(a) [9 marks]

(i) Simplify \( 4f(x) – 2g(x) \):
\( 4f(x) – 2g(x) = 4 \cdot \frac{e^x + e^{-x}}{2} – 2 \cdot \frac{e^x – e^{-x}}{2} = 2(e^x + e^{-x}) – (e^x – e^{-x}) = e^x + 3e^{-x} \) (M1, A1).
Then, \( \frac{1}{4f(x) – 2g(x)} = \frac{1}{e^x + 3e^{-x}} = \frac{e^x}{e^{2x} + 3} \) (multiply by \( \frac{e^x}{e^x} \)) (A1).
(ii) Substitute \( u = e^x \), so \( dx = \frac{du}{u} \), limits \( x = 0 \to u = 1 \), \( x = \ln 3 \to u = 3 \) (A1).
\( \int_0^{\ln 3} \frac{e^x}{e^{2x} + 3} \, dx = \int_1^3 \frac{1}{u^2 + 3} \, du \) (M1).
Integral: \( \int \frac{1}{u^2 + 3} \, du = \frac{1}{\sqrt{3}} \arctan\left(\frac{u}{\sqrt{3}}\right) \) (M1, A1).
Evaluate: \( \left[ \frac{1}{\sqrt{3}} \arctan\left(\frac{u}{\sqrt{3}}\right) \right]_1^3 = \frac{1}{\sqrt{3}} \left( \arctan(\sqrt{3}) – \arctan\left(\frac{\sqrt{3}}{3}\right) \right) \) (M1).
\( = \frac{1}{\sqrt{3}} \left( \frac{\pi}{3} – \frac{\pi}{6} \right) = \frac{\pi \sqrt{3}}{18} \) (A1).

(b) [8 marks]

(i) \( h(x) = n f(x) + g(x) = \frac{(n+1)e^x + (n-1)e^{-x}}{2} \). Set \( h(x) = k \): \( (n+1)e^x + (n-1)e^{-x} = 2k \) (M1).
Let \( u = e^x \): \( (n+1)u + \frac{n-1}{u} = 2k \). Multiply by \( u \): \( (n+1)u^2 – 2ku + (n-1) = 0 \) (A1, A1).
Solve: \( u = \frac{2k \pm \sqrt{4k^2 – 4(n+1)(n-1)}}{2(n+1)} = \frac{k \pm \sqrt{k^2 – n^2 + 1}}{n+1} \) (M1).
Thus, \( x = \ln \left( \frac{k \pm \sqrt{k^2 – n^2 + 1}}{n + 1} \right) \) (A1).
(ii) Discriminant: \( \Delta = 4k^2 – 4(n+1)(n-1) = 4(k^2 – n^2 + 1) > 0 \) (R1).
Since \( k^2 – n^2 + 1 > 0 \), require \( k^2 > n^2 – 1 \implies k > \sqrt{n^2 – 1} \) (R1, A1).

(c) [6 marks]

(i) Method 1: \( t(x) = \frac{e^x – e^{-x}}{e^x + e^{-x}} \). Quotient rule: \( t'(x) = \frac{(e^x + e^{-x})(e^x + e^{-x}) – (e^x – e^{-x})(e^x – e^{-x})}{(e^x + e^{-x})^2} \) (M1).
Numerator: \( (e^x + e^{-x})^2 – (e^x – e^{-x})^2 = 4e^x e^{-x} = 4 \). Thus, \( t'(x) = \frac{4}{(e^x + e^{-x})^2} = \frac{[f(x)]^2 – [g(x)]^2}{[f(x)]^2} \) (A1, A1).
Method 2: Quotient rule: \( t'(x) = \frac{g'(x)f(x) – g(x)f'(x)}{[f(x)]^2} \). Since \( f'(x) = g(x) \), \( g'(x) = f(x) \): \( t'(x) = \frac{f(x)f(x) – g(x)g(x)}{[f(x)]^2} = \frac{[f(x)]^2 – [g(x)]^2}{[f(x)]^2} \) (M1, A1, A1).
Method 3: Rewrite \( t(x) = \frac{\frac{e^x – e^{-x}}{2}}{\frac{e^x + e^{-x}}{2}} \). Product rule: \( t(x) = (e^x – e^{-x})(e^x + e^{-x})^{-1} \), \( t'(x) = (e^x + e^{-x})(e^x + e^{-x})^{-1} – (e^x – e^{-x})^2 (e^x + e^{-x})^{-2} = 1 – \frac{(e^x – e^{-x})^2}{(e^x + e^{-x})^2} = \frac{4}{(e^x + e^{-x})^2} = \frac{[f(x)]^2 – [g(x)]^2}{[f(x)]^2} \) (M1, A1, A1).
Method 4: Quotient rule: \( t'(x) = \frac{g'(x)}{f(x)} – \frac{g(x)f'(x)}{[f(x)]^2} \). Since \( g'(x) = f(x) \), \( f'(x) = g(x) \): \( t'(x) = \frac{f(x)}{f(x)} – \frac{g(x)g(x)}{[f(x)]^2} = 1 – \frac{[g(x)]^2}{[f(x)]^2} = \frac{[f(x)]^2 – [g(x)]^2}{[f(x)]^2} \) (M1, A1, A1).
(ii) Method 1: \( [f(x)]^2 – [g(x)]^2 = \left( \frac{e^x + e^{-x}}{2} \right)^2 – \left( \frac{e^x – e^{-x}}{2} \right)^2 = 1 \) (M1, A1). Since \( [f(x)]^2 > 0 \), \( t'(x) = \frac{1}{[f(x)]^2} > 0 \) (R1).
Method 2: \( [f(x)]^2 – [g(x)]^2 = 1 \) (M1, A1). Thus, \( t'(x) = \frac{1}{[f(x)]^2} > 0 \) since \( [f(x)]^2 > 0 \) (R1).
Method 3: From Method 1 (part i), \( t'(x) = \frac{4}{(e^x + e^{-x})^2} \). Since \( (e^x + e^{-x})^2 > 0 \), \( t'(x) > 0 \) (M1, A1, R1).