JEE Main 2026 Apr 5 Shift 1, Mathematics Q19: Maxima & Minima

JEE Main2026Apr 5, Shift 1Mathematics

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

A
$3 (\frac{e + 1}{e})$
B
$3 (\frac{e - 1}{e})$
C
$\frac{3 - e}{e}$
D
$3 e$

Solution

Setting $x = y = 0$ in the functional equation gives $f (0) = \frac{2 f ( 0 )}{3}$ , so $f (0) = 0$ .

Differentiating with respect to $x$ : $\frac{1}{3} f^{'} (\frac{x + y}{3}) = \frac{f ^{'} ( x )}{3}$ . Setting $x = 0$ : $f^{'} (y /3) = f^{'} (0) = 3$ for every $y$ . Hence $f^{'}$ is constant equal to $3$ , so $f (x) = 3 x$ .

Now $g (x) = 3 + 3 x e^{x}$ and $g^{'} (x) = 3 e^{x} (1 + x)$ , which vanishes only at $x = - 1$ and changes sign from negative to positive there.

So $g_{m i n} = 3 + 3 (- 1) e^{- 1} = 3 - \frac{3}{e} = \frac{3 ( e - 1 )}{e}$ .

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JEE Main 2026 Apr 5 Shift 1, Mathematics Q19: Maxima & Minima

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