11. (a) It is given that \( y \) is directly proportional to \( (x-1)^{4} \) and \( y=20 \) when \( x=3 \). (i) Write down a formula connecting \( y \) and \( x \). (ii) Find the value of \( x \) when \( y=0.406 \), giving your answer to one decimal place. [2]

(a) (i) Since \( y \) is directly proportional to \( (x-1)^{4} \), we can write the formula as: \[ y = k(x-1)^{4} \] where \( k \) is the constant of proportionality. To find the value of \( k \), we use the given condition \( y=20 \) when \( x=3 \): \[ 20 = k(3-1)^{4} \] \[ 20 = k(2)^{4} \] \[ 20 = 16k \] \[ k = \frac{20}{16} \] \[ k = \frac{5}{4} \] Therefore, the formula connecting \( y \) and \( x \) is: \[ y = \frac{5}{4}(x-1)^{4} \] (ii) To find the value of \( x \) when \( y=0.406 \), we substitute \( y \) into the formula and solve for \( x \): \[ 0.406 = \frac{5}{4}(x-1)^{4} \] \[ (x-1)^{4} = \frac{0.406 \times 4}{5} \] \[ (x-1)^{4} = 0.3224 \] \[ x-1 = \sqrt[4]{0.3224} \] \[ x-1 = 0.7 \] \[ x = 1.7 \] Therefore, the value of \( x \) when \( y=0.406 \) is \( x=1.7 \) (rounded to one decimal place).