The function f is such that \( f(x)=4 x-1 \) and the function g is such that \( g(x)=k x^{2} \) where k is a constant. Given that \( \mathrm{fg}(2)=12 \). Find k (correct to 4 dp ). \( \} \mathrm{k}= \)

We are given that \[ f(x)=4x-1 \quad \text{and} \quad g(x)=kx^2, \] and that \[ f(g(2))=12. \] 1. First, compute \( g(2) \): \[ g(2)=k(2)^2=4k. \] 2. Next, evaluate \( f(g(2)) \). Since \( f(x)=4x-1 \), substituting \( x=4k \) gives: \[ f(4k)=4(4k)-1=16k-1. \] 3. Set up the equation using the given condition: \[ 16k-1=12. \] 4. Solve for \( k \): \[ 16k=12+1=13 \quad \Longrightarrow \quad k=\frac{13}{16}=0.8125. \] Thus, correct to 4 decimal places, \[ k=0.8125. \]