JEE Main 2026 — Mathematics Calculus

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

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

$$f(0) = \dfrac{2f(0)}{3} \Rightarrow f(0) = 0$$f(0)=32f(0)​⇒f(0)=0

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

$$f'\left(\dfrac{x+y}{3}\right) \cdot \dfrac{1}{3} = \dfrac{f'(x)}{3}$$f′(3x+y​)⋅31​=3f′(x)​

$$\Rightarrow f'\left(\dfrac{x+y}{3}\right) = f'(x)$$⇒f′(3x+y​)=f′(x)

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

$$f'\left(\dfrac{y}{3}\right) = f'(0) = 3$$f′(3y​)=f′(0)=3

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

Integrating both sides with respect to $x$x:

$$f(x) = 3x + C$$f(x)=3x+C

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

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

$$g(x) = 3 + e^x f(x) = 3 + 3x e^x$$g(x)=3+exf(x)=3+3xex

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

$$g'(x) = 3(e^x + x e^x) = 3e^x(1 + x)$$g′(x)=3(ex+xex)=3ex(1+x)

Setting $g'(x) = 0$g′(x)=0 gives $x = -1$x=−1.

For $x < -1$x<−1, $g'(x) < 0$g′(x)<0 and for $x > -1$x>−1, $g'(x) > 0$g′(x)>0. Therefore, $x = -1$x=−1 is a point of global minimum.

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

$$g(-1) = 3 + 3(-1)e^{-1} = 3 - \dfrac{3}{e} = 3\left(\dfrac{e-1}{e}\right)$$g(−1)=3+3(−1)e−1=3−e3​=3(ee−1​)

Answer: $3\left(\dfrac{e-1}{e}\right)$3(ee−1​)