If f(x) and g(x) are two real-valued function | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(D) Given $f(g(x+y)) = f(g(x)) + f(g(y))$ ... $(i)$ Let $h(x) = f(g(x))$. Then the equation becomes $h(x+y) = h(x) + h(y)$,which is Cauchy's functiona

Vedclass · Accepted Answer

$\phi$

Vedclass · Answer

(D) Given $f(g(x+y)) = f(g(x)) + f(g(y))$ ... $(i)$ Let $h(x) = f(g(x))$. Then the equation becomes $h(x+y) = h(x) + h(y)$,which is Cauchy's functional equation. The solution to this is $h(x) = cx$ for some constant $c$. Given $g(1) = 2$ and $f(2) = 1$,we have $h(1) = f(g(1)) = f(2) = 1$. Since $h(1) = c(1) = 1$,we get $c = 1$. Thus,$h(x) = f(g(x)) = x$. Since $f(g(x)) = x$,$f$ and $g$ are inverse functions of each other. Therefore,$g(f(x)) = x$ for all $x \in R$. The function $g(f(x)) = x$ is a polynomial function,which is continuous everywhere on the set of real numbers $R$. Hence,the set of points where $g(f(x))$ is discontinuous is the empty set,denoted by $\phi$.

If $f(x)$ and $g(x)$ are two real-valued functions such that $f(g(x+y)) = f(g(x)) + f(g(y))$,$g(1) = 2$,and $f(2) = 1$,then the function $g(f(x))$ is discontinuous on the set

Similar Questions

Let $f(x+y)=f(x)f(y)$ for all $x, y$ where $f(0) \neq 0$. If $f(5) = 2$ and $f'(0) = 3$,then $f'(5)$ is equal to

The function $t$ which maps temperature in degree Celsius into temperature in degree Fahrenheit is defined by $t(C) = \frac{9C}{5} + 32$. Find $t(28)$.

Let $f: R \rightarrow R$ be such that $f$ is injective and $f(x) f(y) = f(x+y)$ for $\forall x, y \in R$. If $f(x), f(y), f(z)$ are in $G$.$P$.,then $x, y, z$ are in:

The number of functions $f : \{1, 2, 3, 4\} \rightarrow \{ a \in \mathbb{Z} : |a| \leq 8 \}$ satisfying $f(n) + \frac{1}{n} f(n+1) = 1$ for all $n \in \{1, 2, 3\}$ is

If $f(f(0)) = 0$,where $f(x) = x^2 + ax + b$ and $b \neq 0$,then $a + b =$

Vedclass Test Series

Exam Paper Generator

Online Exam Module