Let $f: \mathbb{R} \rightarrow \mathbb{R}$ be a differentiable function such that $f\left(\frac{x+y}{3}\right)=\frac{f(x)+f(y)}{3}$ for all $x, y \in \mathbb{R}$, and $f^{\prime}(0)=3$. Then the minimum value of the function $g(x)=3+e^x f(x)$ is: [JEE Main 2026, 5 Apr (Shift 1)]

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YESPYQ · Accepted Answer

Correct answer: b) $3\left(\frac{e-1}{e}\right)$. $\mathrm{f}\left(\frac{\mathrm{x}+\mathrm{y}}{3}\right)=\frac{\mathrm{f}(\mathrm{x})+\mathrm{f}(\mathrm{y})}{3}$ Put $\mathrm{x}=\mathrm{y}=0$ $ f(0)=\frac{2 f(0)}{3}$ $\Rightarrow f(0)=0 \ldots(1) $ $\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^{\prime}\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: $f(x)=3 x+C$, and from $f(0)=0, C=0$, so $f(x)=3 x$ 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$ Now, $g^{\prime \prime}(x)=3 e^x(2+x)$ \(g^{\prime \prime}(-1)=3 e^{-1}(1)=\frac{3}{e}&g

Let \(f: \mathbb{R} \rightarrow \mathbb{R}\) be a differentiable function such that \(f\left(\frac{x+y}{3}\right)=\frac{…

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