A differentiable function f : R → R satisfies the functional equation f(x) · f(y) + f(x + y) = ex f(y) + ey f(x) + xy \forall x , y \in R. If f '(0) = 0 and f(0) = 0, then which of the following statements is/are correct?

Engineering

Mathematics

Monotonicity

Question

A differentiable function f : R → R satisfies the functional equation

f(x) · f(y) + f(x + y) = e^x f(y) + e^y f(x) + xy $\forall x, y \in R$ . If f '(0) = 0 and f(0) = 0, then which of the following statements is/are correct?

There exists at least two horizontal tangents to the curve y = f(x) in (– 1, 1)

$\int_{x}^{x^{2}} (f' (t) - f (t)) dt > 0 \forall | x | > 1$

$\underset{x \to 0}{Lim} \frac{f (x)}{x^{2}} = \frac{1}{2}$

F(x₂) > F(x₁) $\forall$ x₂ > x₁, where F(x) = f '(x) – f (x)

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Differentiate w.r.t. y keeping x constant

f(x) f '(y) + f '(x + y) = e^x f '(y) + e^y f(x) + x

Put y = 0 [Given f '(0) = 0]

⇒ f '(x) = f(x) + x

$∴$ f(x) = e^x – x – 1

(A) $\lim_{x \to \infty} \frac{f (x)}{x^{2}} = \lim_{x \to \infty} \frac{e^{x} - x - 1}{x^{2}} = \frac{1}{2}$

(by expansion)

(B) $\int_{x}^{x^{2}} (f' (t) - f (t)) dt = \int_{x}^{x^{2}} t dt = \frac{x^{4} - x^{2}}{2} > \forall | x | \in (1, \infty)$

= (e^x – 1) – (e^x – x – 1) = x

which is an increasing function on R

$∴$ F(x₂) > F(x₁) $\forall$ x₂ > x₁

(D) Now f '(x) = e^x – 1 and f "(x) = e^x > 0 $\forall x \in R$

$∴$ f ' (x) is an strictly increasing function on R

$∵$ f '(–1) < 0 & f '(1) > 0

$∴$ f '(c) = 0 has exactly one root 'c' in (– 1, 1)

i.e. there exists exactly one horizontal tangent to y = f(x) in (– 1, 1).