IBDP Maths SL 5.8 Local maximum and minimum values AA HL Paper 2- Exam Style Questions- New Syllabus
▶️ Answer/Explanation
[21 marks]
(a) For \( x\sqrt{9 – x^2} \), domain \( -3 \leq x \leq 3 \); for \( 2\arcsin\left(\frac{x}{3}\right) \), domain \( -3 \leq x \leq 3 \) (A1).
\( D: -3 \leq x \leq 3 \) (A1).
(b) Volume \( V = \pi \int_0^{2.8} \left( x\sqrt{9 – x^2} + 2\arcsin\left(\frac{x}{3}\right) \right)^2 \, dx = 181 \) (M1, A1, A1).
(c) \( f'(x) = \sqrt{9 – x^2} + x \cdot \frac{-2x}{\sqrt{9 – x^2}} + 2 \cdot \frac{1}{\sqrt{1 – \frac{x^2}{9}}} \cdot \frac{1}{3} \) (M1, A1).
\( = \sqrt{9 – x^2} – \frac{x^2}{\sqrt{9 – x^2}} + \frac{2}{3\sqrt{1 – \frac{x^2}{9}}} = \sqrt{9 – x^2} – \frac{x^2}{\sqrt{9 – x^2}} + \frac{2}{\sqrt{9 – x^2}} \) (A1).
\( = \frac{9 – x^2 – x^2 + 2}{\sqrt{9 – x^2}} = \frac{11 – 2x^2}{\sqrt{9 – x^2}} \) (A1, A1).
(d) \( \int_{-p}^p \frac{11 – 2x^2}{\sqrt{9 – x^2}} \, dx = \left[ x\sqrt{9 – x^2} + 2\arcsin\left(\frac{x}{3}\right) \right]_{-p}^p \) (M1).
\( = (p\sqrt{9 – p^2} + 2\arcsin\left(\frac{p}{3}\right)) – (-p\sqrt{9 – p^2} – 2\arcsin\left(\frac{p}{3}\right)) = 2p\sqrt{9 – p^2} + 4\arcsin\left(\frac{p}{3}\right) \) (A1, AG).
(e) Maximize by setting numerator \( 11 – 2p^2 = 0 \): \( p^2 = \frac{11}{2} \), \( p = \sqrt{\frac{11}{2}} \) (M1, A1).
(f)(i) \( f”(x) = \frac{d}{dx} \left( \frac{11 – 2x^2}{\sqrt{9 – x^2}} \right) = \frac{(-4x)\sqrt{9 – x^2} + (11 – 2x^2) \cdot \frac{-x}{\sqrt{9 – x^2}}}{9 – x^2} \) (M1, A1).
\( = \frac{-4x(9 – x^2) – x(11 – 2x^2)}{(9 – x^2)^{3/2}} = \frac{-36x + 4x^3 – 11x + 2x^3}{(9 – x^2)^{3/2}} = \frac{x(2x^2 – 25)}{(9 – x^2)^{3/2}} \) (A1, A1, AG).
(f)(ii) \( f”(0) = 0 \) (A1).
Sign change: \( f”(x) < 0 \) for \( 0 < x < 3 \), \( f”(x) > 0 \) for \( -3 < x < 0 \), so inflection at \( x = 0 \) (R1).
\( x = \pm \sqrt{\frac{25}{2}} \) outside \( [-3, 3] \), so no inflection (R1).
Markscheme Answers:
(a) Domains \( -3 \leq x \leq 3 \) for both terms (A1), \( D: -3 \leq x \leq 3 \) (A1)
(b) \( V = \pi \int_0^{2.8} \left( x\sqrt{9 – x^2} + 2\arcsin\left(\frac{x}{3}\right) \right)^2 \, dx \) (M1, A1), \( V = 181 \) (A1)
(c) \( \frac{dy}{dx} = \sqrt{9 – x^2} – \frac{x^2}{\sqrt{9 – x^2}} + \frac{2}{3\sqrt{1 – \frac{x^2}{9}}} \) (M1, A1), simplified to \( \frac{11 – 2x^2}{\sqrt{9 – x^2}} \) (A1, A1, A1)
(d) \( \int_{-p}^p \frac{11 – 2x^2}{\sqrt{9 – x^2}} \, dx = \left[ x\sqrt{9 – x^2} + 2\arcsin\left(\frac{x}{3}\right) \right]_{-p}^p \) (M1), \( = 2p\sqrt{9 – p^2} + 4\arcsin\left(\frac{p}{3}\right) \) (A1, AG)
(e) \( 11 – 2p^2 = 0 \) (M1), \( p = \sqrt{\frac{11}{2}} \) (A1)
(f)(i) \( f”(x) = \frac{(-4x)(9 – x^2) + x(11 – 2x^2)}{(9 – x^2)^{3/2}} \) (M1, A1), \( = \frac{x(2x^2 – 25)}{(9 – x^2)^{3/2}} \) (A1, A1, AG)
(f)(ii) \( f”(0) = 0 \) or sign change (A1), inflection at \( x = 0 \) (R1), \( x = \pm \sqrt{\frac{25}{2}} \) outside domain (R1)
[Total 21 marks]
▶️ Answer/Explanation
[19 marks]
(a) \( f'(x) = \frac{b e^x (a e^x + b) – a e^x (a + b e^x)}{(a e^x + b)^2} \) (M1, A1).
\( = \frac{ab e^{2x} + b^2 e^x – a^2 e^x – ab e^{2x}}{(a e^x + b)^2} = \frac{(b^2 – a^2)e^x}{(a e^x + b)^2} \) (A1, AG).
(b) EITHER \( f'(x) = 0 \implies (b^2 – a^2)e^x = 0 \), but \( 0 < b < a \) and \( e^x > 0 \) for all \( x \), so no solution (A1, R1).
OR \( f'(x) < 0 \) for all \( x \in \mathbb{R} \) since \( 0 < b < a \) and \( e^x > 0 \) (A1, R1).
OR \( f'(x) \) cannot be zero because \( e^x \) is never zero (A1, R1).
(c) EITHER \( f”(x) = \frac{(b^2 – a^2)e^x (a e^x + b – 2a e^x)}{(a e^x + b)^3} = \frac{(b^2 – a^2)(b – a e^x)e^x}{(a e^x + b)^3} \) (M1, A1, A1).
\( f”(x) = 0 \implies b – a e^x = 0 \implies x = \ln \frac{b}{a} \) (M1, A1).
\( f\left(\ln \frac{b}{a}\right) = \frac{a^2 + b^2}{2ab} \) (A1).
OR \( f'(x) = (b^2 – a^2)e^x (a e^x + b)^{-2} \).
\( f”(x) = (b^2 – a^2)e^x (a e^x + b)^{-2} + (b^2 – a^2)e^x (-2a e^x)(a e^x + b)^{-3} \) (M1, A1, A1).
\( = (b^2 – a^2)e^x (a e^x + b)^{-3} ((a e^x + b) – 2a e^x) = (b^2 – a^2)e^x (b – a e^x)(a e^x + b)^{-3} \).
\( f”(x) = 0 \implies b – a e^x = 0 \implies x = \ln \frac{b}{a} \) (M1, A1).
\( f\left(\ln \frac{b}{a}\right) = \frac{a^2 + b^2}{2ab} \) (A1).
Coordinates: \( \left( \ln \frac{b}{a}, \frac{a^2 + b^2}{2ab} \right) \).
(d) \( \lim_{x \to -\infty} f(x) = \frac{a}{b} \), horizontal asymptote \( y = \frac{a}{b} \) (A1).
\( \lim_{x \to +\infty} f(x) = \frac{b}{a} \), horizontal asymptote \( y = \frac{b}{a} \) (A1).
\( a e^x + b > 0 \) for all \( x \), so no vertical asymptotes (R1).
(e) For \( a = 4 \), \( b = 1 \), \( y = \frac{1}{2} \implies 4 + e^x = 2(4 e^x + 1) \implies e^x = \frac{7}{2} \), \( x = \ln \frac{7}{2} \) (M1, A1).
\( V = \pi \int_0^{\ln \frac{7}{2}} \left( \left(\frac{4 + e^x}{4 e^x + 1}\right)^2 – \frac{1}{4} \right) dx = 1.09 \) (M1, A1, A1).
Markscheme Answers:
(a) \( f'(x) = \frac{b e^x (a e^x + b) – a e^x (a + b e^x)}{(a e^x + b)^2} \) (M1, A1), \( = \frac{(b^2 – a^2)e^x}{(a e^x + b)^2} \) (A1, AG)
(b) EITHER \( f'(x) = 0 \implies (b^2 – a^2)e^x = 0 \) impossible (A1), since \( 0 < b < a \) and \( e^x > 0 \) (R1)
OR \( f'(x) < 0 \) for all \( x \) (A1, R1)
OR \( f'(x) \neq 0 \) since \( e^x \neq 0 \) (A1, R1)
(c) EITHER \( f”(x) = \frac{(b^2 – a^2)e^x (a e^x + b – 2a e^x)}{(a e^x + b)^3} \) (M1, A1, A1), \( x = \ln \frac{b}{a} \) (M1, A1), \( f\left(\ln \frac{b}{a}\right) = \frac{a^2 + b^2}{2ab} \) (A1)
OR \( f”(x) = (b^2 – a^2)e^x (a e^x + b)^{-2} + (b^2 – a^2)e^x (-2a e^x)(a e^x + b)^{-3} \) (M1, A1, A1), \( x = \ln \frac{b}{a} \) (M1, A1), \( f\left(\ln \frac{b}{a}\right) = \frac{a^2 + b^2}{2ab} \) (A1)
(d) \( \lim_{x \to -\infty} f(x) = \frac{a}{b} \) (A1), \( \lim_{x \to +\infty} f(x) = \frac{b}{a} \) (A1), no vertical asymptotes (R1)
(e) \( y = \frac{1}{2} \implies x = \ln \frac{7}{2} \) (M1, A1), \( V = \pi \int_0^{\ln \frac{7}{2}} \left( \left(\frac{4 + e^x}{4 e^x + 1}\right)^2 – \frac{1}{4} \right) dx = 1.09 \) (M1, A1, A1)
[Total 19 marks]