IB Mathematics SL 5.2 Increasing and decreasing function AI HL Paper 1- Exam Style Questions- New Syllabus
▶️ 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]
▶️ Answer/Explanation
attempt to differentiate \(f(x) = {x^3} – 3{x^2} + 4\) M1
\(f'(x) = 3{x^2} – 6x\) A1
\( = 3x(x – 2)\)
(Critical values occur at) \(x = 0,{\text{ }}x = 2\) (A1)
so \(f\) decreasing on \(x \in ]0,{\text{ }}2[\;\;\;({\text{or }}0 < x < 2)\) A1
\(\boxed{0 < x < 2}\)
\(f”(x) = 6x – 6\) (A1)
setting \(f”(x) = 0\) M1
\( \Rightarrow x = 1\)
coordinate is \((1,{\text{ }}2)\) A1
\(\boxed{(1, 2)}\)