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, ∞) be a function such that g(x+y)=g(x) g(y) for all x, y in mathbb R . If f (-3/5 )=12 and g (-1/3 )=2 , then the value of (f (1/4 )+g(-2)-8 ) g(0) is

Mathematics · Functions

JEE Advanced 2024 — Paper 2 — Question 8

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 $\qquad$

Answer: 51

Numerical answer — enter this value.

Step-by-step solution

$\quad f(x)=k x, g(x)=a^{x}$

$f\left(-\frac{3}{5}\right)=12, g\left(-\frac{1}{3}\right)=2$

$-\frac{3 k}{5}=12, a^{-\frac{1}{3}}=2$

$\mathrm{k}=-20, \frac{1}{\mathrm{a}}=8 \Rightarrow \mathrm{a}=\frac{1}{8}$

$f(x)=-20 x, g(x)=\left(\frac{1}{8}\right)^{x}, g(0)=1$

$f\left(\frac{1}{4}\right)=-20 \times \frac{1}{4}=-5, g(-2)=\left(\frac{1}{8}\right)^{-2}=64$

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

$\left[f\left(\frac{1}{4}\right)+g(-2)-8\right]=51$

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Exam: JEE Advanced 2024
Paper: Paper 2
Subject: Mathematics
Chapter: Functions
Topic: Functional Equations

Answer: 51

Step-by-step solution

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