(a) Values of \( p \) and \( q \):

Solve: \[ 4x^2 – 1 = 0 \quad (M1) \]

\[ (2x – 1)(2x + 1) = 0 \implies p = \frac{1}{2}, q = -\frac{1}{2} \quad (A1) \]

[2 marks]

(b) Derivative \( f'(x) \):

Using quotient rule: \[ f'(x) = \frac{u’v – uv’}{v^2} \], where \( u = 3x + 2 \), \( v = 4x^2 – 1 \quad (M1) \)

\( u’ = 3 \), \(v’ = 8x \quad (A1) \)

\[ f'(x) = \frac{3(4x^2 – 1) – 8x(3x + 2)}{(4x^2 – 1)^2} = \frac{-12x^2 – 16x – 3}{(4x^2 – 1)^2} \quad (A1) \]

Note: Award A1 for each correct term in the numerator with correct signs, provided the denominator is correct.

Alternatively, using product rule: \[ f(x) = (3x + 2)(4x^2 – 1)^{-1} \]

\[ f'(x) = -8x(3x + 2)(4x^2 – 1)^{-2} + 3(4x^2 – 1)^{-1} \quad (A1)(A1) \]

[3 marks]

(c) Point of Inflection:

Solve: \[ f”(x) = 0 \quad (M1) \]

\[ x = -1.60 \quad (A1) \]

[2 marks]

(d) Sketch of \( y = f(x) \):

Vertical asymptotes: \[ x = \frac{1}{2}, x = -\frac{1}{2} \quad (A1) \]

Horizontal asymptote: \[ y = 0 \quad (A1) \]

Correct branches, asymptotic behavior, coordinates of local maxima/minima, and axes intercepts as shown in the diagram: \[ \text{(A1 for each branch, A1 for intercepts and extrema)} \]

Note: Award A1 for both vertical asymptotes with equations, A1 for horizontal asymptote with equation, A1 for each correct branch including asymptotic behavior, coordinates of minimum and maximum points, and axes intercepts. If vertical asymptotes are absent or not vertical, award maximum A0A1A0A1A1.

[5 marks]

(e) Asymptotes of \( y = g(x) \):

Vertical asymptote: \[ x = -\frac{2}{3} \quad (A1) \]

Oblique asymptote gradient: \[ \frac{4}{3} \quad (A1) \]

Find oblique asymptote by polynomial division: \[ g(x) = \frac{4x^2 – 1}{3x + 2} \quad (M1) \]

\[ g(x) = \frac{4}{3}x – \frac{8}{9} + \text{remainder} \implies y = \frac{4}{3}x – \frac{8}{9} \quad (A1) \]

Note: Do not award the final A1 if the answer is not given as an equation.

[4 marks]

(f) Solve \( f(x) < g(x) \):

Consider: \[ g(x) – f(x) = 0 \quad (M1) \]

Critical values:

\[ \begin{cases} x = -0.568729 \\ x = 1.31872 \end{cases} \quad (A1) \]

Solution intervals:

\[ \begin{cases} -\frac{2}{3} < x < -0.569 \\ -0.5 < x < 0.5 \\ x > 1.32 \end{cases} \quad (A1)(A1)(A1) \]

Note: Penalize once for use of \( \leq \) instead of \( < \).

[4 marks]