Let f: R -> R be a differentiable function such that f((x+y)/3) = (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:

The correct answer is Option 2. Full mathematical solution steps and hints are accessible in the interactive StemCET workspace.

Maths Q19 – JEE-MAIN 2026

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Home Maths Relations and FunctionsQuestion #8

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:

Clue 1 of 3: Functional Equation Analysis

The given functional equation $f (\frac{x + y}{3}) = \frac{f (x) + f (y)}{3}$ for a differentiable function implies a specific linear form for $f (x)$ .

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Step 1: Determine the form of

f (x)

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The given functional equation is $f (\frac{x + y}{3}) = \frac{f (x) + f (y)}{3}$ . For a differentiable function $f : R \to R$ , this type of equation implies that $f (x)$ must be a linear function of the form $f (x) = a x + b$ . Substituting this into the equation:

a (\frac{x + y}{3}) + b = \frac{(a x + b) + (a y + b)}{3}

\frac{a (x + y)}{3} + b = \frac{a (x + y) + 2 b}{3}

a (x + y) + 3 b = a (x + y) + 2 b

3 b = 2 b ⟹ b = 0

Thus, $f (x)$ must be of the form $f (x) = a x$ .

Step 2: Find the specific function

f (x)

○ Expand

We are given $f^{'} (0) = 3$ . For $f (x) = a x$ , the derivative is $f^{'} (x) = a$ . Therefore, $a = 3$ . So, the function is $f (x) = 3 x$ .

Step 3: Find the minimum value of

g (x)

○ Expand

Substitute $f (x) = 3 x$ into the expression for $g (x)$ : $g (x) = 3 + e^{x} f (x) = 3 + e^{x} (3 x) = 3 + 3 x e^{x}$ . To find the minimum value, we calculate the first derivative $g^{'} (x)$ and set it to zero:

g^{'} (x) = \frac{d}{d x} (3 + 3 x e^{x}) = 0 + 3 (1 \cdot e^{x} + x \cdot e^{x}) = 3 e^{x} (1 + x)

g^{'} (x) = 0 ⟹ 3 e^{x} (1 + x) = 0

Since $e^{x} > 0$ for all $x$ , we must have $1 + x = 0$ , which means $x = - 1$ . To confirm this is a minimum, we check the second derivative:

g^{″} (x) = \frac{d}{d x} (3 e^{x} (1 + x)) = 3 (e^{x} (1 + x) + e^{x} (1)) = 3 e^{x} (x + 2)

g^{″} (- 1) = 3 e^{- 1} (- 1 + 2) = 3 e^{- 1} = \frac{3}{e}

Since $g^{″} (- 1) > 0$ , $x = - 1$ corresponds to a local minimum. The minimum value of $g (x)$ is:

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

💡 Teacher's Secret Hint

Remember to use the product rule when differentiating $x e^{x}$ .

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Maths Question 19 – JEE-MAIN 2026

Clue 1 of 3: Functional Equation Analysis

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