A differentiable function f:R→R satisfies the functional equation f(x).f(y)+f(x+y)=exf(y)+eyf(x)+xy ∀ x,y∈R. If f'(0)=0 and f(0)=0 then which of the following statements is/are correct?

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A differentiable function $f : R \to R$ satisfies the functional equation $f (x) . f (y) + f (x + y) = e^{x} f (y) + e^{y} f (x) + x y \forall x, y \in R .$ If $f' (0) = 0$ and $f (0) = 0$ then which of the following statements is/are correct?

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There exists atleast one horizontal tangents to the curve $y = f (x) i n (- 1, 1)$

$\underset{x \to 0}{l i m} \frac{f (x)}{x^{2}} = - \frac{1}{2}$

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

$F (x_{2}) > F (x_{1}), \forall x_{2} > x_{1}, F (x) = f' (x) - f (x)$

answer is B, C, D.

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Detailed Solution

$\begin{array}{l} f (x) . f (y) + f (x + y) = e^{x} f (y) + e^{y} f (x) + x y \\ ∴ f (x) . f' (y) + f' (x + y) = e^{x} f' (y) + e^{y} f (x) + x \\ P u t y = 0 \\ f (x) . f' (0) + f' (x) = e^{x} f' (0) + f (x) + x \\ \Rightarrow f' (x) - f (x) = x \Rightarrow f (x) = e^{x} - x - 1 \\ ∴ (A) \lim_{x \to 0} \frac{f (x)}{x^{2}} = \lim_{x \to 0} \frac{e^{x} - x - 1}{x^{2}} = \frac{1}{2} \\ (B) \int_{x}^{x^{2}} t d t = {(\frac{t 2}{2})}_{x}^{x^{2}} = \frac{x^{4} - x^{2}}{2} \leq 0 \forall | x | \leq 1 \end{array}$

$\begin{array}{l} (C) F (x) = f' (x) - f (x) = (e^{x} - 1) - (e^{x} - x - 1) = x w h i c h i s i n c r e a \sin g f u n c t i o n s o \\ F (x_{2}) > F (x_{1}) i f x_{2} > x_{1} \\ (D) N o w f' (x) = e^{x} - 1 a n d f'' (x) = e^{x} > 0 \forall x \in R \\ ∴ f' (x) i s increasing f u n c t i o n o r R \\ ∵ f' (- 1) = - v e a n d f' (1) = + v e \\ ∵ f' (c) = 0 h a s e x a c t l y a r e r o o t i n (- 1, 1) . \\ i . e . o n e h o r i z o n t a l \tan g e n t . \ln (- 1, 1) \end{array}$