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MathematicsSingle correct+4 / -1Question 19 of 75

JEE Main 2026 · 5 Apr · Shift 1, Mathematics question 19

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 the function g(x)=3+exf(x)g(x)=3+e^xf(x), is:

Choose one option
Answer and worked solution

Official answer: B

Find f.

  • x=y=0x=y=0: f(0)=23f(0)f(0)=\tfrac23f(0), so f(0)=0f(0)=0.
  • y=0y=0: f(x/3)=f(x)/3f(x/3)=f(x)/3, so the equation becomes f(x+y)=f(x)+f(y)f(x+y)=f(x)+f(y).

A differentiable additive function is linear: f(x)=f′(0)x=3xf(x)=f'(0)x=3x.

Minimum. g=3+3xexg=3+3xe^x, g′=3ex(1+x)=0g'=3e^x(1+x)=0 at x=−1x=-1, a minimum:
g(−1)=3−3e=3(e−1e).g(-1)=3-\frac3e=3\left(\frac{e-1}e\right).

(B)

Topics:Types of functionsMaxima and minima
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