Let f : R rightarrow R be a differentiable function such that f (x+y/3 )=f(x)+f(y)/3 for all x, y in R and f prime (0)=3 . Then the minimum value of the function g(x)=3+e x f(x) , is

Mathematics · Functions

JEE Main 2026 — 5 April, Morning Shift — Question 42

Let $\mathrm{f}: \mathrm{R} \rightarrow \mathrm{R}$ be a differentiable function such that $f\left(\frac{x+y}{3}\right)=\frac{f(x)+f(y)}{3}$ for all $x, y \in R$ and $f^{\prime}(0)=3$ . Then the minimum value of the function $g(x)=3+e^{x} f(x)$ , is

Option A:
$3\left(\frac{\mathrm{e}+1}{\mathrm{e}}\right)$
Option B:
$3\left(\frac{\mathrm{e}-1}{\mathrm{e}}\right)$
Correct
Option C:
$\frac{3-e}{e}$
Option D:
3 e

Answer: B

Step-by-step solution

$\mathrm{f}\left(\frac{\mathrm{x}+\mathrm{y}}{3}\right)=\frac{\mathrm{f}(\mathrm{x})+\mathrm{f}(\mathrm{y})}{3}$ , put $\mathrm{x}=\mathrm{y}=0$ $f(0)=\frac{2 f(0)}{3}$

\Rightarrow \mathrm{f}(0)=0 \end{gathered}$$ $\mathrm{f}^{\prime}\left(\frac{\mathrm{x}+\mathrm{y}}{3}\right) \cdot \frac{1}{3}=\frac{1}{3} \mathrm{f}^{\prime}(\mathrm{x})$ put $\mathrm{x}=0$ f' $\left(\frac{\mathrm{y}}{3}\right) \frac{1}{3}=\frac{1}{3} \times 3$ $\mathrm{f}^{\prime}\left(\frac{\mathrm{y}}{3}\right)=3$ put $\mathrm{y}=3 \mathrm{x}$ $\mathrm{f}^{\prime}(\mathrm{x})=3$ integrate both sides $\mathrm{f}(\mathrm{x})=3 \mathrm{x}+\mathrm{c}$ from eq. (1) $\mathrm{f}(0)=0$ $\Rightarrow \mathrm{f}(\mathrm{x})=3 \mathrm{x}$ Now $\mathrm{g}(\mathrm{x})=3+\mathrm{e}^{\mathrm{x}} \cdot 3 \mathrm{x}$ $\mathrm{g}^{\prime}(\mathrm{x})=3\left[\mathrm{e}^{\mathrm{x}}+\mathrm{x} \cdot \mathrm{e}^{\mathrm{x}}\right]$ $=3 \mathrm{e}^{\mathrm{x}}(\mathrm{x}+1)$ $\mathrm{g}^{\prime}(\mathrm{x})=0$, at $\mathrm{x}=-1$ $(\mathrm{g}(\mathrm{x}))_{\text {min }}=3+\mathrm{e}^{-1}(-3)$ $=3\left[1-\frac{1}{\mathrm{e}}\right]=\frac{3(\mathrm{e}-1)}{\mathrm{e}}$

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Exam: JEE Main 2026
Paper: 5 April, Morning Shift
Subject: Mathematics
Chapter: Functions
Topic: Functional Equations