Application of DerivativesmediumPYQ · JEE Main · 5 Apr 2026 · Shift 1 (Morning)Free

Minimum of g(x)=3+e^x f(x): 3(e-1)/e | JEE Main 2026

JEE Maths question with a full step-by-step solution.

Question

Let

f:\mathbb{R}\to\mathbb{R}

be a differentiable function such that

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

for all

x,y\in\mathbb{R}

and

f'(0)=3

. Then the minimum value of

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

3\left(\dfrac{e+1}{e}\right)

3\left(\dfrac{e-1}{e}\right)

correct

\dfrac{3-e}{e}

3e

Solution

Step 1: Put

x=y=0

f(0)=\dfrac{2f(0)}{3}\Rightarrow\dfrac{f(0)}{3}=0\Rightarrow f(0)=0

. Step 2: Differentiate w.r.t.

x

f'\!\left(\dfrac{x+y}{3}\right)\cdot\dfrac13=\dfrac{f'(x)}{3}\Rightarrow f'\!\left(\dfrac{x+y}{3}\right)=f'(x)

. Set

x=0

f'\!\left(\dfrac{y}{3}\right)=f'(0)=3

for all

y

. Integrate with

f(0)=0

f(x)=3x.

Step 3:

g(x)=3+e^x(3x)=3+3xe^x

g'(x)=3(e^x+xe^x)=3e^x(1+x)

g'(x)=0\Rightarrow x=-1

;

g'<0

for

x<-1

g'>0

for

x>-1

, so

x=-1

is a minimum. Step 4:

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

. Correct answer: (2)

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