\( \frac{1}{x-4} + 1 = x-3 \) (M1)

\( \frac{1}{x-4} = x-4 \implies (x-4)^2 = 1 \implies x^2 – 8x + 15 = 0 \) (A1)

Solve: \( (x-3)(x-5) = 0 \implies x = 3, 5 \) (M1)

At \( x = 5 \), \( y = x – 3 = 5 – 3 = 2 \) (A1)

Answer: \( B(5, 2) \) (A1) [5 marks]

Recognize regions: triangle from \( x = 3 \) to \( x = 5 \), and \( f(x) \) from \( x = 5 \) to \( x = k \) (R1)

Triangle area: \( \int_3^5 (x – 3) \, \mathrm{d}x = \left[ \frac{x^2}{2} – 3x \right]_3^5 = (12.5 – 15) – (4.5 – 9) = 2 \) (A1)

Integrate \( f(x) = \frac{1}{x-4} + 1 \): \( \int \left( \frac{1}{x-4} + 1 \right) \mathrm{d}x = \ln |x-4| + x (+C) \) (A1)(A1)

Evaluate: \( \left[ \ln |x-4| + x \right]_5^k = \ln |k-4| + k – 5 \) (M1)

Total area: \( 2 + \ln |k-4| + k – 5 = \ln |k-4| + k – 3 \) (M1)

Equate to \( \ln p + 8 \): \( \ln |k-4| + k – 3 = \ln p + 8 \) (M1)

Non-log terms: \( k – 3 = 8 \implies k = 11 \) (A1)

Log terms: \( \ln |11-4| = \ln 7 \implies p = 7 \) (A1)

Answer: \( k = 11 \), \( p = 7 \) (A1) [10 marks]