(a) [5 marks]

Method 1:
For \( f(x) = \sqrt{\frac{x}{1 – x}} \), compute the derivative.
Rewrite \( f(x) = \left( \frac{x}{1 – x} \right)^{\frac{1}{2}} \). Differentiate \( \frac{x}{1 – x} \):
\( \frac{d}{dx} \left( \frac{x}{1 – x} \right) = \frac{(1 – x) \cdot 1 – x \cdot (-1)}{(1 – x)^2} = \frac{1 – x + x}{(1 – x)^2} = \frac{1}{(1 – x)^2} \) (M1A1).
Chain rule for \( f(x) \): \( f'(x) = \frac{1}{2} \left( \frac{x}{1 – x} \right)^{-\frac{1}{2}} \cdot \frac{1}{(1 – x)^2} = \frac{1}{2} \left( \frac{1 – x}{x} \right)^{\frac{1}{2}} \cdot \frac{1}{(1 – x)^2} \) (M1A1).
Simplify: \( f'(x) = \frac{1}{2} \cdot \frac{(1 – x)^{\frac{1}{2}}}{x^{\frac{1}{2}}} \cdot \frac{1}{(1 – x)^2} = \frac{1}{2} x^{-\frac{1}{2}} (1 – x)^{\frac{1}{2} – 2} = \frac{1}{2} x^{-\frac{1}{2}} (1 – x)^{-\frac{3}{2}} \) (AG).
For \( 0 < x < 1 \), \( x^{-\frac{1}{2}} > 0 \), \( (1 – x)^{-\frac{3}{2}} > 0 \), so \( f'(x) > 0 \), hence \( f \) is increasing (R1).
Method 2:
Rewrite \( f(x) = \frac{x^{\frac{1}{2}}}{(1 – x)^{\frac{1}{2}}} \). Quotient rule: Let \( u = x^{\frac{1}{2}} \), \( v = (1 – x)^{\frac{1}{2}} \), so \( u’ = \frac{1}{2} x^{-\frac{1}{2}} \), \( v’ = \frac{1}{2} (1 – x)^{-\frac{1}{2}} (-1) = -\frac{1}{2} (1 – x)^{-\frac{1}{2}} \).
\( f'(x) = \frac{u’ v – u v’}{v^2} = \frac{\left( \frac{1}{2} x^{-\frac{1}{2}} \right) (1 – x)^{\frac{1}{2}} – x^{\frac{1}{2}} \left( -\frac{1}{2} (1 – x)^{-\frac{1}{2}} \right)}{(1 – x)} \) (M1A1).
Numerator: \( \frac{1}{2} x^{-\frac{1}{2}} (1 – x)^{\frac{1}{2}} + \frac{1}{2} x^{\frac{1}{2}} (1 – x)^{-\frac{1}{2}} \).
Factor: \( \frac{1}{2} x^{-\frac{1}{2}} (1 – x)^{-\frac{1}{2}} [ (1 – x) + x ] = \frac{1}{2} x^{-\frac{1}{2}} (1 – x)^{-\frac{1}{2}} \cdot 1 \) (A1, M1).
So, \( f'(x) = \frac{1}{2} x^{-\frac{1}{2}} (1 – x)^{-\frac{1}{2}} \cdot (1 – x)^{-1} = \frac{1}{2} x^{-\frac{1}{2}} (1 – x)^{-\frac{3}{2}} \) (AG).
Since \( f'(x) > 0 \) for \( 0 < x < 1 \), \( f \) is increasing (R1).

(b) [6 marks]

Find the second derivative to locate the point of inflexion.
Given \( f'(x) = \frac{1}{2} x^{-\frac{1}{2}} (1 – x)^{-\frac{3}{2}} \), compute \( f”(x) \):
Product rule: Let \( u = \frac{1}{2} x^{-\frac{1}{2}} \), \( v = (1 – x)^{-\frac{3}{2}} \), so \( u’ = \frac{1}{2} \cdot \left( -\frac{1}{2} \right) x^{-\frac{3}{2}} = -\frac{1}{4} x^{-\frac{3}{2}} \), \( v’ = -\frac{3}{2} (1 – x)^{-\frac{5}{2}} (-1) = \frac{3}{2} (1 – x)^{-\frac{5}{2}} \).
\( f”(x) = u’ v + u v’ = \left( -\frac{1}{4} x^{-\frac{3}{2}} \right) (1 – x)^{-\frac{3}{2}} + \left( \frac{1}{2} x^{-\frac{1}{2}} \right) \left( \frac{3}{2} (1 – x)^{-\frac{5}{2}} \right) \) (M1A1).
Simplify: \( f”(x) = -\frac{1}{4} x^{-\frac{3}{2}} (1 – x)^{-\frac{3}{2}} + \frac{3}{4} x^{-\frac{1}{2}} (1 – x)^{-\frac{5}{2}} \).
Factor: \( f”(x) = x^{-\frac{3}{2}} (1 – x)^{-\frac{5}{2}} \left[ -\frac{1}{4} (1 – x) + \frac{3}{4} x \right] = x^{-\frac{3}{2}} (1 – x)^{-\frac{5}{2}} \cdot \frac{-1 + x + 3x}{4} = -\frac{1}{4} x^{-\frac{3}{2}} (1 – x)^{-\frac{5}{2}} (1 – 4x) \).
Set \( f”(x) = 0 \): \( 1 – 4x = 0 \implies x = \frac{1}{4} \) (M1A1).
Check for inflexion: Test sign change of \( f”(x) \). For \( x < \frac{1}{4} \), e.g., \( x = 0.2 \), \( 1 – 4 \cdot 0.2 = 0.2 > 0 \), so \( f”(x) < 0 \); for \( x > \frac{1}{4} \), e.g., \( x = 0.3 \), \( 1 – 4 \cdot 0.3 = -0.2 < 0 \), so \( f”(x) > 0 \). Sign change confirms a point of inflexion (R1).
Find \( y \)-coordinate: \( f\left( \frac{1}{4} \right) = \sqrt{\frac{\frac{1}{4}}{1 – \frac{1}{4}}} = \sqrt{\frac{\frac{1}{4}}{\frac{3}{4}}} = \sqrt{\frac{1}{3}} = \frac{1}{\sqrt{3}} \) (A1).
Coordinates: \( \left( \frac{1}{4}, \frac{1}{\sqrt{3}} \right) \).

(c) [11 marks]

Compute \( \int f(x) \, dx = \int \sqrt{\frac{x}{1 – x}} \, dx \) using \( x = \sin^2 \theta \).
If \( x = \sin^2 \theta \), then \( \frac{dx}{d\theta} = 2 \sin \theta \cos \theta \), so \( dx = 2 \sin \theta \cos \theta \, d\theta \) (M1A1).
Substitute: \( \frac{x}{1 – x} = \frac{\sin^2 \theta}{1 – \sin^2 \theta} = \frac{\sin^2 \theta}{\cos^2 \theta} = \tan^2 \theta \), so \( f(x) = \sqrt{\tan^2 \theta} = \tan \theta \).
Integral: \( \int \sqrt{\frac{x}{1 – x}} \, dx = \int \tan \theta \cdot 2 \sin \theta \cos \theta \, d\theta \) (M1A1).
Simplify: \( \tan \theta \cdot 2 \sin \theta \cos \theta = 2 \cdot \frac{\sin \theta}{\cos \theta} \cdot \sin \theta \cos \theta = 2 \sin^2 \theta \) (A1).
So, \( \int 2 \sin^2 \theta \, d\theta \). Use identity: \( \sin^2 \theta = \frac{1 – \cos 2\theta}{2} \).
Integral: \( \int 2 \cdot \frac{1 – \cos 2\theta}{2} \, d\theta = \int (1 – \cos 2\theta) \, d\theta \) (M1A1).
Integrate: \( \int 1 \, d\theta – \int \cos 2\theta \, d\theta = \theta – \frac{1}{2} \sin 2\theta + c \) (A1).
Back-substitute: \( \theta = \arcsin \sqrt{x} \) (A1).
For \( \sin 2\theta = \sin (2 \arcsin \sqrt{x}) \), use \( \sin 2\theta = 2 \sin \theta \cos \theta \), where \( \sin \theta = \sqrt{x} \), \( \cos \theta = \sqrt{1 – x} \).
So, \( \frac{1}{2} \sin 2\theta = \sin \theta \cos \theta = \sqrt{x} \sqrt{1 – x} = \sqrt{x – x^2} \) (M1A1).
Thus, \( \int f(x) \, dx = \theta – \frac{1}{2} \sin 2\theta + c = \arcsin \sqrt{x} – \sqrt{x – x^2} + c \) (AG).