Date | May 2012 | Marks available | 5 | Reference code | 12M.1.hl.TZ1.12 |
Level | HL only | Paper | 1 | Time zone | TZ1 |
Command term | Deduce and Show that | Question number | 12 | Adapted from | N/A |
Question
Let f(x)=√x1−x, 0<x<1.
Show that f′(x)=12x−12(1−x)−32 and deduce that f is an increasing function.
Show that the curve y=f(x) has one point of inflexion, and find its coordinates.
Use the substitution x=sin2θ to show that ∫f(x)dx=arcsin√x−√x−x2+c .
Markscheme
EITHER
derivative of x1−x is (1−x)−x(−1)(1−x)2 M1A1
f′(x)=12(x1−x)−121(1−x)2 M1A1
=12x−12(1−x)−32 AG
f′(x)>0 (for all 0<x<1) so the function is increasing R1
OR
f(x)=x12(1−x)12
f′(x)=(1−x)12(12x−12)−12x12(1−x)−12(−1)1−x M1A1
=12x−12(1−x)−12+12x12(1−x)−32 A1
=12x−12(1−x)−32[1−x+x] M1
=12x−12(1−x)−32 AG
f′(x)>0 (for all 0<x<1) so the function is increasing R1
[5 marks]
f′(x)=12x−12(1−x)−32
⇒f″ 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]
Examiners report
Part (a) was generally well done, although few candidates made the final deduction asked for. Those that lost other marks in this part were generally due to mistakes in algebraic manipulation. In part (b) whilst many students found the second derivative and set it equal to zero, few then confirmed that it was a point of inflexion. There were several good attempts for part (c), even though there were various points throughout the question that provided stopping points for other candidates.
Part (a) was generally well done, although few candidates made the final deduction asked for. Those that lost other marks in this part were generally due to mistakes in algebraic manipulation. In part (b) whilst many students found the second derivative and set it equal to zero, few then confirmed that it was a point of inflexion. There were several good attempts for part (c), even though there were various points throughout the question that provided stopping points for other candidates.
Part (a) was generally well done, although few candidates made the final deduction asked for. Those that lost other marks in this part were generally due to mistakes in algebraic manipulation. In part (b) whilst many students found the second derivative and set it equal to zero, few then confirmed that it was a point of inflexion. There were several good attempts for part (c), even though there were various points throughout the question that provided stopping points for other candidates.