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:R→Rf:\mathbb{R}\to\mathbb{R} be a differentiable function such that f(x+y3)=f(x)+f(y)3f\left(\dfrac{x+y}{3}\right)=\dfrac{f(x)+f(y)}{3} for all x,y∈Rx,y\in\mathbb{R} and f′(0)=3f'(0)=3. Then the minimum value of g(x)=3+exf(x)g(x)=3+e^x f(x) is
A3(e+1e)3\left(\dfrac{e+1}{e}\right)
B3(e−1e)3\left(\dfrac{e-1}{e}\right)correct
C3−ee\dfrac{3-e}{e}
D3e3e
Solution
Step 1: Put x=y=0x=y=0: f(0)=2f(0)3⇒f(0)3=0⇒f(0)=0f(0)=\dfrac{2f(0)}{3}\Rightarrow\dfrac{f(0)}{3}=0\Rightarrow f(0)=0. Step 2: Differentiate w.r.t. xx: f′ ⁣(x+y3)⋅13=f′(x)3⇒f′ ⁣(x+y3)=f′(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=0x=0: f′ ⁣(y3)=f′(0)=3f'\!\left(\dfrac{y}{3}\right)=f'(0)=3 for all yy. Integrate with f(0)=0f(0)=0:
f(x)=3x.f(x)=3x.
 Step 3: g(x)=3+ex(3x)=3+3xexg(x)=3+e^x(3x)=3+3xe^x. g′(x)=3(ex+xex)=3ex(1+x)g'(x)=3(e^x+xe^x)=3e^x(1+x). g′(x)=0⇒x=−1g'(x)=0\Rightarrow x=-1; g′<0g'<0 for x<−1x<-1, g′>0g'>0 for x>−1x>-1, so x=−1x=-1 is a minimum. Step 4: g(−1)=3+3(−1)e−1=3−3e=3(e−1)eg(-1)=3+3(-1)e^{-1}=3-\dfrac{3}{e}=\dfrac{3(e-1)}{e}. Correct answer: (2)
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