Let $f: \mathbb{R} \rightarrow \mathbb{R}$ be a function such that $f(x+y)=f(x)+f(y)$ for all $x, y

Question

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

Professor PC Thomas Classes · Answer

(51) : $f(x+y)=f(x)+f(y)$ This gives $f(x)=k x \forall x \in Q$\ As $f(-3 / 5)=12=k(-3 / 5)$ $$ \begin{array}{ll} 	herefore & k=-20 \ & f(x)=-20 x \forall x \in Q \end{array} $$ Similarly, $g(x+y)=g(x) g(y)$\ Given, $g(x)=e^{\lambda x}, x \in Q$ $$ \begin{aligned} & g(-1 / 3)=2=e^{-\lambda / 3} \ 	herefore \quad & g(x)=e^{\lambda x}=\left(e^{-\lambda / 3}ight)^{-3 x}=2^{-3 x}=\frac{1}{8^{x}} \end{aligned} $$ Now, $f(1 / 4)=-5$ $$ g(-2)=\frac{1}{8^{-2}}=64, g(0)=1 $$ Hence, $\left(f\left(\frac{1}{4}ight)+g(-2)-8ight) g(0)=51$