JEE Main 2026 — Application of Derivatives Question with Solution

Given $f\left(\frac{x+y}{3}\right)=\frac{f(x)+f(y)}{3}$

Substituting $x=0$ and $y=0$, we get:

$f(0)=\frac{2f(0)}{3}\Rightarrow f(0)=0$

Differentiating the given equation partially with respect to $x$, treating $y$ as a constant:

$f^{\prime }\left(\frac{x+y}{3}\right)\cdot \frac{1}{3}=\frac{f^{\prime }(x)}{3}$

$\Rightarrow f^{\prime }\left(\frac{x+y}{3}\right)=f^{\prime }(x)$

Substituting $x=0$, we get:

$f^{\prime }\left(\frac{y}{3}\right)=f^{\prime }(0)=3$

Since this is true for all $y\in \mathbb{R}$, $f^{\prime }(x)=3$ for all $x\in \mathbb{R}$.

Integrating both sides with respect to $x$:

$f(x)=3x+C$

Using $f(0)=0$, we get $C=0$. Thus, $f(x)=3x$.

Now, the function $g(x)$ is given by:

$g(x)=3+e^{x}f(x)=3+3x{e}^x$

To find the minimum value, we differentiate $g(x)$ with respect to $x$:

$g^{\prime }(x)=3(e^x+x{e}^x)=3e^x(1+x)$

Setting $g^{\prime }(x)=0$ gives $x=-1$.

For $x<-1$, $g^{\prime }(x)<0$ and for $x>-1$, $g^{\prime }(x)>0$. Therefore, $x=-1$ is a point of global minimum.

The minimum value of $g(x)$ is:

$g(-1)=3+3(-1)e^{-1}=3-\frac{3}{e}=3\left(\frac{e-1}{e}\right)$

Answer: $3\left(\frac{e-1}{e}\right)$