20}e^{-\frac{x^2}{400}} \) (M1, A1).
\( f”(x) = -\frac{3}{20}e^{-\frac{x^2}{400}} + (-\frac{3x}{20}) \cdot (-\frac{2x}{400})e^{-\frac{x^2}{400}} = -\frac{3}{20}e^{-\frac{x^2}{400}} + \frac{3x^2}{4000}e^{-\frac{x^2}{400}} = \frac{3}{20}e^{-\frac{x^2}{400}} \left( \frac{x^2}{200} – 1 \right) \) (M1, A1).
(b) Maximum gradient when \( f”(x) = 0 \): \( \frac{3}{20}e^{-\frac{x^2}{400}} \left( \frac{x^2}{200} – 1 \right) = 0 \implies \frac{x^2}{200} – 1 = 0 \implies x^2 = 200 \implies x = \pm \sqrt{200} \) (M1, A1).
Check sign change: \( f”(-15) > 0 \), \( f”(-14) < 0 \) indicates maximum at \( x = -\sqrt{200} \) (R1, AG).
(c) Area \( A = 2a \cdot 30e^{-\frac{a^2}{400}} = 60a e^{-\frac{a^2}{400}} \) (M1, A1).
\( \frac{dA}{da} = 60e^{-\frac{a^2}{400}} – 60a \cdot \frac{2a}{400}e^{-\frac{a^2}{400}} = 0 \implies 1 – \frac{a^2}{200} = 0 \implies a = \sqrt{200} \) (M1, A1).
\( A_{max} = 60 \cdot \sqrt{200} \cdot e^{-\frac{200}{400}} = 600\sqrt{2} e^{-\frac{1}{2}} \) (M1, AG).
(d)(i) Perimeter \( P = 4a + 60e^{-\frac{a^2}{400}} \) (A1, A1).
(d)(ii) \( I(a) = \frac{4a + 60e^{-\frac{a^2}{400}}}{60a e^{-\frac{a^2}{400}}} \) (A1).
Minimize \( I(a) \) via graphing or calculus, yielding \( a = 12.6 \) (M1, A1).
(d)(iii) Area under roof \( \int_{-20}^{20} 30e^{-\frac{x^2}{400}} \, dx = 896.18\ldots \) (M1, A1).
Area of living space \( 60 \cdot 12.6 \cdot e^{-\frac{(12.6)^2}{400}} = 508.56\ldots \).
Percentage unused \( \frac{896.18 – 508.56}{896.18} \cdot 100 \approx 43.3\% \) (A1).
Markscheme Answers:
(a) \( f'(x) = -\frac{3x}{20}e^{-\frac{x^2}{400}} \) (M1, A1), \( f”(x) = \frac{3}{20}e^{-\frac{x^2}{400}} \left( \frac{x^2}{200} – 1 \right) \) (M1, A1)
(b) \( f”(x) = 0 \) (M1), \( x = \pm \sqrt{200} \) (A1), sign change at \( x = -\sqrt{200} \) (R1, AG)
(c) \( A = 60a e^{-\frac{a^2}{400}} \) (M1, A1), \( \frac{dA}{da} = 0 \implies a = \sqrt{200} \) (M1, A1), \( A_{max} = 600\sqrt{2} e^{-\frac{1}{2}} \) (M1, AG)
(d)(i) \( P = 4a + 60e^{-\frac{a^2}{400}} \) (A1, A1)
(d)(ii) \( I(a) = \frac{4a + 60e^{-\frac{a^2}{400}}}{60a e^{-\frac{a^2}{400}}} \) (A1), minimize \( I(a) \) (M1), \( a = 12.6 \) (A1)
(d)(iii) \( \int_{-20}^{20} 30e^{-\frac{x^2}{400}} \, dx = 896.18\ldots \) (M1, A1), percentage \( 43.3\% \) (A1)
[Total 21 marks]
▶️ Answer/Explanation
[9 marks]
(a) Differentiate \( 4x^2 + y^2 = 7 \): \( 8x + 2y \frac{dy}{dx} = 0 \) (M1, A1).
\( \frac{dy}{dx} = -\frac{4x}{y} \).
Gradient of normal \( = -\frac{1}{\frac{dy}{dx}} = \frac{y}{4x} \) (M1).
At \( \left(1, \sqrt{3}\right) \), gradient \( = \frac{\sqrt{3}}{4} \) (A1).
Equation: \( y – \sqrt{3} = \frac{\sqrt{3}}{4}(x – 1) \) (M1), \( y = \frac{\sqrt{3}}{4}x + \frac{3\sqrt{3}}{4} \) (A1).
(b) Volume \( V = \pi \int_0^{\frac{\sqrt{7}}{2}} y^2 \, dx \), where \( y^2 = 7 – 4x^2 \) (M1, A1).
\( V = \pi \int_0^{\frac{\sqrt{7}}{2}} (7 – 4x^2) \, dx = \pi \left[ 7x – \frac{4x^3}{3} \right]_0^{\frac{\sqrt{7}}{2}} = \pi \left( 7 \cdot \frac{\sqrt{7}}{2} – \frac{4}{3} \cdot \frac{7\sqrt{7}}{8} \right) = \frac{7\sqrt{7}\pi}{3} \) (A1).
Markscheme Answers:
(a) \( 8x + 2y \frac{dy}{dx} = 0 \) (M1, A1), \( \frac{dy}{dx} = -\frac{4x}{y} \), normal gradient \( \frac{y}{4x} \) (M1), \( \frac{\sqrt{3}}{4} \) at \( \left(1, \sqrt{3}\right) \) (A1), \( y – \sqrt{3} = \frac{\sqrt{3}}{4}(x – 1) \) (M1), \( y = \frac{\sqrt{3}}{4}x + \frac{3\sqrt{3}}{4} \) (A1)
(b) \( V = \pi \int_0^{\frac{\sqrt{7}}{2}} (7 – 4x^2) \, dx \) (M1, A1), \( V = \frac{7\sqrt{7}\pi}{3} \) (A1)
[Total 9 marks]