IBDP Maths AHL 5.15 Derivatives of secx , cscx , cotx AA HL Paper 1- Exam Style Questions- New Syllabus
▶️ Answer/Explanation
METHOD 1
\(\frac{{2x}}{{1 + {x^4}}} + \frac{{2y}}{{1 + {y^4}}}\frac{{{\text{d}}y}}{{{\text{d}}x}} = 0\) M1A1A1
Note: Award M1 for implicit differentiation, A1 for LHS and A1 for RHS.
\(\frac{{{\text{d}}y}}{{{\text{d}}x}} = – \frac{{x\left( {1 + {y^4}} \right)}}{{y\left( {1 + {x^4}} \right)}}\) A1
METHOD 2
\({y^2} = \tan \left( {\frac{\pi }{4} – \arctan {x^2}} \right)\)
\( = \frac{{\tan \frac{\pi }{4} – \tan \left( {\arctan {x^2}} \right)}}{{1 + \left( {\tan \frac{\pi }{4}} \right)\left( {\tan \left( {\arctan {x^2}} \right)} \right)}}\) (M1)
\( = \frac{{1 – {x^2}}}{{1 + {x^2}}}\) A1
\(2y\frac{{{\text{d}}y}}{{{\text{d}}x}} = \frac{{ – 2x\left( {1 + {x^2}} \right) – 2x\left( {1 – {x^2}} \right)}}{{{{\left( {1 + {x^2}} \right)}^2}}}\) M1
\(2y\frac{{{\text{d}}y}}{{{\text{d}}x}} = \frac{{ – 4x}}{{{{\left( {1 + {x^2}} \right)}^2}}}\)
\(\frac{{{\text{d}}y}}{{{\text{d}}x}} = – \frac{{2x}}{{y{{\left( {1 + {x^2}} \right)}^2}}}\) A1
\(\left( { = \frac{{2x\sqrt {1 + {x^2}} }}{{\sqrt {1 – {x^2}} {{\left( {1 + {x^2}} \right)}^2}}}} \right)\)
[4 marks]
\({y^2} = \tan \left( {\frac{\pi }{4} – \arctan \frac{1}{2}} \right)\) (M1)
\( = \frac{{\tan \frac{\pi }{4} – \tan \left( {\arctan \frac{1}{2}} \right)}}{{1 + \left( {\tan \frac{\pi }{4}} \right)\left( {\tan \left( {\arctan \frac{1}{2}} \right)} \right)}}\) (M1)
Note: The two M1s may be awarded for working in part (a).
\( = \frac{{1 – \frac{1}{2}}}{{1 + \frac{1}{2}}} = \frac{1}{3}\) A1
\(y = – \frac{1}{{\sqrt 3 }}\) A1
substitution into \(\frac{{{\text{d}}y}}{{{\text{d}}x}}\)
\( = \frac{{4\sqrt 6 }}{9}\) A1
Note: Accept \(\frac{{8\sqrt 3 }}{{9\sqrt 2 }}\) etc.
[5 marks]
▶️ Answer/Explanation
EITHER
Derivative of \( \frac{x}{1 – x} \) is \( \frac{(1 – x) – x(-1)}{{(1 – x)}^2} \). M1A1
\( f'(x) = \frac{1}{2} {\left( \frac{x}{1 – x} \right)}^{-\frac{1}{2}} \frac{1}{{(1 – x)}^2} \). M1A1
\( = \frac{1}{2} x^{-\frac{1}{2}} (1 – x)^{-\frac{3}{2}} \). AG
\( f'(x) > 0 \) (for all \( 0 < x < 1 \)) so the function is increasing. R1
OR
\( f(x) = \frac{{x^{\frac{1}{2}}}}{{(1 – x)^{\frac{1}{2}}}} \)
\( f'(x) = \frac{{(1 – x)^{\frac{1}{2}} \left( \frac{1}{2} x^{-\frac{1}{2}} \right) – \frac{1}{2} x^{\frac{1}{2}} (1 – x)^{-\frac{1}{2}} (-1)}}{{1 – x}} \). M1A1
\( = \frac{1}{2} x^{-\frac{1}{2}} (1 – x)^{-\frac{1}{2}} + \frac{1}{2} x^{\frac{1}{2}} (1 – x)^{-\frac{3}{2}} \). A1
\( = \frac{1}{2} x^{-\frac{1}{2}} (1 – x)^{-\frac{3}{2}} [1 – x + x] \). M1
\( = \frac{1}{2} x^{-\frac{1}{2}} (1 – x)^{-\frac{3}{2}} \). AG
\( f'(x) > 0 \) (for all \( 0 < x < 1 \)) so the function is increasing. R1
[5 marks]
\( f'(x) = \frac{1}{2} x^{-\frac{1}{2}} (1 – x)^{-\frac{3}{2}} \)
\( \Rightarrow f”(x) = -\frac{1}{4} x^{-\frac{3}{2}} (1 – x)^{-\frac{3}{2}} + \frac{3}{4} x^{-\frac{1}{2}} (1 – x)^{-\frac{5}{2}} \). M1A1
\( = -\frac{1}{4} x^{-\frac{3}{2}} (1 – x)^{-\frac{5}{2}} [1 – 4x] \)
\( f”(x) = 0 \Rightarrow x = \frac{1}{4} \). M1A1
\( f”(x) \) changes sign at \( x = \frac{1}{4} \) hence there is a point of inflexion. R1
\( x = \frac{1}{4} \Rightarrow y = \frac{1}{\sqrt{3}} \). A1
the coordinates are \( \left( \frac{1}{4}, \frac{1}{\sqrt{3}} \right) \)
[6 marks]
\( x = \sin^2 \theta \Rightarrow \frac{\text{d}x}{\text{d}\theta} = 2 \sin \theta \cos \theta \). M1A1
\( \int \sqrt{\frac{x}{1 – x}} \text{d}x = \int \sqrt{\frac{\sin^2 \theta}{1 – \sin^2 \theta}} 2 \sin \theta \cos \theta \text{d}\theta \). M1A1
\( = \int 2 \sin^2 \theta \text{d}\theta \). A1
\( = \int 1 – \cos 2\theta \text{d}\theta \). M1A1
\( = \theta – \frac{1}{2} \sin 2\theta + c \). A1
\( \theta = \arcsin \sqrt{x} \). A1
\( \frac{1}{2} \sin 2\theta = \sin \theta \cos \theta = \sqrt{x} \sqrt{1 – x} = \sqrt{x – x^2} \). M1A1
hence \( \int \sqrt{\frac{x}{1 – x}} \text{d}x = \arcsin \sqrt{x} – \sqrt{x – x^2} + c \). AG
[11 marks]