Let f : R → R be a function such that f(x + y) = f(x) + f(y) for all x, y ∈ R and g : R → (0, ∞)

Question

Let \(f: \mathbb{R} \rightarrow \mathbb{R}\) be a function such that \(f(x+y)=f(x)+f(y)\) for all \(x, y \in \mathbb{R}\), and \(g: \mathbb{R} \rightarrow(0, \infty)\) be a function such that \(g(x+y)=g(x) g(y)\) for all \(x, y \in \mathbb{R}\). If \(f\left(\frac{-3}{5}\right)=12\) and \(g\left(\frac{-1}{3}\right)=2\), then the value of \(\left(f\left(\frac{1}{4}\right)+g(-2)-8\right) g(0)\) is _______.

Answer 1

Correct answer: 51

\(f(x+y)=f(x)+f(y)\)

\(\Rightarrow f(x)=k x\)

\(f\left(\frac{-3}{5}\right)=12 \Rightarrow k=-20\)

\(\therefore f(x)=-20 x\)

\(g(x+y)=g(x) g(y)\)

\( \Rightarrow g(x)=a^{x}\)

\(g\left(\frac{-1}{3}\right)=2 \Rightarrow a=\frac{1}{8}\)

\(\therefore g(x)=\left(\frac{1}{8}\right)^{x}\)

\(\left(f\left(\frac{1}{4}\right)+g(-2)-8\right) g(0)=(-5+64-8) \times 1=51\)