Show that $f'(x) = \frac{1}{2}{x^{ - \frac{1}{2}}}{(1 - x)^{ - \frac{3}{2}}}$ 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 $x = {\sin ^2}\theta$ to show that $\int {f(x){\text{d}}x}  = \arcsin \sqrt x  - \sqrt {x - {x^2}}  + c$ .

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] 

 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.