IBDP Maths AHL 5.15 Derivatives of secx , cscx , cotx AA HL Paper 2- Exam Style Questions- New Syllabus
▶️ Answer/Explanation
[9 marks]
(a) Method 1: \( \frac{2x}{1 + x^4} + \frac{2y}{1 + y^4} \frac{dy}{dx} = 0 \) (M1, A1, A1).
\( \frac{dy}{dx} = -\frac{x(1 + y^4)}{y(1 + x^4)} \) (A1).
Method 2: \( y^2 = \tan \left( \frac{\pi}{4} – \arctan x^2 \right) \) (M1).
\( = \frac{\tan \frac{\pi}{4} – \tan (\arctan x^2)}{1 + (\tan \frac{\pi}{4})(\tan (\arctan x^2))} \) (M1).
\( = \frac{1 – x^2}{1 + x^2} \) (A1).
\( 2y \frac{dy}{dx} = \frac{-2x(1 + x^2) – 2x(1 – x^2)}{(1 + x^2)^2} \) (M1).
\( 2y \frac{dy}{dx} = \frac{-4x}{(1 + x^2)^2} \) (A1).
\( \frac{dy}{dx} = -\frac{2x}{y(1 + x^2)^2} \) (A1).
(b) \( y^2 = \tan \left( \frac{\pi}{4} – \arctan \frac{1}{2} \right) \) (M1).
\( = \frac{\tan \frac{\pi}{4} – \tan (\arctan \frac{1}{2})}{1 + (\tan \frac{\pi}{4})(\tan (\arctan \frac{1}{2}))} \) (M1).
\( = \frac{1 – \frac{1}{2}}{1 + \frac{1}{2}} = \frac{1}{3} \) (A1).
\( y = -\frac{1}{\sqrt{3}} \) (A1).
Substitution into \( \frac{dy}{dx} \) (A1).
\( = \frac{4\sqrt{6}}{9} \) (A1).
Markscheme Answers:
(a) Method 1: \( \frac{2x}{1 + x^4} + \frac{2y}{1 + y^4} \frac{dy}{dx} = 0 \) (M1, A1, A1), \( \frac{dy}{dx} = -\frac{x(1 + y^4)}{y(1 + x^4)} \) (A1)
Method 2: \( y^2 = \tan \left( \frac{\pi}{4} – \arctan x^2 \right) \) (M1), \( = \frac{\tan \frac{\pi}{4} – \tan (\arctan x^2)}{1 + (\tan \frac{\pi}{4})(\tan (\arctan x^2))} \) (M1), \( = \frac{1 – x^2}{1 + x^2} \) (A1), \( 2y \frac{dy}{dx} = \frac{-2x(1 + x^2) – 2x(1 – x^2)}{(1 + x^2)^2} \) (M1), \( 2y \frac{dy}{dx} = \frac{-4x}{(1 + x^2)^2} \) (A1), \( \frac{dy}{dx} = -\frac{2x}{y(1 + x^2)^2} \) (A1)
(b) \( y^2 = \tan \left( \frac{\pi}{4} – \arctan \frac{1}{2} \right) \) (M1), \( = \frac{\tan \frac{\pi}{4} – \tan (\arctan \frac{1}{2})}{1 + (\tan \frac{\pi}{4})(\tan (\arctan \frac{1}{2}))} \) (M1), \( = \frac{1 – \frac{1}{2}}{1 + \frac{1}{2}} = \frac{1}{3} \) (A1), \( y = -\frac{1}{\sqrt{3}} \) (A1), substitution into \( \frac{dy}{dx} \) (A1), \( = \frac{4\sqrt{6}}{9} \) (A1)
[Total 9 marks]
▶️ Answer/Explanation
[19 marks]
a.i \( f'(x) = \frac{x \cdot \frac{1}{x} – \ln x}{x^2} \) (M1, A1).
\( = \frac{1 – \ln x}{x^2} \).
So \( f'(x) = 0 \) when \( \ln x = 1 \), i.e. \( x = e \) (A1).
a.ii \( f'(x) > 0 \) when \( x < e \) and \( f'(x) < 0 \) when \( x > e \) (R1).
Hence local maximum (AG).
a.iii \( y \leq \frac{1}{e} \) (A1).
b \( f”(x) = \frac{x^2 \cdot \frac{-1}{x} – (1 – \ln x) \cdot 2x}{x^4} \) (M1).
\( = \frac{-x – 2x + 2x \ln x}{x^4} \).
\( = \frac{-3 + 2 \ln x}{x^3} \) (A1).
\( f”(x) = 0 \) (M1).
\( -3 + 2 \ln x = 0 \).
\( x = e^{\frac{3}{2}} \).
Since \( f”(x) < 0 \) when \( x < e^{\frac{3}{2}} \) and \( f”(x) > 0 \) when \( x > e^{\frac{3}{2}} \) (R1).
Then point of inflexion \( \left( e^{\frac{3}{2}}, \frac{3}{2e^{\frac{3}{2}}} \right) \) (A1).
c 
(A1, A1, A1).
d.i 
(A1, A1).
d.ii All real values (A1).
d.iii 
(M1, A1).
\( -e^2 < x < -1 \) (accept \( x < -1 \)) (A1).
Markscheme Answers:
a.i \( f'(x) = \frac{x \cdot \frac{1}{x} – \ln x}{x^2} \) (M1, A1), \( = \frac{1 – \ln x}{x^2} \), \( f'(x) = 0 \) when \( \ln x = 1 \), i.e. \( x = e \) (A1)
a.ii \( f'(x) > 0 \) when \( x < e \) and \( f'(x) < 0 \) when \( x > e \) (R1), hence local maximum (AG)
a.iii \( y \leq \frac{1}{e} \) (A1)
b \( f”(x) = \frac{x^2 \cdot \frac{-1}{x} – (1 – \ln x) \cdot 2x}{x^4} \) (M1), \( = \frac{-x – 2x + 2x \ln x}{x^4} \), \( = \frac{-3 + 2 \ln x}{x^3} \) (A1), \( f”(x) = 0 \) (M1), \( -3 + 2 \ln x = 0 \), \( x = e^{\frac{3}{2}} \), since \( f”(x) < 0 \) when \( x < e^{\frac{3}{2}} \) and \( f”(x) > 0 \) when \( x > e^{\frac{3}{2}} \) (R1), then point of inflexion \( \left( e^{\frac{3}{2}}, \frac{3}{2e^{\frac{3}{2}}} \right) \) (A1)
c (A1, A1, A1)
d.i (A1, A1)
d.ii All real values (A1)
d.iii (M1, A1), \( -e^2 < x < -1 \) (accept \( x < -1 \)) (A1)
[Total 19 marks]