Let f: R rightarrow R be a function such that | vedclass

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

Question

(C) Given $f(x+y)=f(x)+f(y)$. This is Cauchy's functional equation,which implies $f(x)=ax$ for all $x \in Q$.Given $f\left(\frac{-3}{5}\right)=12$,we

Vedclass · Accepted Answer

$51$

Vedclass · Answer

(C) Given $f(x+y)=f(x)+f(y)$. This is Cauchy's functional equation,which implies $f(x)=ax$ for all $x \in Q$.Given $f\left(\frac{-3}{5}\right)=12$,we have $a\left(\frac{-3}{5}\right)=12$,so $a=-20$. Thus,$f(x)=-20x$.Then $f\left(\frac{1}{4}\right)=-20\left(\frac{1}{4}\right)=-5$.Given $g(x+y)=g(x)g(y)$,this implies $g(x)=b^x$ for some $b>0$.Given $g\left(\frac{-1}{3}\right)=2$,we have $b^{-1/3}=2$,which means $b^{-1}=2^3=8$,so $b=\frac{1}{8}$.Thus,$g(x)=\left(\frac{1}{8}\right)^x=8^{-x}=2^{-3x}$.Then $g(-2)=2^{-3(-2)}=2^6=64$.Also,$g(0)=b^0=1$.Substituting these values into the expression: $\left(f\left(\frac{1}{4}\right)+g(-2)-8\right)g(0) = (-5+64-8)(1) = 51$.

Similar Questions

If $f(x)$ is a function such that $f(x+y)=f(x)+f(y)$ and $f(1)=7$, then $\sum_{r=1}^n f(r)=$

If $f: R-\{0\} \rightarrow R$ is defined by $3 f(x)+4 f\left(\frac{1}{x}\right)=\frac{2-x}{x}$, then find the value of $f(3)$.

If $f(x)$ is a quadratic function such that $f(x) f\left(\frac{1}{x}\right) = f(x) + f\left(\frac{1}{x}\right)$,then $\sqrt{f\left(\frac{2}{3}\right) + f\left(\frac{3}{2}\right)} = $

Let $f$ be a differentiable function satisfying the relation $f(xy) = xf(y) + yf(x) - 2xy$ for all $x, y > 0$ and $f'(1) = 3$. Which of the following statements is true?

Let $f: R \rightarrow R$ be a function which satisfies $f(x+y)=f(x)+f(y)$ for all $x, y \in R$. If $f(1)=2$ and $g(n)=\sum_{k=1}^{n-1} f(k)$ for $n \in N$,then the value of $n$ for which $g(n)=20$ is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module