Let be a differentiable function such that for all , and . Then the minimum value of the function , is: [2026]

Question

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: [2026]

Smart Achievers · Accepted Answer

(2)

Given,

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

$f' (x) = \lim_{h \to 0} \frac{f (x + h) - f (x)}{h}$

$= \lim_{h \to 0} \frac{f (\frac{3 x + 3 h}{3}) - f (\frac{3 x + 0}{3})}{h}$

$= \lim_{h \to 0} \frac{\frac{f (3 x) + f (3 h)}{3} - \frac{f (3 x) + f (0)}{3}}{h}$

$= \lim_{h \to 0} \frac{f (3 h) - f (0)}{3 h}$

$\Rightarrow f' (x) = f' (0)$

$\Rightarrow f' (x) = 3$ [ $∵ f' (0) = 3$ ]

Integrating on both sides,

$f (x) = 3 x + λ$ ... (ii)

Put x = y = 0 in (i)

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

$\Rightarrow 3 f (0) = 2 f (0) \Rightarrow f (0) = 0$

Put x = 0 in (ii)

$\Rightarrow 0 = 0 + λ \Rightarrow λ = 0 \Rightarrow f (x) = 3 x$

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

$\Rightarrow g (x) = 3 + e^{x} \cdot 3 x$

$\Rightarrow g' (x) = 3 e^{x} + 3 x e^{x} = 0$

$\Rightarrow 3 e^{x} (1 + x) = 0 \Rightarrow x = - 1$

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

$= 6 e^{x} + 3 x e^{x} = 3 e^{x} (2 + x)$

Minima at x = –1

$∴ g {(x)}_{\min} = g (- 1) = 3 + e^{- 1} 3 (- 1)$

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

Solution

Q.
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: [2026]

1 $3 (\frac{e + 1}{e})$

2 $3 (\frac{e - 1}{e})$

3 $\frac{3 - e}{e}$

4 3e

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Solution

Q. Let f:R→R be a differentiable function such that f(x+y3)=f(x)+f(y)3 for all x,y∈R, and f'(0)=3. Then the minimum value of the function g(x)=3+exf(x), is: [2026]

1 3(e+1e)

2 3(e–1e)