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 $\mathrm{g}(\mathrm{x})=3+\mathrm{e}^{\mathrm{x}} \mathrm{f}(\mathrm{x})$, is

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

(A) $3\left(\frac{\mathrm{e}+1}{\mathrm{e}}\right)$

JEE Main · 2026 · 05 Apr 2026 · Shift 1 · Application of Derivatives

Let

f : R \to R

be a differentiable function such that

f (\frac{x + y}{3}) = \frac{f ( x ) + f ( y )}{3}

for all

x, y \in R

and

f^{'} (0) = 3

. Then the minimum value of the function

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

, is

Answer: (B)

Setting

x = y = 0

f (\frac{x + y}{3}) = \frac{f ( x ) + f ( y )}{3}

gives

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

, so

f (0) = 0

. Differentiating the relation with respect to

x

and then putting

x = 0

gives

\frac{1}{3} f^{'} (\frac{y}{3}) = \frac{f ^{'} ( 0 )}{3} = 1

, hence

f^{'} (\frac{y}{3}) = 3

for every

y

; thus

f^{'} (x) = 3

throughout and

f (x) = 3 x

. Therefore

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

and

g^{'} (x) = 3 e^{x} (1 + x)

, which vanishes only at

x = - 1

, where

g^{''} (- 1) = 3 e^{- 1} > 0

. The minimum value of

g

g (- 1) = 3 - \frac{3}{e} = 3 (\frac{e - 1}{e})

JEE Main · 2026 · 05 Apr 2026 · Shift 1 · Single correct option