a. [3 marks]

For \( f(x) \) to be a valid PDF, \(\int_0^{2k} f(x) \, dx = 1\) (M1).
Compute:
\[ \int_0^k kx \, dx + \int_k^{2k} (2kx – x^2) \, dx = 1 \]
– \(\int_0^k kx \, dx = \left[ \frac{kx^2}{2} \right]_0^k = \frac{k k^2}{2} = \frac{k^3}{2}\)
– \(\int_k^{2k} (2kx – x^2) \, dx = \left[ kx^2 – \frac{x^3}{3} \right]_k^{2k} = \left( k (2k)^2 – \frac{(2k)^3}{3} \right) – \left( k k^2 – \frac{k^3}{3} \right) = \left( 4k^3 – \frac{8k^3}{3} \right) – \left( k^3 – \frac{k^3}{3} \right) = 4k^3 – \frac{8k^3}{3} – k^3 + \frac{k^3}{3} = 3k^3 – \frac{7k^3}{3} = \frac{2k^3}{3}\) (M1)
Total: \(\frac{k^3}{2} + \frac{2k^3}{3} = \frac{3k^3}{6} + \frac{4k^3}{6} = \frac{7k^3}{6} = 1\)
\[ 7k^3 = 6 \] (A1)

b. [4 marks]

The median \( m \) satisfies \(\int_0^m f(x) \, dx = 0.5\) or \(\int_m^{2k} f(x) \, dx = 0.5\) (M1).
From part (a), \( k = \left( \frac{6}{7} \right)^{1/3} \approx 0.949914 \).
Check: \(\int_0^k kx \, dx = \left[ \frac{kx^2}{2} \right]_0^k = \frac{k^3}{2} = \frac{\frac{6}{7}}{2} = \frac{3}{7} \approx 0.428571 < 0.5\), so \( m > k \) (A1).
EITHER
\(\int_0^k kx \, dx + \int_k^m (2kx – x^2) \, dx = 0.5\), so \(\int_k^m (2kx – x^2) \, dx = 0.5 – \frac{3}{7} = \frac{1}{14}\)
\[ \left[ kx^2 – \frac{x^3}{3} \right]_k^m = k m^2 – \frac{m^3}{3} – k k^2 + \frac{k^3}{3} = k m^2 – \frac{m^3}{3} – \frac{2k^3}{3} = \frac{1}{14} \] (A1)
OR
\(\int_m^{2k} (2kx – x^2) \, dx = 0.5\), since \(\int_k^{2k} (2kx – x^2) \, dx = \frac{2k^3}{3} = \frac{2 \cdot \frac{6}{7}}{3} = \frac{4}{7}\), so \(\int_k^m (2kx – x^2) \, dx = \frac{4}{7} – 0.5 = \frac{4}{7} – \frac{3.5}{7} = \frac{0.5}{7} = \frac{1}{14}\)
\[ \left[ kx^2 – \frac{x^3}{3} \right]_k^m = \frac{1}{14} \] (A1)
THEN
Solve: \( k m^2 – \frac{m^3}{3} – \frac{2k^3}{3} = \frac{1}{14} \), with \( k^3 = \frac{6}{7} \), so \(\frac{2k^3}{3} = \frac{4}{7}\):
\[ k m^2 – \frac{m^3}{3} – \frac{4}{7} = \frac{1}{14} \]
\[ k m^2 – \frac{m^3}{3} = \frac{1}{14} + \frac{4}{7} = \frac{9}{14} \]
Numerically, \( m \approx 1.02925 \), so \( m = 1.03 \) (to 3 s.f.) (A1).